How to transform Columns to rows in R?

问题:

I kind of have the same problem. I have data in this kind of order: ;=column

D1 ;hurs

1  ;0.12

1  ;0.23

1  ;0.34

1  ;0.01

2  ;0.24

2  ;0.67

2  ;0.78

2  ;0.98

and I like to have it like this:

D1; X; X; X; X    
1;0.12; 0.23; 0.34; 0.01; 
2;0.24; 0.67; 0.78; 0.98;

I would like to sort it with respect to D1 and like to reshape it? Does anyone have an idea? I need to do this for 7603 values of D1.

回答1:

I would look into Hadley's reshape package. It does all sorts of great stuff. The code below will work with your toy example, but there is probably a more elegant way of doing this. Simply, your data already appear to be in the ?melt form, so you can simply ?cast it.

Also, check out these links

http://www.statmethods.net/management/reshape.html

http://had.co.nz/reshape/

library(reshape)

help(package=reshape)
?melt

D1 <- c(1,1,1,1,2,2,2,2)
hurs <- c(.12, .23, .34, .01, .24, .67, .78, .98)
var <- rep(paste("X", 1:4, sep=""), 2)

foo <- data.frame(D1, var, hurs)
foo

cast(foo, D1~var)

回答2:

Digging up skeletons not likely to ever be claimed, why not use aggregate()?

dat = read.table(header = TRUE, sep = ";", text = "D1 ;hurs
1  ;0.12
1  ;0.23
1  ;0.34
1  ;0.01
2  ;0.24
2  ;0.67
2  ;0.78
2  ;0.98")
aggregate(hurs ~ D1, dat, c)
#   D1 hurs.1 hurs.2 hurs.3 hurs.4
# 1  1   0.12   0.23   0.34   0.01
# 2  2   0.24   0.67   0.78   0.98

If the lengths of each id in D1 are not the same, you can also use base R reshape() after first creating a "time" variable:

dat2 <- dat[-8, ]
dat2$timeSeq <- ave(dat2$D1, dat2$D1, FUN = seq_along)
reshape(dat2, direction="wide", idvar="D1", timevar="timeSeq")
#   D1 hurs.1 hurs.2 hurs.3 hurs.4
# 1  1   0.12   0.23   0.34   0.01
# 5  2   0.24   0.67   0.78     NA

回答3:

I have assumed that there are unequal number of hurs per D1 (7603 values)

txt = 'D1 ;hurs
 1 ;0.12
 1 ;0.23
 1 ;0.34
 1 ;0.01
 2 ;0.24
 2 ;0.67
 2 ;0.78
 2 ;0.98'

dat <- read.table(textConnection(txt),header=T,sep=";")
dat$Lp <- 1:nrow(dat)
dat <- dat[order(dat$D1,dat$Lp),]
out <- split(dat$hurs,dat$D1)
out <- sapply(names(out),function(x) paste(paste(c(x,out[[x]]),collapse=";"),";",sep="",collapse=""))

回答4:

reshape2 is actually better than reshape. Using reshape uses significantly more memory and time than reshape2 (at least for my specific example using something like 9million rows).

回答5:

You might check Hadley Wickham's reshape package and its cast() function

http://had.co.nz/reshape/