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  • Connect R to Quickbooks

    - by Btibert3
    Has anyone connected the R package to QuickBooks? I know there is an ODBC driver than can be bought. Just wondering if anyone has already gone down this road. Any insight will be much appreciated! ~ Brock

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  • Adding multiple vectors in R

    - by Elais
    I have a problem where I have to add thirty-three integer vectors of equal length from a dataset in R. I know the simple solution would be Vector1 + Vector2 + Vector3 +VectorN But I am sure there is a way to code this. Also some vectors have NA in place of integers so I need a way to skip those. I know this may be very basic but I am new to this.

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  • What is the optimal way to run a set of regressions in R.

    - by stevejb
    Assume that I have sources of data X and Y that are indexable, say matrices. And I want to run a set of independent regressions and store the result. My initial approach would be results = matrix(nrow=nrow(X), ncol=(2)) for(i in 1:ncol(X)) { matrix[i,] = coefficients(lm(Y[i,] ~ X[i,]) } But, loops are bad, so I could do it with lapply as out <- lapply(1:nrow(X), function(i) { coefficients(lm(Y[i,] ~ X[i,])) } ) Is there a better way to do this?

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  • Smooth Error in qplot from ggplot2

    - by Jared
    I have some data that I am trying to plot faceted by its Type with a smooth (Loess, LM, whatever) superimposed. Generation code is below: testFrame <- data.frame(Time=sample(20:60,50,replace=T),Dollars=round(runif(50,0,6)),Type=sample(c("First","Second","Third","Fourth"),50,replace=T,prob=c(.33,.01,.33,.33))) I have no problem either making a faceted plot, or plotting the smooth, but I cannnot do both. The first three lines of code below work fine. The fourth line is where I have trouble: qplot(Time,Dollars,data=testFrame,colour=Type) qplot(Time,Dollars,data=testFrame,colour=Type) + geom_smooth() qplot(Time,Dollars,data=testFrame) + facet_wrap(~Type) qplot(Time,Dollars,data=testFrame) + facet_wrap(~Type) + geom_smooth() It gives the following error: Error in [<-.data.frame(*tmp*, var, value = list(NA = NULL)) : missing values are not allowed in subscripted assignments of data frames What am I missing to overlay a smooth in a faceted plot? I could have sworn I had done this before, possibly even with the same data.

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  • how to do introspection in R (stat package)

    - by Lebron James
    Hi all, I am somewhat new to R, and i have this piece of code which generates a variable that i don't know the type for. Are there any introspection facility in R which will tell me which type this variable belongs to? The following illustrates the property of this variable: I am working on linear model selection, and the resource I have is lm result from another model. Now I want to retrieve the lm call by the command summary(model)$call so that I don't need to hardcode the model structure. However, since I have to change the dataset, I need to do a bit of modification on the "string", but apparently it is not a simple string. I wonder if there is any command similar to string.replace so that I can manipulate this variable from the variable $call. Thanks > str<-summary(rdnM)$call > str lm(formula = y ~ x1, data = rdndat) > str[1] lm() > str[2] y ~ x1() > str[3] rdndat() > str[3] <- data Warning message: In str[3] <- data : number of items to replace is not a multiple of replacement length > str lm(formula = y ~ x1, data = c(10, 20, 30, 40)) > str<-summary(rdnM)$call > str lm(formula = y ~ x1, data = rdndat) > str[3] <- 'data' > str lm(formula = y ~ x1, data = "data") > str<-summary(rdnM)$call > type str Error: unexpected symbol in "type str" >

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  • Parallel processing in R 2.11 Windows 64-bit using SNOW not quite working

    - by Abhijit
    I'm running R 2.11 64-bit on a WinXP64 machine with 8 processors. With R 2.10.1 the following code spawned 6 R processes for parallel processing: require(foreach) require(doSNOW) cl = makeCluster(6, type='SOCK') registerDoSNOW(cl) bl2 = foreach(i=icount(length(unqmrno))) %dopar% { (Some code here) } stopCluster(cl) When I run the same code in R 2.11 Win64, the 6 R processes are not spawning, and the code hangs. I'm wondering if this is a problem with the port of SNOW to 2.11-64bit, or if any additional code is required on my part. Thanks

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  • Plotting 3-tuple data points in a surface / contour plot using matplotlib

    - by morpheous
    I have some surface data that is generated by an external program as XYZ values. I want to create the following graphs, using matplotlib: Surface plot Contour plot Contour plot overlayed with a surface plot I have looked at several examples for plotting surfaces and contours in matplotlib - however, the Z values seems to be a function of X and Y i.e. Y ~ f(X,Y). I assume that I will somehow need to transform my Y variables, but I have not seen any example yet, that shows how to do this. So, my question is this: given a set of (X,Y,Z) points, how may I generate Surface and contour plots from that data? BTW, just to clarify, I do NOT want to create scatter plots. Also although I mentioned matplotlib in the title, I am not averse to using rpy(2), if that will allow me to create these charts.

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  • What is an efficient method for partitioning and aggregating intervals from timestamped rows in a da

    - by mattrepl
    From a data frame with timestamped rows (strptime results), what is the best method for aggregating statistics for intervals? Intervals could be an hour, a day, etc. I've found the aggregate function, but that doesn't help with assigning each row to an interval. I'm planning on adding a column to the data frame that denotes interval and using that with aggregate, but if there's a better solution it'd be great to hear it. Thanks for any pointers!

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  • What is the best way to run a loop of regressions in R?

    - by stevejb
    Assume that I have sources of data X and Y that are indexable, say matrices. And I want to run a set of independent regressions and store the result. My initial approach would be results = matrix(nrow=nrow(X), ncol=(2)) for(i in 1:ncol(X)) { matrix[i,] = coefficients(lm(Y[i,] ~ X[i,]) } But, loops are bad, so I could do it with lapply as out <- lapply(1:nrow(X), function(i) { coefficients(lm(Y[i,] ~ X[i,])) } ) Is there a better way to do this?

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  • Creating a Large Matrix in ff

    - by Ryan Rosario
    I am trying to create a huge matrix in ff, and I know that ff is good for this sort of thing. But, there is a major problem. The dimensions of the matrix exceed .Machine$max_integer! I am running on a 64 bit machine, using 64bit R and 64bit ff. Is there any way to get around this problem? It's been suggested that R is using the MAXINT value from stdint.h. Is there any way to fix this without changing that file and possibly breaking build? > ffMatrix <- ff(vmode="boolean", dim=c(1e10,1e10)) Error in if (length < 0 || length > .Machine$integer.max) stop("length must be between 1 and .Machine$integer.max") : missing value where TRUE/FALSE needed In addition: Warning message: In ff(vmode = "boolean", dim = c(1e+10, 1e+10)) : NAs introduced by coercion > 1e+10 > .Machine$integer.max [1] TRUE

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  • editing Rnw in Emacs, gets confused if in math mode or not

    - by stevejb
    When editing .Rnw files with emacs, sometimes it gets confused as to if I am in math mode or not. Then, the syntax highlighting gets messed up, and C-f-i inserts \textit{} and \mathit{} opposite to how it normally should. Is seems like there is some bool storing the state of math mode or not, and it gets inadvertently flipped. Is there a way I can manually flip it back?

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  • R library for discrete Markov chain simulation

    - by stevejb
    Hello, I am looking for something like the 'msm' package, but for discrete Markov chains. For example, if I had a transition matrix defined as such Pi <- matrix(c(1/3,1/3,1/3, 0,2/3,1/6, 2/3,0,1/2)) for states A,B,C. How can I simulate a Markov chain according to that transition matrix? Thanks,

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  • can lapply not modify variables in a higher scope

    - by stevejb
    I often want to do essentially the following: mat <- matrix(0,nrow=10,ncol=1) lapply(1:10, function(i) { mat[i,] <- rnorm(1,mean=i)}) But, I would expect that mat would have 10 random numbers in it, but rather it has 0. (I am not worried about the rnorm part. Clearly there is a right way to do that. I am worry about affecting mat from within an anonymous function of lapply) Can I not affect matrix mat from inside lapply? Why not? Is there a scoping rule of R that is blocking this?

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  • how to wrap a function that only takes individual elements to make it take a list

    - by stevejb
    Hello, Say I have a function handed to me that I cannot change and must use as is. This function takes several objects in the form of oldFunction( object1, object2, object3, ...) where ... are other arguments. I want to write a wrapper to take a list of objects. My idea was this. sjb.ListWrapper <- function(myList,...) { lLen <- length(myList) myStr <- "" for( i in 1:lLen) { myStr <- paste(myStr, "myList[[", i , "]],",sep="") } myCode <- paste("oldFunction(", myStr, "...)") eval({myCode}) } However, the issue is that I want to use this from Sweave and I need the output of oldFunction to be printed. What is the right way to do this? Thanks.

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  • What is the simplest method to fill the area under a geom_freqpoly line?

    - by mattrepl
    The x-axis is time broken up into time intervals. There is an interval column in the data frame that specifies the time for each row. The column is a factor, where each interval is a different factor level. Plotting a histogram or line using geom_histogram and geom_freqpoly works great, but I'd like to have a line, like that provided by geom_freqpoly, with the area filled. Currently I'm using geom_freqpoly like this: ggplot(quake.data, aes(interval, fill=tweet.type)) + geom_freqpoly(aes(group = tweet.type, colour = tweet.type)) + opts(axis.text.x=theme_text(angle=-60, hjust=0, size = 6)) I would prefer to have a filled area, such as provided by geom_density, but without smoothing the line: UPDATE: The geom_area has been suggested, is there any way to use a ggplot2-generated statistic, such as ..count.., for the geom_area's y-values? Or, does the count aggregation need to occur prior to using ggplot2?

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  • Is it possible to plot a single density over a discrete variable?

    - by mattrepl
    The x-axis is time broken up into time intervals. There is an interval column in the data frame that specifies the time for each row. Plotting a histogram or line using geom_histogram and geom_freqpoly works great, but I'd like to use geom_density to get a filled area. Perhaps there is a better way to achieve this. Right now, if I use geom_density, curves are created for each discrete factor level instead of smoothing over all of them.

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  • Using R to download zipped data file, extract, and import data

    - by Jeromy Anglim
    @EZGraphs on Twitter writes: "Lots of online csvs are zipped. Is there a way to download, unzip the archive, and load the data to a data.frame using R? #Rstats" I was also trying to do this today, but ended up just downloading the zip file manually. I tried something like: fileName <- "http://www.newcl.org/data/zipfiles/a1.zip" con1 <- unz(fileName, filename="a1.dat", open = "r") but I feel as if I'm a long way off. Any thoughts?

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  • Programming R/Sweave for proper \Sexpr output

    - by deoksu
    Hi I'm having a bit of a problem programming R for Sweave, and the #rstats twitter group often points here, so I thought I'd put this question to the SO crowd. I'm an analyst- not a programmer- so go easy on me my first post. Here's the problem: I am drafting a survey report in Sweave with R and would like to report the marginal returns in line using \Sexpr{}. For example, rather than saying: Only 14% of respondents said 'X'. I want to write the report like this: Only \Sexpr{p.mean(variable)}$\%$ of respondents said 'X'. The problem is that Sweave() converts the results of the expression in \Sexpr{} to a character string, which means that the output from expression in R and the output that appears in my document are different. For example, above I use the function 'p.mean': p.mean<- function (x) {options(digits=1) mmm<-weighted.mean(x, weight=weight, na.rm=T) print(100*mmm) } In R, the output looks like this: p.mean(variable) >14 but when I use \Sexpr{p.mean(variable)}, I get an unrounded character string (in this case: 13.5857142857143) in my document. I have tried to limit the output of my function to 'digits=1' in the global environment, in the function itself, and and in various commands. It only seems to contain what R prints, not the character transformation that is the result of the expression and which eventually prints in the LaTeX file. as.character(p.mean(variable)) >[1] 14 >[1] "13.5857142857143" Does anyone know what I can do to limit the digits printed in the LaTeX file, either by reprogramming the R function or with a setting in Sweave or \Sexpr{}? I'd greatly appreciate any help you can give. Thanks, David

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  • R: manipulating data.frames containing strings and booleans.

    - by Mike Dewar
    Hello. I have a data.frame in R; it's called p. Each element in the data.frame is either True or False. My variable p has, say, m rows and n columns. For every row there is strictly only one TRUE element. It also has column names, which are strings. What I would like to do is the following: For every row in p I see a TRUE I would like to replace with the name of the corresponding column I would then like to collapse the data.frame, which now contains FALSEs and column names, to a single vector, which will have m elements. I would like to do this in an R-thonic manner, so as to continue my enlightenment in R and contribute to a world without for-loops. I can do step 1 using the following for loop: for (i in seq(length(colnames(p)))) { p[p[,i]==TRUE,i]=colnames(p)[i] } but theres's no beauty here and I have totally subscribed to this for-loops-in-R-are-probably-wrong mentality. Maybe wrong is too strong but they're certainly not great. I don't really know how to do step 2. I kind of hoped that the sum of a string and FALSE would return the string but it doesn't. I kind of hoped I could use an OR operator of some kind but can't quite figure that out (Python responds to False or 'bob' with 'bob'). Hence, yet again, I appeal to you beautiful Rstats people for help!

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