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  • I am validating for RegEx errors -- how can I return which keyword, it failed on?

    - by Trip
    I'm passing new objects through this set of regex : (?i)exp\s|(?i)expire\s|(?i)print|(?i)mention|(?i)spring|(?i)summer|(?i)winter|(?i)jan(\s|\.)|(?i)january|(?i)february|(?i)feb(\.|\s)|(?i)march|mar(\.|\s)|(?i)april|(?i)june|(?i)july|(?i)august|(?i)aug(\s|\.)|(?i)september|(?i)sept(\.|\s)|(?i)november|(?i)nov(\.|\s)|(?i)december|(?i)dec(\.|\s)|(?i)holiday|(?i)christmas|(?i)holloween|(?i)easter|(?i)season|(?i)ends|(?i)end If it errors, for example on the word christmas , how can I dynamically pull the word it errors on, and display it as the cause of the error?

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  • Size of an Array

    - by Florent
    Hi all ! I'm new in objective C and iphone and it's hard from me. How do you get the size of an NSarray ? ? ? and print it in the console using NSLog ? thanks now it's basic but hard for me.

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  • Python process will not exit

    - by oneself
    Hi, I'm use nosetests to run some tests. However, after the tests have finished running, the nosetests process just sits there, and will not exit. Is there anyway to diagnose this? Does Python have a facility similar to sending Java a kill -QUIT which will print a stack trace?

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  • Python's string.translate() doesn't fully work?

    - by Rhubarb
    Given this example, I get the error that follows: print u'\2033'.translate({2033:u'd'}) C:\Python26\lib\encodings\cp437.pyc in encode(self, input, errors) 10 11 def encode(self,input,errors='strict'): ---> 12 return codecs.charmap_encode(input,errors,encoding_map) 13 14 def decode(self,input,errors='strict'): UnicodeEncodeError: 'charmap' codec can't encode character u'\x83' in position 0

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  • removing elements incrementally from a list

    - by Javier
    Dear all, I've a list of float numbers and I would like to delete incrementally a set of elements in a given range of indexes, sth. like: for j in range(beginIndex, endIndex+1): print ("remove [%d] => val: %g" % (j, myList[j])) del myList[j] However, since I'm iterating over the same list, the indexes (range) are not valid any more for the new list. Does anybody has some suggestions on how to delete the elements properly? Best wishes

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  • C++ problem with string stream istringstream

    - by user69514
    I am reading a file in the following format 1001 16000 300 12.50 2002 24000 360 10.50 3003 30000 300 9.50 where the items are: loan id, principal, months, interest rate. I'm not sure what it is that I am doing wrong with my input string stream, but I am not reading the values correctly because only the loan id is read correctly. Everything else is zero. Sorry this is a homework, but I just wanted to know if you could help me identify my error. if( inputstream.is_open() ){ /** print the results **/ cout << fixed << showpoint << setprecision(2); cout << "ID " << "\tPrincipal" << "\tDuration" << "\tInterest" << "\tPayment" <<"\tTotal Payment" << endl; cout << "---------------------------------------------------------------------------------------------" << endl; /** assign line read while we haven't reached end of file **/ string line; istringstream instream; while( inputstream >> line ){ instream.clear(); instream.str(line); /** assing values **/ instream >> loanid >> principal >> duration >> interest; /** compute monthly payment **/ double ratem = interest / 1200.0; double expm = (1.0 + ratem); payment = (ratem * pow(expm, duration) * principal) / (pow(expm, duration) - 1.0); /** computer total payment **/ totalPayment = payment * duration; /** print out calculations **/ cout << loanid << "\t$" << principal <<"\t" << duration << "mo" << "\t" << interest << "\t$" << payment << "\t$" << totalPayment << endl; } }

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  • python how to find the median of a list

    - by user3450574
    I'm trying to write a function named median that takes a list as an input and returns the median value of the list. I'm working with Python 2.7.2 The list can be of any size and the numbers are not guaranteed to be in any particular order. If the list contains an even number of elements, the function should return the average of the middle two. This is the code I'm starting with: def median(list): print(median([7,12,3,1,6,9]))

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  • Lambda recursive PHP functions.

    - by Kendall Hopkins
    Is it possible to have a PHP function that is both recursive and anonymous (lambda). This is my attempt to get it to work, but it doesn't pass in the function name. $factorial = function( $n ) use ( $factorial ) { if( $n == 1 ) return 1; return $factorial( $n - 1 ) * $n; }; print $factorial( 5 ); I'm also aware that this is a bad way to implement factorial, it's just an example.

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  • Tried http post doesn't work

    - by Rebol Tutorial
    I wanted to try the example here http://www.codeconscious.com/rebol/rebol-net.html#HTTP print read/custom http://babelfish.altavista.com/translate.dyn reduce ['POST {text=REBOL+Rules&lp=en_fr}] Since the page has changed since I modified it to write clipboard:// read/custom http://babelfish.altavista.com/translate.dyn reduce ['POST {trtext=hello+world&lp=en_fr&btnTrTxt=Translate}] It does return an html page but it doesn't contain any translation. What did I miss thanks ?

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  • Properties of mbox message in mbox module in Python

    - by Rajasankar
    I trying my luck to manage my mailbox with python. My example code is for eachmail in mailbox.mbox(mboxfile): print eachmail['From'] I got following by printing entire content. Delivered-To Subject To Content-Type MIME-Version Message-Id Is there any full document showing what are all the properties I can get from the mbox message instance? Python docs doesn't specify any of these http://docs.python.org/library/mailbox.html#mailbox.mbox

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  • problem in getting desired date format in loop

    - by Rishi2686
    Hi there, I have a date format like this : $date1 = "Sun May 09 20:07:50 +0000 2010"; and I have to convert it to: 09-05-2010 I am using date("d-m-Y", $date1)); When I print this individually it gives proper result but I am using it in loop it gives me results like: 31-12-1969 I don't understand why this is so? Can you help, please?

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  • problem with list return type??

    - by kaushik
    my list has value such as m=[['na','1','2']['ka','31','45']['ra','3','5'] d=0 r=2 t=m[d][r] print t # this is givin number i.e 2 Now when I use this value u=[] u=m[t] I am getting an err msg saying type error list does take str values... i want to use like this how can i convert that t into a integer?? please suggest.. thanks..

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  • python-nth perfect square

    - by kasyap
    Write a program that computes the sum of the logarithms of all the primes from 2 to some number n, and print out the sum of the logs of the primes, the number n, and the ratio of these two quantities. Test this for different values of n.

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  • pagination - 10 pages per page

    - by arthur
    I have a pagination script that displays a list of all pages like so: prev [1][2][3][4][5][6][7][8][9][10][11][12][13][14] next But I would like to only show ten of the numbers at a time: prev [3][4][5][6][7][8][9][10][11][12] next How can I accomplish this? Here is my code so far: <?php /* Set current, prev and next page */ $page = (!isset($_GET['page']))? 1 : $_GET['page']; $prev = ($page - 1); $next = ($page + 1); /* Max results per page */ $max_results = 2; /* Calculate the offset */ $from = (($page * $max_results) - $max_results); /* Query the db for total results. You need to edit the sql to fit your needs */ $result = mysql_query("select title from topics"); $total_results = mysql_num_rows($result); $total_pages = ceil($total_results / $max_results); $pagination = ''; /* Create a PREV link if there is one */ if($page > 1) { $pagination .= '< a hr_ef="?page='.$prev.'">Previous</a> '; } /* Loop through the total pages */ for($i = 1; $i <= $total_pages; $i++) { if(($page) == $i) { $pagination .= $i; } else { $pagination .= '< a hr_ef="index.php?page='.$i.'">'.$i.'</a>'; } } /* Print NEXT link if there is one */ if($page < $total_pages) { $pagination .= '< a hr_ef="?page='.$next.'"> Next</a>'; } /* Now we have our pagination links in a variable($pagination) ready to print to the page. I pu it in a variable because you may want to show them at the top and bottom of the page */ /* Below is how you query the db for ONLY the results for the current page */ $result=mysql_query("select * from topics LIMIT $from, $max_results "); while ($i = mysql_fetch_array($result)) { echo $i['title'].'<br />'; } echo $pagination; ?>

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  • Is there a Silverlight equivalent to "Application.OpenForms"?

    - by lightmeetsdark
    Basically, I'm trying to take information entered by the user on one page and print it out to another page via a "printer friendly" version, or report, of something. I have a MainPage.xaml that is, as the name suggests, my main page, but in a window there is the subpage AdCalculator.xaml where the user enters the information and PrintEstimate.xaml that is navigated to via a button on AdCalculator. I would like to be able to transfer the information entered in the textboxes from AdCalculator and print it out via text blocks in PrintEstimate. So in order to do that I have the following code: Views.AdCalculator AdCalc = new Views.AdCalculator(); string PrintCompanyName = AdCalc.CompanyName; string PrintContactName = AdCalc.txt_CustomerName.Text; string PrintBillingAddress1 = AdCalc.txt_BillingAddress.Text; string PrintBillingAddress2 = AdCalc.txt_BillingAddressLine2.Text; string PrintPhoneNumber = AdCalc.txt_PhoneNumber.Text; string PrintNumOfAds = AdCalc.txt_NumofAds.Text; string PrintRateOfPlay = AdCalc.Cmb_Rate.SelectedValue.ToString(); string PrintNumOfMonths = AdCalc.txt_NumofMonths.Text; string PrintTotalDue = AdCalc.txt_InvoiceSummary_TotalDue.Text; PrintEstimate PrintEstimatePage = new PrintEstimate(); PrintEstimatePage.txt_CompanyName.Text = PrintCompanyName; PrintEstimatePage.txt_CustomerName.Text = PrintContactName; PrintEstimatePage.txt_BillingAddress.Text = PrintBillingAddress1; PrintEstimatePage.txt_BillingAddressLine2.Text = PrintBillingAddress2; PrintEstimatePage.txt_PhoneNumber.Text = PrintPhoneNumber; PrintEstimatePage.txt_InvoiceSummary_NumofAds.Text = PrintNumOfAds; PrintEstimatePage.txt_InvoiceSummary_RateofPlay.Text = PrintRateOfPlay; PrintEstimatePage.txt_InvoiceSummary_NumOfMonths.Text = PrintNumOfMonths; PrintEstimatePage.txt_EstimateTotal.Text = PrintTotalDue; Only problem is, when I instantiate the new AdCalculator page, it clears the values, so nothing is actually retained as far as user-input goes. Following a lead from a colleague, I believe all I need to do is change the line Views.AdCalculator AdCalc = new Views.AdCalculator(); to Views.AdCalculator AdCalc = (AdCalculator)Application.OpenForms["AdCalculator"]; except the "Apllication.OpenForms" doesn't register. I know there are a lot of differences in the way C# code-behind is laid out for silverlight applications, so I didn't know if there was an equivalent that anyone knew about to "Application.OpenForms" that would help solve my issue or if there was any other way to go about getting my task done. Thank you very much!

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  • Create lags with a for-loop in R

    - by cptn
    I've got a data.frame with stock data of several companies (here it's only two). I want 10 additional columns in my stock data.frame df with lagged dates (from -5 days to +5 days) for both companies in my event data.frame. I'm using a for loop which is probably not the best solution, but it works partially. DATE <- c("01.01.2000","02.01.2000","03.01.2000","06.01.2000","07.01.2000","09.01.2000","10.01.2000","01.01.2000","02.01.2000","04.01.2000","06.01.2000","07.01.2000","09.01.2000","10.01.2000") RET <- c(-2.0,1.1,3,1.4,-0.2, 0.6, 0.1, -0.21, -1.2, 0.9, 0.3, -0.1,0.3,-0.12) COMP <- c("A","A","A","A","A","A","A","B","B","B","B","B","B","B") df <- data.frame(DATE, RET, COMP, stringsAsFactors=F) df # DATE RET COMP # 1 01.01.2000 -2.00 A # 2 02.01.2000 1.10 A # 3 03.01.2000 3.00 A # 4 06.01.2000 1.40 A # 5 07.01.2000 -0.20 A # 6 09.01.2000 0.60 A # 7 10.01.2000 0.10 A # 8 01.01.2000 -0.21 B # 9 02.01.2000 -1.20 B # 10 04.01.2000 0.90 B # 11 06.01.2000 0.30 B # 12 07.01.2000 -0.10 B # 13 09.01.2000 0.30 B # 14 10.01.2000 -0.12 B this loop works fine comp <- as.vector(unique(df$COMP)) mylist <- vector('list', length(comp)) # create lags in DATE for(i in 1:length(comp)) { print(i) comp_i <- comp[i] df_k <- df[df$COMP %in% comp_i, ] # all trading days of one firm df_k <- transform(df_k, DATEm1 = c(NA, head(DATE, -1)), DATEm2 = c(NA, NA, head(DATE, -2)), DATEm3 = c(NA, NA, NA, head(DATE, -3)), DATEm4 = c(NA, NA, NA, NA,head(DATE, -4)), DATEm5 = c(NA, NA, NA, NA, NA, head(DATE, -5)), DATEp1 = c(DATE[-1], NA)) #DATEp2 = c(DATE[-2], NA, NA), #DATEp3 = c(DATE[-3], NA, NA, NA), #DATEp4 = c(DATE[-4], NA, NA, NA, NA), #DATEp5 = c(DATE[-5], NA, NA, NA, NA, NA)) mylist[[i]] = df_k } df1 <- do.call(rbind, mylist) But if I add the lines with DATEp2, DATEp3, DATEp4, DATEp5. the code doesn't work. Can anybody tell me what I'm doing wrong here? Here the code with all the lagged dates. # create lags in DATE for(i in 1:length(comp)) { print(i) comp_i <- comp[i] df_k <- df[df$COMP %in% comp_i, ] # all trading days of one firm df_k <- transform(df_k, DATEm1 = c(NA, head(DATE, -1)), DATEm2 = c(NA, NA, head(DATE, -2)), DATEm3 = c(NA, NA, NA, head(DATE, -3)), DATEm4 = c(NA, NA, NA, NA,head(DATE, -4)), DATEm5 = c(NA, NA, NA, NA, NA, head(DATE, -5)), DATEp1 = c(DATE[-1], NA), DATEp2 = c(DATE[-2], NA, NA), DATEp3 = c(DATE[-3], NA, NA, NA), DATEp4 = c(DATE[-4], NA, NA, NA, NA), DATEp5 = c(DATE[-5], NA, NA, NA, NA, NA)) mylist[[i]] = df_k } df1 <- do.call(rbind, mylist)

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  • What's the non brute force way to filter a Python dictionary?

    - by Thierry Lam
    I can filter the following dictionary like: data = { 1: {'name': 'stackoverflow', 'traffic': 'high'}, 2: {'name': 'serverfault', 'traffic': 'low'}, 3: {'name': 'superuser', 'traffic': 'low'}, 4: {'name': 'mathoverflow', 'traffic': 'low'}, } traffic = 'low' for k, v in data.items(): if v['traffic'] == traffic: print k, v Is there an alternate way to do the above filtering?

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  • Clarification needed about Python CSV file format parsing

    - by HH
    Format is like: CHINA;2002-06-25 00:00:00.000;5,60 CHINA;2002-06-26 00:00:00.000;5,32 CHINA;2002-06-27 00:00:00.000;5,31 and I try to use Python's CSV tools to parse it but cannot understand the paragraph, source: And while the module doesn’t directly support parsing strings, it can easily be done: import csv for row in csv.reader(['one,two,three']): print row Could someone clarify the line ['one,two,three']? How would you use it with format A;B;C?

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  • decorating a function and adding functionalities preserving the number of argument

    - by pygabriel
    I'd like to decorate a function, using a pattern like this: def deco(func): def wrap(*a,**kw): print "do something" return func(*a,**kw) return wrap The problem is that if the function decorated has a prototype like that: def function(a,b,c): return When decorated, the prototype is destroyed by the varargs, for example, calling function(1,2,3,4) wouldn't result in an exception. Is that a way to avoid that? How can define the wrap function with the same prototype as the decorated (func) one? There's something conceptually wrong?

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