Replace a value in a data frame based on a conditional if statement

Question

In the R data frame coded for below  I would like to replace all of the times that B  appears with b     junk  lt - data frame x  lt - rep LETTERS 1 4   3   y  lt - letters 1 12   colnames junk   lt - c  nm    val     this provides      nm val 1   A   a 2   B   b 3   C   c 4   D   d 5   A   e 6   B   f 7   C   g 8   D   h 9   A   i 10  B   j 11  C   k 12  D   l   My initial attempt was to use a for and if statements like so    for i in junk nm  if i  in   B   junk nm  lt -  b    but as I am sure you can see  this replaces ALL of the values of junk nm with b   I can see why this is doing this but I can t seem to get it to replace only those cases of junk nm where the original value was B   NOTE   I managed to solve the problem with gsub but in the interest of learning R I still would like to know how to get my original approach to work  if it is possible

User · Accepted Answer

Easier to convert nm to characters and then make the change:

junk$nm <- as.character(junk$nm)
junk$nm[junk$nm == "B"] <- "b"

EDIT: And if indeed you need to maintain nm as factors, add this in the end:

junk$nm <- as.factor(junk$nm)

User · Answer

another useful way to replace values   library plyr  junk nm  lt - revalue junk nm  c  B   b

User · Answer

stata replace lt -function data replacevar replacevalue ifs      ifs parse text ifs    yy as numeric eval ifs data parent frame       x sum yy    data cbind data yy    data yy  1 replacevar  replacevalue   message noquote paste0 x    replacement are made      print message    return data  1  ncol data -1        Call this function using below line   d stata replace d  under20  1  age lt 20

User · Answer

The easiest way to do this in one command is to use which command and also need not to change the factors into character by doing this   junk nm which junk nm   B    lt - b

User · Answer

Short answer is   junk nm junk nm  in   B    lt -  b    Take a look at Index vectors in R Introduction  if you don t read it yet      EDIT  As noticed in comments this solution works for character vectors so fail on your data   For factor best way is to change level   levels junk nm  levels junk nm    B    lt -  b

User · Answer

If you are working with character variables  note that stringsAsFactors is false here  you can use replace    junk  lt - data frame x  lt - rep LETTERS 1 4   3   y  lt - letters 1 12   stringsAsFactors   FALSE  colnames junk   lt - c  nm    val    junk nm  lt - replace junk nm  junk nm     B    b   junk      nm val   1   A   a   2   b   b   3   C   c   4   D   d

User · Answer

You have created a factor variable  in nm so you either need to avoid doing so or add an additional level to the factor attributes  You should also avoid using  lt - in the arguments to data frame    Option 1   junk  lt - data frame x   rep LETTERS 1 4   3   y  letters 1 12   stringsAsFactors FALSE  junk nm junk nm     B    lt -  b    Option 2   levels junk nm   lt - c levels junk nm    b   junk nm junk nm     B    lt -  b  junk

User · Answer

As the data you show are factors  it complicates things a little bit   diliop s Answer approaches the problem by converting to nm to a character variable  To get back to the original factors a further step is required   An alternative is to manipulate the levels of the factor in place    gt  lev  lt - with junk  levels nm    gt  lev lev     B    lt -  b   gt  junk2  lt - within junk  levels nm   lt - lev   gt  junk2    nm val 1   A   a 2   b   b 3   C   c 4   D   d 5   A   e 6   b   f 7   C   g 8   D   h 9   A   i 10  b   j 11  C   k 12  D   l   That is quite simple and I often forget that there is a replacement function for levels     Edit  As noted by  Seth in the comments  this can be done in a one-liner  without loss of clarity   within junk  levels nm  levels nm      B    lt -  b

[r] Replace a value in a data frame based on a conditional (`if`) statement

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