how about gdata functions?
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2
= sample(LETTERS[2:6], 10, replace = TRUE)))
tmp.orig=tmp
library(gdata)
bigMap=mapLevels(x=list(factor(tmp[,"R1"]),factor(tmp[,"R2"])),combine =
T,codes=FALSE)
mapLevels(tmp[,"
No, that would remap B to A. Convert to character before doing this, then back
to factors.
--
Sent from my phone. Please excuse my brevity.
On July 6, 2017 4:43:00 PM PDT, Ista Zahn wrote:
>Untested, but I expect that setting the levels to be the same across
>the
>two factors
>
>levels(tmp$R1)
Untested, but I expect that setting the levels to be the same across the
two factors
levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6]
and proceeding as before should be fine.
Best,
Ista
On Jul 6, 2017 6:54 PM, "Gang Chen" wrote:
Thanks a lot, Ista! I really appreciate it.
How about a slightl
Thanks a lot, Ista! I really appreciate it.
How about a slightly different case as the following:
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace =
TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE)))
x R1 R2
1 C B
2 B B
3 C E
4 E C
5
How about
foo <- with(list(r1 = tmp$R1,
r2 = tmp$R2,
swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 != 0),
{
tmp[swapme, "R1"] <- r2[swapme]
tmp[swapme, "R2"] <- r1[swapme]
tmp
})
Best,
Ista
On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen wrote:
Suppose that we have the following dataframe:
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace =
TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE)))
x R1 R2
1 1 B B
2 2 B A
3 3 C D
4 4 E B
5 5 B D
6 6 E C
7 7 E D
8 8 D E
9 9
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