Re: [R] dataframe operation
In conversing offline with Indermaur it seems that the elements of b are supposed to correspond to the rows rather than columns. In that case we can have the simpler solution: 0 * DF + b On 1/24/07, Gabor Grothendieck [EMAIL PROTECTED] wrote: Here is a slight variation on Marc's idea: isna - is.na(DF) DF[] - replace(100 * col(isna), isna, NA) On 1/24/07, Marc Schwartz [EMAIL PROTECTED] wrote: On Wed, 2007-01-24 at 14:16 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 14:10 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 20:27 +0100, Indermaur Lukas wrote: hi i have a dataframe a which looks like: column1, column2, column3 10,12, 0 NA, 0,1 12,NA,50 i want to replace all values in column1 to column3 which do not contain NA with values of vector b (100,200,300). any idea i can do it? i appreciate any hint regards lukas Here is one possibility: sapply(seq(along = colnames(DF)), function(x) ifelse(is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 10 120 [2,] 10001 [3,] 12 200 50 Note that the returned object will be a matrix, so if you need a data frame, just coerce the result with as.data.frame(). OKthat's what I get for pulling the trigger too fast. Just reverse the logic in the function: sapply(seq(along = colnames(DF)), function(x) ifelse(!is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 100 200 300 [2,] NA 200 300 [3,] 100 NA 300 I misread the query initially. Here is another possibility, which may be faster depending upon the actual size and dims of your initial data frame. Preallocate a matrix of replacement values: Mat - matrix(rep(seq(along = colnames(DF)) * 100, each = nrow(DF)), ncol = ncol(DF)) Mat [,1] [,2] [,3] [1,] 100 200 300 [2,] 100 200 300 [3,] 100 200 300 Now do the replacement: ifelse(!is.na(DF), Mat, NA) column1 column2 column3 1 100 200 300 2 NA 200 300 3 100 NA 300 In doing some testing, the above may be about 10 times faster than using sapply() in my first solution, again depending upon the structure of your DF. HTH, Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe operation
On Wed, 2007-01-24 at 20:27 +0100, Indermaur Lukas wrote: hi i have a dataframe a which looks like: column1, column2, column3 10,12, 0 NA, 0,1 12,NA,50 i want to replace all values in column1 to column3 which do not contain NA with values of vector b (100,200,300). any idea i can do it? i appreciate any hint regards lukas Here is one possibility: sapply(seq(along = colnames(DF)), function(x) ifelse(is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 10 120 [2,] 10001 [3,] 12 200 50 Note that the returned object will be a matrix, so if you need a data frame, just coerce the result with as.data.frame(). HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe operation
On Wed, 2007-01-24 at 14:10 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 20:27 +0100, Indermaur Lukas wrote: hi i have a dataframe a which looks like: column1, column2, column3 10,12, 0 NA, 0,1 12,NA,50 i want to replace all values in column1 to column3 which do not contain NA with values of vector b (100,200,300). any idea i can do it? i appreciate any hint regards lukas Here is one possibility: sapply(seq(along = colnames(DF)), function(x) ifelse(is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 10 120 [2,] 10001 [3,] 12 200 50 Note that the returned object will be a matrix, so if you need a data frame, just coerce the result with as.data.frame(). OKthat's what I get for pulling the trigger too fast. Just reverse the logic in the function: sapply(seq(along = colnames(DF)), function(x) ifelse(!is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 100 200 300 [2,] NA 200 300 [3,] 100 NA 300 I misread the query initially. HTH, Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe operation
Hint: Try ?subset at the R prompt Indermaur Lukas [EMAIL PROTECTED] wrote: hi i have a dataframe a which looks like: column1, column2, column3 10,12, 0 NA, 0,1 12,NA,50 i want to replace all values in column1 to column3 which do not contain NA with values of vector b (100,200,300). any idea i can do it? i appreciate any hint -- Mike Prager, NOAA, Beaufort, NC * Opinions expressed are personal and not represented otherwise. * Any use of tradenames does not constitute a NOAA endorsement. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe operation
On Wed, 2007-01-24 at 14:16 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 14:10 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 20:27 +0100, Indermaur Lukas wrote: hi i have a dataframe a which looks like: column1, column2, column3 10,12, 0 NA, 0,1 12,NA,50 i want to replace all values in column1 to column3 which do not contain NA with values of vector b (100,200,300). any idea i can do it? i appreciate any hint regards lukas Here is one possibility: sapply(seq(along = colnames(DF)), function(x) ifelse(is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 10 120 [2,] 10001 [3,] 12 200 50 Note that the returned object will be a matrix, so if you need a data frame, just coerce the result with as.data.frame(). OKthat's what I get for pulling the trigger too fast. Just reverse the logic in the function: sapply(seq(along = colnames(DF)), function(x) ifelse(!is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 100 200 300 [2,] NA 200 300 [3,] 100 NA 300 I misread the query initially. Here is another possibility, which may be faster depending upon the actual size and dims of your initial data frame. Preallocate a matrix of replacement values: Mat - matrix(rep(seq(along = colnames(DF)) * 100, each = nrow(DF)), ncol = ncol(DF)) Mat [,1] [,2] [,3] [1,] 100 200 300 [2,] 100 200 300 [3,] 100 200 300 Now do the replacement: ifelse(!is.na(DF), Mat, NA) column1 column2 column3 1 100 200 300 2 NA 200 300 3 100 NA 300 In doing some testing, the above may be about 10 times faster than using sapply() in my first solution, again depending upon the structure of your DF. HTH, Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dataframe operation
Here is a slight variation on Marc's idea: isna - is.na(DF) DF[] - replace(100 * col(isna), isna, NA) On 1/24/07, Marc Schwartz [EMAIL PROTECTED] wrote: On Wed, 2007-01-24 at 14:16 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 14:10 -0600, Marc Schwartz wrote: On Wed, 2007-01-24 at 20:27 +0100, Indermaur Lukas wrote: hi i have a dataframe a which looks like: column1, column2, column3 10,12, 0 NA, 0,1 12,NA,50 i want to replace all values in column1 to column3 which do not contain NA with values of vector b (100,200,300). any idea i can do it? i appreciate any hint regards lukas Here is one possibility: sapply(seq(along = colnames(DF)), function(x) ifelse(is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 10 120 [2,] 10001 [3,] 12 200 50 Note that the returned object will be a matrix, so if you need a data frame, just coerce the result with as.data.frame(). OKthat's what I get for pulling the trigger too fast. Just reverse the logic in the function: sapply(seq(along = colnames(DF)), function(x) ifelse(!is.na(DF[[x]]), 100 * x, DF[[x]])) [,1] [,2] [,3] [1,] 100 200 300 [2,] NA 200 300 [3,] 100 NA 300 I misread the query initially. Here is another possibility, which may be faster depending upon the actual size and dims of your initial data frame. Preallocate a matrix of replacement values: Mat - matrix(rep(seq(along = colnames(DF)) * 100, each = nrow(DF)), ncol = ncol(DF)) Mat [,1] [,2] [,3] [1,] 100 200 300 [2,] 100 200 300 [3,] 100 200 300 Now do the replacement: ifelse(!is.na(DF), Mat, NA) column1 column2 column3 1 100 200 300 2 NA 200 300 3 100 NA 300 In doing some testing, the above may be about 10 times faster than using sapply() in my first solution, again depending upon the structure of your DF. HTH, Marc __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
