Replace rows in a data frame based on criteria

Question

I have the following data frame

I would like to replace all rows which have f1>3 with a row (id = x, f1 = 0, f2 = 0) So the above would map to

But when I tried

replace(x,which(x$f1>3),data.frame(id = 'x',f1=0,f2=0))

It didn't do it right, it gave

   id f1 f2
1  a  1  x
2  b  3  x
3  c  4  x
Warning message:
In `[<-.data.frame`(`*tmp*`, list, value = list(id = 1L, f1 = 0,  :
  provided 3 variables to replace 1 variables

Could someone suggest a way to do this at scale? Thanks.

I see no content relating to either 'plyr' or 'apply'. Suggest you remove those tags. `replace` only takes a vector as first argument, not a data.frame. — IRTFM, Sep 13 '12 at 17:31

Justin · Accepted Answer · 2012-09-13T17:42:16.593

10

Something like this:

 x[x$f1 > 3,] <- data.frame('x', 0, 0)

should do the trick!

as per @DWin's comment, this won't work with a factor id column. Using this same technique can work like this though:

levels(x$id) <- c(levels(x$id), 'x')
x[x$f1 > 3,] <- data.frame('x', 0, 0)
droplevels(x$id)

edited Sep 13 '12 at 17:42

answered Sep 13 '12 at 17:03

Justin

42,475
9
93
111

1

This appears to be an untested idea, since in the typical case where'id' is a factor it will fail. – IRTFM Sep 13 '12 at 17:26
Ah, touche! I have `options(stringsAsFactors=FALSE)` and it always bites me in the butt. – Justin Sep 13 '12 at 17:33
Ah, indeed. Probably a very wise move for personal use, but not so wise for SO testing. – IRTFM Sep 13 '12 at 17:35
@Dwin updated with `levels` fiddling, thanks for the reminder – Justin Sep 13 '12 at 17:42
Yep. and unlike the other answer it will replace multiple instances of matched criteria. – IRTFM Sep 13 '12 at 17:45

Jilber Urbina · Answer 2 · 2012-09-13T18:02:29.940

Here's another approach (taking into account @DWin's comment)

dfrm <- data.frame(id=letters[1:3], f1=c(1,3,4), f2=c(3,5,7))

dfrm[dfrm$f1>3, -1] <- 0
levels(dfrm$id) <- c(levels(dfrm$id), 'x')
dfrm[,1]<- with(dfrm, replace(id, f1==0, 'x')) ; dfrm

     id f1 f2
  1  a  1  3
  2  b  3  5
  3  x  0  0

Using some more data:

set.seed(007); N <- 12
 f1 <- sample(1:9, N, replace=TRUE)
 f2 <- sample(1:9, N, replace=TRUE)
 dfrm2 <- data.frame(id=letters[1:N], f1=f1, f2=f2)

 dfrm2[dfrm2$f1>3, -1] <- 0
 levels(dfrm2$id) <- c(levels(dfrm2$id), 'x')
 dfrm2[,1]<- with(dfrm2, replace(id, f1==0, 'x')) ; dfrm2
   id f1 f2
1   x  0  0
2   x  0  0
3   c  2  5
4   d  1  1
5   e  3  6
6   x  0  0
7   x  0  0
8   x  0  0
9   i  2  6
10  x  0  0
11  k  2  9
12  l  3  9

I deleted my previous code, because they weren't useful for this question.

Neither of these will replace all the instances when there are multiple such rows. Try it with rbind(dfrm,dfrm). — IRTFM, Sep 13 '12 at 17:33
Thank you @DWin, I've just edited my answer, it's not as elegant as Justin's but it works (I hope). — Jilber Urbina, Sep 13 '12 at 18:10

Replace rows in a data frame based on criteria

2 Answers2

Linked