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  • Calculating Growth-Rates by applying log-differences

    - by mropa
    I am trying to transform my data.frame by calculating the log-differences of each column and controlling for the rows id. So basically I like to calculate the growth rates for each id's variable. So here is a random df with an id column, a time period colum p and three variable columns: df <- data.frame (id = c("a","a","a","c","c","d","d","d","d","d"), p = c(1,2,3,1,2,1,2,3,4,5), var1 = rnorm(10, 5), var2 = rnorm(10, 5), var3 = rnorm(10, 5) ) df id p var1 var2 var3 1 a 1 5.375797 4.110324 5.773473 2 a 2 4.574700 6.541862 6.116153 3 a 3 3.029428 4.931924 5.631847 4 c 1 5.375855 4.181034 5.756510 5 c 2 5.067131 6.053009 6.746442 6 d 1 3.846438 4.515268 6.920389 7 d 2 4.910792 5.525340 4.625942 8 d 3 6.410238 5.138040 7.404533 9 d 4 4.637469 3.522542 3.661668 10 d 5 5.519138 4.599829 5.566892 Now I have written a function which does exactly what I want BUT I had to take a detour which is possibly unnecessary and can be removed. However, somehow I am not able to locate the shortcut. Here is the function and the output for the posted data frame: fct.logDiff <- function (df) { df.log <- dlply (df, "code", function(x) data.frame (p = x$p, log(x[, -c(1,2)]))) list.nalog <- llply (df.log, function(x) data.frame (p = x$p, rbind(NA, sapply(x[,-1], diff)))) ldply (list.nalog, data.frame) } fct.logDiff(df) id p var1 var2 var3 1 a 1 NA NA NA 2 a 2 -0.16136569 0.46472004 0.05765945 3 a 3 -0.41216720 -0.28249264 -0.08249587 4 c 1 NA NA NA 5 c 2 -0.05914281 0.36999681 0.15868378 6 d 1 NA NA NA 7 d 2 0.24428771 0.20188025 -0.40279188 8 d 3 0.26646102 -0.07267311 0.47041227 9 d 4 -0.32372771 -0.37748866 -0.70417351 10 d 5 0.17405309 0.26683625 0.41891802 The trouble is due to the added NA-rows. I don't want to collapse the frame and reduce it, which would be automatically done by the diff() function. So I had 10 rows in my original frame and am keeping the same amount of rows after the transformation. In order to keep the same length I had to add some NAs. I have taken a detour by transforming the data.frame into a list, add the NAs, and afterwards transform the list back into a data.frame. That looks tedious. Any ideas to avoid the data.frame-list-data.frame class transformation and optimize the function?

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  • Coding the R-ight way - avoiding the for loop

    - by mropa
    I am going through one of my .R files and by cleaning it up a little bit I am trying to get more familiar with writing the code the r-ight way. As a beginner, one of my favorite starting points is to get rid of the for() loops and try to transform the expression into a functional programming form. So here is the scenario: I am assembling a bunch of data.frames into a list for later usage. dataList <- list (dataA, dataB, dataC, dataD, dataE ) Now I like to take a look at each data.frame's column names and substitute certain character strings. Eg I like to substitute each "foo" and "bar" with "baz". At the moment I am getting the job done with a for() loop which looks a bit awkward. colnames(dataList[[1]]) [1] "foo" "code" "lp15" "bar" "lh15" colnames(dataList[[2]]) [1] "a" "code" "lp50" "ls50" "foo" matchVec <- c("foo", "bar") for (i in seq(dataList)) { for (j in seq(matchVec)) { colnames (dataList[[i]])[grep(pattern=matchVec[j], x=colnames (dataList[[i]]))] <- c("baz") } } Since I am working here with a list I thought about the lapply function. My attempts handling the job with the lapply function all seem to look alright but only at first sight. If I write f <- function(i, xList) { gsub(pattern=c("foo"), replacement=c("baz"), x=colnames(xList[[i]])) } lapply(seq(dataList), f, xList=dataList) the last line prints out almost what I am looking for. However, if i take another look at the actual names of the data.frames in dataList: lapply (dataList, colnames) I see that no changes have been made to the initial character strings. So how can I rewrite the for() loop and transform it into a functional programming form? And how do I substitute both strings, "foo" and "bar", in an efficient way? Since the gsub() function takes as its pattern argument only a character vector of length one.

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