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  • Optimal two variable linear regression SQL statement (censoring outliers)

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here (with five outliers highlighted): Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

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  • Combining aggregate functions in an (ANSI) SQL statement

    - by morpheous
    I have aggregate functions foo(), foobar(), fredstats(), barneystats() I want to create a domain specific query language (DSQL) above my DB, to facilitate using using a domain language to query the DB. The 'language' comprises of boolean expressions (or more specifically SQL like criteria) which I then 'translate' back into pure (ANSI) SQL and send to the underlying Db. The following lines are examples of what the language statements will look like, and hopefully, it will help further clarify the concept: **Example 1** DQL statement: foobar('yellow') between 1 and 3 and fredstats('weight') > 42 Translation: fetch all rows in an underlying table where computed values for aggregate function foobar() is between 1 and 3 AND computed value for AGG FUNC fredstats() is greater than 42 **Example 2** DQL statement: fredstats('weight') < barneystats('weight') AND foo('fighter') in (9,10,11) AND foobar('green') <> 42 Translation: Fetch all rows where the specified criteria matches **Example 3** DQL statement: foobar('green') / foobar('red') <> 42 Translation: Fetch all rows where the specified criteria matches **Example 4** DQL statement: foobar('green') - foobar('red') >= 42 Translation: Fetch all rows where the specified criteria matches Given the following information: The table upon which the queries above are being executed is called 'tbl' table 'tbl' has the following structure (id int, name varchar(32), weight float) The result set returns only the tbl.id, tbl.name and the names of the aggregate functions as columns in the result set - so for example the foobar() AGG FUNC column will be called foobar in the result set. So for example, the first DQL query will return a result set with the following columns: id, name, foobar, fredstats Given the above, my questions then are: What would be the underlying SQL required for Example1 ? What would be the underlying SQL required for Example3 ? Given an algebraic equation comprising of AGGREGATE functions, Is there a way of generalizing the algorithm needed to generate the required ANSI SQL statement(s)? I am using PostgreSQL as the db, but I would prefer to use ANSI SQL wherever possible.

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  • Performance question: Inverting an array of pointers in-place vs array of values

    - by Anders
    The background for asking this question is that I am solving a linearized equation system (Ax=b), where A is a matrix (typically of dimension less than 100x100) and x and b are vectors. I am using a direct method, meaning that I first invert A, then find the solution by x=A^(-1)b. This step is repated in an iterative process until convergence. The way I'm doing it now, using a matrix library (MTL4): For every iteration I copy all coeffiecients of A (values) in to the matrix object, then invert. This the easiest and safest option. Using an array of pointers instead: For my particular case, the coefficients of A happen to be updated between each iteration. These coefficients are stored in different variables (some are arrays, some are not). Would there be a potential for performance gain if I set up A as an array containing pointers to these coefficient variables, then inverting A in-place? The nice thing about the last option is that once I have set up the pointers in A before the first iteration, I would not need to copy any values between successive iterations. The values which are pointed to in A would automatically be updated between iterations. So the performance question boils down to this, as I see it: - The matrix inversion process takes roughly the same amount of time, assuming de-referencing of pointers is non-expensive. - The array of pointers does not need the extra memory for matrix A containing values. - The array of pointers option does not have to copy all NxN values of A between each iteration. - The values that are pointed to the array of pointers option are generally NOT ordered in memory. Hopefully, all values lie relatively close in memory, but *A[0][1] is generally not next to *A[0][0] etc. Any comments to this? Will the last remark affect performance negatively, thus weighing up for the positive performance effects?

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  • If Then Else Statement Condition Being Ignored?

    - by Matma
    I think im going mad but can some show me what im missing, it must be some stupidly simple i just cant see the wood for the trees. BOTH side of this if then else statement are being executed? Ive tried commenting out the true side and moving the condition to a seperate variable with the same result. However if i explicitly set the condition to 1=0 or 1=1 then the if then statement is operating as i would expect. i.e. only executing one side of the equation... The only time ive seen this sort of thing is when the compiler has crashed and is no longer compiling (without visible indication that its not) but ive restarted studio with the same results, ive cleaned the solution, built and rebuilt with no change? please show me the stupid mistake im making using vs2005 if it matters. Dim dset As DataSet = New DataSet If (CboCustomers.SelectedValue IsNot Nothing) AndAlso (CboCustomers.SelectedValue <> "") Then Dim Sql As String = "Select sal.SalesOrderNo As SalesOrder,cus.CustomerName,has.SerialNo, convert(varchar,sal.Dateofpurchase,103) as Date from [dbo].[Customer_Table] as cus " & _ " inner join [dbo].[Hasp_table] as has on has.CustomerID=cus.CustomerTag " & _ " inner join [dbo].[salesorder_table] as sal On sal.Hasp_ID =has.Hasp_ID Where cus.CustomerTag = '" & CboCustomers.SelectedValue.ToString & "'" Dim dap As SqlDataAdapter = New SqlDataAdapter(Sql, FormConnection) dap.Fill(dset, "dbo.Customer_Table") DGCustomer.DataSource = dset.Tables("dbo.Customer_Table") Else Dim erm As String = "wtf" End If

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  • Optimal two variable linear regression SQL statement

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 5 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; and insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

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  • Optimal two variable linear regression calculation

    - by Dave Jarvis
    Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT, FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < 15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Question The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Related Sites Least absolute deviations Robust regression Thank you!

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  • Explaining the forecasts from an ARIMA model

    - by Samik R.
    I am trying to explain to myself the forecasting result from applying an ARIMA model to a time-series dataset. The data is from the M1-Competition, the series is MNB65. For quick reference, I have a google doc spreadsheet with the data. I am trying to fit the data to an ARIMA(1,0,0) model and get the forecasts. I am using R. Here are some output snippets: > arima(x, order = c(1,0,0)) Series: x ARIMA(1,0,0) with non-zero mean Call: arima(x = x, order = c(1, 0, 0)) Coefficients: ar1 intercept 0.9421 12260.298 s.e. 0.0474 202.717 > predict(arima(x, order = c(1,0,0)), n.ahead=12) $pred Time Series: Start = 53 End = 64 Frequency = 1 [1] 11757.39 11786.50 11813.92 11839.75 11864.09 11887.02 11908.62 11928.97 11948.15 11966.21 11983.23 11999.27 I have a few questions: (1) How do I explain that although the dataset shows a clear downward trend, the forecast from this model trends upward. This also happens for ARIMA(2,0,0), which is the best ARIMA fit for the data using auto.arima (forecast package) and for an ARIMA(1,0,1) model. (2) The intercept value for the ARIMA(1,0,0) model is 12260.298. Shouldn't the intercept satisfy the equation: C = mean * (1 - sum(AR coeffs)), in which case, the value should be 715.52. I must be missing something basic here. (3) This is clearly a series with non-stationary mean. Why is an AR(2) model still selected as the best model by auto.arima? Could there be an intuitive explanation? Thanks.

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  • How to get bit rotation function to accept any bit size?

    - by calccrypto
    i have these 2 functions i got from some other code def ROR(x, n): mask = (2L**n) - 1 mask_bits = x & mask return (x >> n) | (mask_bits << (32 - n)) def ROL(x, n): return ROR(x, 32 - n) and i wanted to use them in a program, where 16 bit rotations are required. however, there are also other functions that require 32 bit rotations, so i wanted to leave the 32 in the equation, so i got: def ROR(x, n, bits = 32): mask = (2L**n) - 1 mask_bits = x & mask return (x >> n) | (mask_bits << (bits - n)) def ROL(x, n, bits = 32): return ROR(x, bits - n) however, the answers came out wrong when i tested this set out. yet, the values came out correctly when the code is def ROR(x, n): mask = (2L**n) - 1 mask_bits = x & mask return (x >> n) | (mask_bits << (16 - n)) def ROL(x, n,bits): return ROR(x, 16 - n) what is going on and how do i fix this?

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  • How to handle alpha in a manual "Overlay" blend operation?

    - by quixoto
    I'm playing with some manual (walk-the-pixels) image processing, and I'm recreating the standard "overlay" blend. I'm looking at the "Photoshop math" macros here: http://www.nathanm.com/photoshop-blending-math/ (See also here for more readable version of Overlay) Both source images are in fairly standard RGBA (8 bits each) format, as is the destination. When both images are fully opaque (alpha is 1.0), the result is blended correctly as expected: But if my "blend" layer (the top image) has transparency in it, I'm a little flummoxed as to how to factor that alpha into the blending equation correctly. I expect it to work such that transparent pixels in the blend layer have no effect on the result, opaque pixels in the blend layer do the overlay blend as normal, and semitransparent blend layer pixels have some scaled effect on the result. Can someone explain to me the blend equations or the concept behind doing this? Bonus points if you can help me do it such that the resulting image has correctly premultiplied alpha (which only comes into play for pixels that are not opaque in both layers, I think.) Thanks! // factor in blendLayerA, (1-blendLayerA) somehow? resultR = ChannelBlend_Overlay(baseLayerR, blendLayerR); resultG = ChannelBlend_Overlay(baseLayerG, blendLayerG); resultB = ChannelBlend_Overlay(baseLayerB, blendLayerB); resultA = 1.0; // also, what should this be??

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  • Problems sorting by date

    - by Nathan
    I'm attempting to only display data added to my database 24 hours ago or less. However, for some reason, the code I've written isn't working, and both entries in my database, one from 1 hour ago, one from 2 days ago, show up. Is there something wrong with my equation? Thanks! public void UpdateValues() { double TotalCost = 0; double TotalEarned = 0; double TotalProfit = 0; double TotalHST = 0; for (int i = 0; i <= Program.TransactionList.Count - 1; i++) { DateTime Today = DateTime.Now; DateTime Jan2013 = DateTime.Parse("01-01-2013"); //Hours since Jan12013 int TodayHoursSince2013 = Convert.ToInt32(Math.Round(Today.Subtract(Jan2013).TotalHours)); //7 int ItemHoursSince2013 = Program.TransactionList[i].HoursSince2013; //Equals 7176, and 7130 if (ItemHoursSince2013 - TodayHoursSince2013 <= 24) { TotalCost += Program.TransactionList[i].TotalCost; TotalEarned += Program.TransactionList[i].TotalEarned; TotalProfit += Program.TransactionList[i].TotalEarned - Program.TransactionList[i].TotalCost; TotalHST += Program.TransactionList[i].TotalHST; } } label6.Text = "$" + String.Format("{0:0.00}", TotalCost); label7.Text = "$" + String.Format("{0:0.00}", TotalEarned); label8.Text = "$" + String.Format("{0:0.00}", TotalProfit); label10.Text = "$" + String.Format("{0:0.00}", TotalHST); }

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  • Remove redundant SQL code

    - by Dave Jarvis
    Code The following code calculates the slope and intercept for a linear regression against a slathering of data. It then applies the equation y = mx + b against the same result set to calculate the value of the regression line for each row. Can the two separate sub-selects be joined so that the data and its slope/intercept are calculated without executing the data gathering part of the query twice? SELECT AVG(D.AMOUNT) as AMOUNT, Y.YEAR * ymxb.SLOPE + ymxb.INTERCEPT as REGRESSION_LINE, Y.YEAR as YEAR, MAKEDATE(Y.YEAR,1) as AMOUNT_DATE FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D, (SELECT ((avg(t.AMOUNT * t.YEAR)) - avg(t.AMOUNT) * avg(t.YEAR)) / (stddev( t.AMOUNT ) * stddev( t.YEAR )) as CORRELATION, ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM ( SELECT AVG(D.AMOUNT) as AMOUNT, Y.YEAR as YEAR, MAKEDATE(Y.YEAR,1) as AMOUNT_DATE FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ) t ) ymxb WHERE $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR Question How do I execute the duplicate bits only once per query, instead of twice? The duplicate bit is the WHERE clause: $X{ IN, C.ID, CityCode } AND SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < $P{Radius} AND S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND Y.YEAR BETWEEN 1900 AND 2009 AND M.YEAR_REF_ID = Y.ID AND M.CATEGORY_ID = $P{CategoryCode} AND M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' Related http://stackoverflow.com/questions/1595659/how-to-eliminate-duplicate-calculation-in-sql Thank you!

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  • Getting the most recent post based on date

    - by camcim
    Hi guys, How do I go about displaying the most recent post when I have two tables, both containing a column called creation_date This would be simple if all I had to do was get the most recent post based on posts created_on value however if a post contains replies I need to factor this into the equation. If a post has a more recent reply I want to get the replies created_on value but also get the posts post_id and subject. The posts table structure: CREATE TABLE `posts` ( `post_id` bigint(20) unsigned NOT NULL auto_increment, `cat_id` bigint(20) NOT NULL, `user_id` bigint(20) NOT NULL, `subject` tinytext NOT NULL, `comments` text NOT NULL, `created_on` datetime NOT NULL, `status` varchar(10) NOT NULL default 'INACTIVE', `private_post` varchar(10) NOT NULL default 'PUBLIC', `db_location` varchar(10) NOT NULL, PRIMARY KEY (`post_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ; The replies table structure: CREATE TABLE `replies` ( `reply_id` bigint(20) unsigned NOT NULL auto_increment, `post_id` bigint(20) NOT NULL, `user_id` bigint(20) NOT NULL, `comments` text NOT NULL, `created_on` datetime NOT NULL, `notify` varchar(5) NOT NULL default 'YES', `status` varchar(10) NOT NULL default 'INACTIVE', `db_location` varchar(10) NOT NULL, PRIMARY KEY (`reply_id`) ) ENGINE=MyISAM DEFAULT CHARSET=latin1 AUTO_INCREMENT=5 ; Here is my query so far. I've removed my attempt of extracting the dates. $strQuery = "SELECT posts.post_id, posts.created_on, replies.created_on, posts.subject "; $strQuery = $strQuery."FROM posts ,replies "; $strQuery = $strQuery."WHERE posts.post_id = replies.post_id "; $strQuery = $strQuery."AND posts.cat_id = '".$row->cat_id."'";

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  • Showing latex commands in text string using mathjax

    - by robezy
    I have a text string, for ex. 'A vehicle travels from A to B, distance {$d} km at constant speed. While returning back to A on same path it {$variation} its speed by {$v} km/hr. The total time of journey is {$t} hours. Find the original speed of vehicle.' The variables in the curly brackets are to be replaced by appropriate latex equation. I'm using php's preg_replace to replace the variables with latex commands. Unfortunately, my latex commands are coming as it is. It is not processed by mathjax. For ex, above text becomes A vehicle travels from A to B, distance 1 km at constant speed. While returning back to A on same path it increased its speed by (\frac{3}{2}) km/hr. The total time of journey is 1 hours. Find the original speed of vehicle. The frac is show as it is. What is wrong here? Please ask me if you need any more info. Thanks

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  • Assign sage variable values into R objects via sagetex and Sweave

    - by sheed03
    I am writing a short Sweave document that outputs into a Beamer presentation, in which I am using the sagetex package to solve an equation for two parameters in the beta binomial distribution, and I need to assign the parameter values into the R session so I can do additional processing on those values. The following code excerpt shows how I am interacting with sage: <<echo=false,results=hide>>= mean.raw <- c(5, 3.5, 2) theta <- 0.5 var.raw <- mean.raw + ((mean.raw^2)/theta) @ \begin{frame}[fragile] \frametitle{Test of Sage 2} \begin{sagesilent} var('a1, b1, a2, b2, a3, b3') eqn1 = [1000*a1/(a1+b1)==\Sexpr{mean.raw[1]}, ((1000*a1*b1)*(1000+a1+b1))/((a1+b1)^2*(a1+b1+1))==\Sexpr{var.raw[1]}] eqn2 = [1000*a2/(a2+b2)==\Sexpr{mean.raw[2]}, ((1000*a2*b2)*(1000+a2+b2))/((a2+b2)^2*(a2+b2+1))==\Sexpr{var.raw[2]}] eqn3 = [1000*a3/(a3+b3)==\Sexpr{mean.raw[3]}, ((1000*a3*b3)*(1000+a3+b3))/((a3+b3)^2*(a3+b3+1))==\Sexpr{var.raw[3]}] s1 = solve(eqn1, a1,b1) s2 = solve(eqn2, a2,b2) s3 = solve(eqn3, a3,b3) \end{sagesilent} Solutions of Beta Binomial Parameters: \begin{itemize} \item $\sage{s1[0]}$ \item $\sage{s2[0]}$ \item $\sage{s3[0]}$ \end{itemize} \end{frame} Everything compiles just fine, and in that slide I am able to see the solutions to the three equations respective parameters in that itemized list (for example the first item in the itemized list from that beamer slide is outputted as [a1=(328/667), b1=(65272/667)] (I am not able to post an image of the beamer slide but I hope you get the idea). I would like to save the parameter values a1,b1,a2,b2,a3,b3 into R objects so that I can use them in simulations. I cannot find any documentation in the sagetex package on how to save output from sage commands into variables for use with other programs (in this case R). Any suggestions on how to get these values into R?

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  • Keeping updated with the latest technologies.

    - by Prashanth
    I am a software developer and I have been programming since the past six years. I simply love the mental challenge involved in trying to come up with solutions to hard problems, reading up programming literature, blogs by prominent developers and so on. I work on Microsoft platform and I have trouble keeping up with the pace at which various frameworks are rolled out. Remoting,WCF,ASP.NET,ASP.NET MVC, LINQ, WPF, WWF, OSLO, ADO.NET data services, DSL tools etc etc. Even understanding all these frameworks at an abstract level and see how they are all tied up with MS vision itself is a major hurdle. Now when you add other non microsoft technologies, programming languages etc to the equation, I wonder how do people manage? Given that there are only 24 hours in a day, how does one keep himself updated about so many technology changes that happen everyday? My question is , is it even worth doing that? The thing is, I am also interested in other fields such as literature, science. I try my best to at least gain a superficial understanding of what is happening in other fields of my interest and don't want to give up on that :)

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  • mysql query help, take total sum from a table, and based on discount value on another table calcula

    - by vegatron
    hi I have a table called invoices: CREATE TABLE IF NOT EXISTS `si_invoices` ( `id` int(10) NOT NULL AUTO_INCREMENT, `biller_id` int(10) NOT NULL DEFAULT '0', `customer_id` int(10) NOT NULL DEFAULT '0', `type_id` int(10) NOT NULL DEFAULT '0', `inv_tax_id` int(10) NOT NULL, `date` date NOT NULL DEFAULT '0000-00-00', `unreg_customer` tinyint(1) NOT NULL DEFAULT '0', `discount` decimal(10,2) NOT NULL DEFAULT '0.00', `discount_type` tinyint(1) NOT NULL DEFAULT '0', PRIMARY KEY (`id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=20 ; each invoice has items that are stored in invoice_items table : CREATE TABLE IF NOT EXISTS `si_invoice_items` ( `id` int(10) NOT NULL AUTO_INCREMENT, `invoice_id` int(10) NOT NULL DEFAULT '0', `quantity` int(10) unsigned NOT NULL DEFAULT '0', `product_id` int(10) DEFAULT '0', `warehouse_id` int(10) NOT NULL, `unit_price` decimal(25,2) DEFAULT '0.00', `total` decimal(25,2) DEFAULT '0.00', `description` text, PRIMARY KEY (`id`), KEY `invoice_id` (`invoice_id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 AUTO_INCREMENT=56 ; and tax table CREATE TABLE IF NOT EXISTS `si_tax` ( `tax_id` int(11) NOT NULL AUTO_INCREMENT, `tax_description` varchar(50) COLLATE utf8_unicode_ci DEFAULT NULL, `tax_percentage` decimal(25,6) DEFAULT '0.000000', `type` varchar(1) COLLATE utf8_unicode_ci DEFAULT NULL, `tax_enabled` varchar(1) COLLATE utf8_unicode_ci NOT NULL DEFAULT '1', PRIMARY KEY (`tax_id`) ) ENGINE=MyISAM DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=5 ; here is what I want to do step 1: get the sum_total of the invoice Items for a speciefic invoice step 2: calculate the discount, in the invoice table I have a discount_type field : if its equal to 0 , then there will be no discount if its equal to 1 , the discount value will be stored in the discount field if its equal to 2 , the discount is a percentage of sum_total step 3: calculate the taxes based on inv_tax_id based on the tax id , I will look in the tax table , get the tax_percentage and multiply it by the (sum_total - discount) in short here is the equation $gross_total = $sum_total - $disount + taxes

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  • How to run OpenGL code with out compiling?

    - by Ole Jak
    So I have some openGL code (such code for example) /* FUNCTION: YCamera :: CalculateWorldCoordinates ARGUMENTS: x mouse x coordinate y mouse y coordinate vec where to store coordinates RETURN: n/a DESCRIPTION: Convert mouse coordinates into world coordinates */ void YCamera :: CalculateWorldCoordinates(float x, float y, YVector3 *vec) { // START GLint viewport[4]; GLdouble mvmatrix[16], projmatrix[16]; GLint real_y; GLdouble mx, my, mz; glGetIntegerv(GL_VIEWPORT, viewport); glGetDoublev(GL_MODELVIEW_MATRIX, mvmatrix); glGetDoublev(GL_PROJECTION_MATRIX, projmatrix); real_y = viewport[3] - (GLint) y - 1; // viewport[3] is height of window in pixels gluUnProject((GLdouble) x, (GLdouble) real_y, 1.0, mvmatrix, projmatrix, viewport, &mx, &my, &mz); /* 'mouse' is the point where mouse projection reaches FAR_PLANE. World coordinates is intersection of line(camera->mouse) with plane(z=0) (see LaMothe 306) Equation of line in 3D: (x-x0)/a = (y-y0)/b = (z-z0)/c Intersection of line with plane: z = 0 x-x0 = a(z-z0)/c <=> x = x0+a(0-z0)/c <=> x = x0 -a*z0/c y = y0 - b*z0/c */ double lx = fPosition.x - mx; double ly = fPosition.y - my; double lz = fPosition.z - mz; double sum = lx*lx + ly*ly + lz*lz; double normal = sqrt(sum); double z0_c = fPosition.z / (lz/normal); vec->x = (float) (fPosition.x - (lx/normal)*z0_c); vec->y = (float) (fPosition.y - (ly/normal)*z0_c); vec->z = 0.0f; } I want to run It but with out precompiling. Is there any way to do such thing

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  • jquery calculation sum two different type of item

    - by st4n
    I'm writing a script like the example shown in the demo where All of the "Total" values (Including the "Grand Total") are Automatically Calculated using the calc () method. at this link: But I have some fields in which to apply the equation qty * price, and others where I want to do other operations .. you can tell me how? thank you very much i try with this, but it is a very stupid code .. and the grandTotal .. not sum the two different fields: function recalc() { $("[id^=total_item]").calc("qty * price", { qty: $("input[name^=qty_item_]"), price: $("[id^=price_item_]") }, function (s){ // return the number as a dollar amount return "$" + s.toFixed(2); }, function ($this){ // sum the total of the $("[id^=total_item]") selector var sum = $this.sum(); $("#grandTotal").text( // round the results to 2 digits "$" + sum.toFixed(2) ); }); $("[id^=total_otheritem]").calc("qty1 / price1", { qty1: $("input[name^=qty_other_]"), price1: $("[id^=price_other_]") }, function (s){ // return the number as a dollar amount return "$" + s.toFixed(2); }, function ($this){ var sum = $this.sum(); $("#grandTotal").text( // round the results to 2 digits "$" + sum.toFixed(2) ); }); }

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  • why i^=j^=i^=j isn't equal to *i^=*j^=*i^=*j

    - by klvoek
    In c , when there is variables (assume both as int) i less than j , we can use the equation i^=j^=i^=j to exchange the value of the two variables. For example, let int i = 3, j = 5; after computed i^=j^=i^=j, I got i = 5, j = 3 . What is so amazing to me. But, if i use two int pointers to re-do this , with *i^=*j^=*i^=*j , use the example above what i got will be i = 0 and j = 3. Then, describe it simply: In C 1 int i=3, j=5; i^=j^=i^=j; // after this i = 5, j=3 2 int i = 3, j= 5; int *pi = &i, *pj = &j; *pi^=*pj^=*pi^=*pj; // after this, $pi = 0, *pj = 5 In JavaScript var i=3, j=5; i^=j^=i^=j; // after this, i = 0, j= 3 the result in JavaScript makes this more interesting to me my sample code , on ubuntu server 11.0 & gcc #include <stdio.h> int main(){ int i=7, j=9; int *pi=&i, *pj=&j; i^=j^=i^=j; printf("i=%d j=%d\n", i, j); i=7, j==9; *pi^=*pj^=*pi^=*pj printf("i=%d j=%d\n", *pi, *pj); } however, i had spent hours to test and find out why, but nothing means. So, please help me. Or, just only i made some mistake???

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  • Solving for the coefficent of linear equations with one known coefficent

    - by CppLearner
    clc; clear all; syms y a2 a3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % [ 0.5 0.25 0.125 ] [ a2 ] [ y ] % [ 1 1 1 ] [ a3 ] = [ 3 ] % [ 2 4 8 ] [ 6 ] [ 2 ] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% M = [0.5 0.25 0.125; 1 1 1; 2 4 8]; t = [a2 a3 6]; r = [y 3 2]; sol = M * t' s1 = solve(sol(1), a2) % solve for a2 s2 = solve(sol(2), a3) % solve for a3 This is what I have so far. These are my output sol = conj(a2)/2 + conj(a3)/4 + 3/4 conj(a2) + conj(a3) + 6 2*conj(a2) + 4*conj(a3) + 48 s1 = - conj(a3)/2 - 3/2 - Im(a3)*i s2 = - conj(a2) - 6 - 2*Im(a2)*i sol looks like what we would have if we put them back into equation form: 0.5 * a2 + 0.25 * a3 + 0.125 * a4 a2 + a3 + a4 = 3 2*a2 + 4*a3 + 8*a4 = 2 where a4 is known == 6. My problem is, I am stuck with how to use solve to actually solve these equations to get the values of a2 and a3. s2 solve for a3 but it doesn't match what we have on paper (not quite). a2 + a3 + 6 = 3 should yield a3 = -3 - a2. because of the imaginary. Somehow I need to equate the vector solution sol to the values [y 3 2] for each row.

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  • facebook graph api does not return all feed items on facebook page

    - by Nick Franceschina
    at the time of this question, if you go here: http://www.facebook.com/realplayer you'll see six posts down, I have posted a photo with a message of "#highfive Cincinnati, OH" but if you to either of these: http://graph.facebook.com/realplayer/feed http://graph.facebook.com/realplayer/tagged the JSON that is returned seemingly includes everything on the wall, except for MY post. there is another photo post from someone else down below mine, and it is showing up (and both my photo and his photo are in the "Fan photos" section) obviously, since I can see everything with these links already, it appears that access_token is not a part of the equation... BUT, some more info: if I use an access_token from a session that isn't me, I can't see the post in the JSON if I use an access_token from MY logged in session, then I DO see the post in the JSON so I'm very confused. if everyone in the world can see those posts on the wall without even authenticating, then I expect all of them to come back in the graph api as well. anyone have thoughts on this? I am aware of the "manage_page" permission... which I can use to get a list of accounts and special offline access tokens for those pages... and that's something I can explore... but it seems like alot of work when my post seemingly SHOULD be there in the graph

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  • Combining two data sets and plotting in matlab

    - by bautrey
    I am doing experiments with different operational amplifier circuits and I need to plot my measured results onto a graph. I have two data sets: freq1 = [.1 .2 .5 .7 1 3 4 6 10 20 35 45 60 75 90 100]; %kHz Vo1 = [1.2 1.6 1.2 2 2 2.4 14.8 20.4 26.4 30.4 53.6 68.8 90 114 140 152]; %mV V1 = 19.6; Acm = Vo1/(1000*V1); And: freq2 = [.1 .5 1 30 60 70 85 100]; %kHz Vo1 = [3.96 3.96 3.96 3.84 3.86 3.88 3.88 3.88]; %V V1 = .96; Ad = Vo1/(2*V1); (I would show my plots but apparently I need more reps for that) I need to plot the equation, CMRR vs freq: CMRR = 20*log10(abs(Ad/Acm)); The size of Ad and Acm are different and the frequency points do not match up, but the boundaries of both of these is the same, 100Hz to 100kHz (x-axis). On the line of CMRR, Matlab says that Ad and Acm matrix dimensions do not agree. How I think I would solve this is using freq1 as the x-axis for CMRR and then taking approximated points from Ad according to the value on freq1. Or I could do function approximations of Ad and Acm and then do the divide operator on those. I do not know how I would code up this two ideas. Any other ideas would helpful, especially simpler ones. Thanks

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  • Collision detection problems...

    - by thyrgle
    Hi, I have written the following: -(void) checkIfLineCollidesWithAll { float slope = ((160-L1Circle1.position.y)-(160-L1Circle2.position.y))/((240-L1Circle1.position.x)-(240-L1Circle2.position.x)); float b = (160-L1Circle1.position.y) - slope * (240-L1Circle1.position.x); if ((240-L1Sensor1.position.x) < (240-L1Circle1.position.x) && (240-L1Sensor1.position.x) < (240-L1Circle2.position.x) || ((240-L1Sensor1.position.x) > (240-L1Circle1.position.x) && (240-L1Sensor1.position.x) > (240-L1Circle2.position.x))) { [L1Sensor1 setTexture:[[CCTextureCache sharedTextureCache] addImage:@"SensorOk.png"]]; } else if (slope == INFINITY || slope == -INFINITY) { if (L1Circle1.position.y + 16 >= L1Sensor1.position.y || L1Circle1.position.y - 16 <= L1Sensor1.position.y) { [L1Sensor1 setTexture:[[CCTextureCache sharedTextureCache] addImage:@"SensorBad.png"]]; } else { [L1Sensor1 setTexture:[[CCTextureCache sharedTextureCache] addImage:@"SensorOk.png"]]; } } else if (160-L1Sensor1.position.y + 12 >= slope*(240-L1Sensor1.position.x) + b && 160-L1Sensor1.position.y - 12 <= slope*(240-L1Sensor1.position.x) + b) { [L1Sensor1 setTexture:[[CCTextureCache sharedTextureCache] addImage:@"SensorBad.png"]]; } else { [L1Sensor1 setTexture:[[CCTextureCache sharedTextureCache] addImage:@"SensorOk.png"]]; } } Basically what this does is finds m and b in the well known equation: y = mx + b and then substitutes coordinates of the L1Sensor1 (the circle I'm trying to detect if it it intersects with the line segment) to see if y = mx + b hold true. But, there are two problems, first, when slope approaches infinity the range of what the L1Sensor1 should "react" to (it "reacts" by changing its image) becomes smaller. Also, the code that should handle infinity is not working. Thanks for the help in advanced.

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  • Solving quadratic programming using R

    - by user702846
    I would like to solve the following quadratic programming equation using ipop function from kernlab : min 0.5*x'*H*x + f'*x subject to: A*x <= b Aeq*x = beq LB <= x <= UB where in our example H 3x3 matrix, f is 3x1, A is 2x3, b is 2x1, LB and UB are both 3x1. edit 1 My R code is : library(kernlab) H <- rbind(c(1,0,0),c(0,1,0),c(0,0,1)) f = rbind(0,0,0) A = rbind(c(1,1,1), c(-1,-1,-1)) b = rbind(4.26, -1.73) LB = rbind(0,0,0) UB = rbind(100,100,100) > ipop(f,H,A,b,LB,UB,0) Error in crossprod(r, q) : non-conformable arguments I know from matlab that is something like this : H = eye(3); f = [0,0,0]; nsamples=3; eps = (sqrt(nsamples)-1)/sqrt(nsamples); A=ones(1,nsamples); A(2,:)=-ones(1,nsamples); b=[nsamples*(eps+1); nsamples*(eps-1)]; Aeq = []; beq = []; LB = zeros(nsamples,1); UB = ones(nsamples,1).*1000; [beta,FVAL,EXITFLAG] = quadprog(H,f,A,b,Aeq,beq,LB,UB); and the answer is a vector of 3x1 equals to [0.57,0.57,0.57]; However when I try it on R, using ipop function from kernlab library ipop(f,H,A,b,LB,UB,0)) and I am facing Error in crossprod(r, q) : non-conformable arguments I appreciate any comment

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  • Vertical line on HxW canvas of pixels

    - by bobby williams
    I searched and found nothing. I'm trying to draw lines (simple y=mx+b ones) on a canvas of black pixels. It works fine, but no line occurs when it is vertical. I'm not sure why. My first if statement checks if the denominator is zero, therefore m is undefined and no need for a line equation. My second and third if statement check how steep it is and based on that, calculate the points in between. I don't think there is a need for other classes, since I think there is a bug in my code or I'm just not translating the mathematics into code properly. If more is needed, I'll be happy to post more. /** * Returns an collection of points that connects p1 and p2 */ public ArrayList getPoints() { ArrayList points = new ArrayList(); // checks to see if denominator in m is zero. if zero, undefined. if ((p2.getX() - p1.getX()) == 0) { for (int y = p1.getY(); y<p2.getY(); y++) { points.add(new Point(p1.getX(), y, getColor())); } } double m = (double)(p2.getY()-p1.getY())/(double)(p2.getX()-p1.getX()); int b = (int)(p1.getY() - (m * p1.getX())); // checks to see if slope is steep. if (m > -1 || m < 1) { for (int x = p1.getX(); x<p2.getX(); x++) { int y = (int) ((m*x)+b); points.add(new Point(x, y, getColor())); } } // checks to see if slope is not steep. if (m <= -1 || m >= 1) { for (int y = p1.getY(); y<p2.getY(); y++) { int x = (int) ((y-b)/m); points.add(new Point(x, y, getColor())); } } return points; }

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