问题
I have a bunch of columns in a dataframe which I want to paste together (seperated by \"-\") as follows:
data <- data.frame(\'a\' = 1:3,
\'b\' = c(\'a\',\'b\',\'c\'),
\'c\' = c(\'d\', \'e\', \'f\'),
\'d\' = c(\'g\', \'h\', \'i\'))
i.e.
a b c d
1 a d g
2 b e h
3 c f i
Which I want to become:
a x
1 a-d-g
2 b-e-h
3 c-f-i
I could normally do this with:
within(data, x <- paste(b,c,d,sep=\'-\'))
and then removing the old columns, but unfortunately I do not know the names of the columns specifically, only a collective name for all of the columns, e.g. I would know that cols <- c(\'b\',\'c\',\'d\')
Does anyone know a way of doing this?
回答1:
# your starting data..
data <- data.frame('a' = 1:3, 'b' = c('a','b','c'), 'c' = c('d', 'e', 'f'), 'd' = c('g', 'h', 'i'))
# columns to paste together
cols <- c( 'b' , 'c' , 'd' )
# create a new column `x` with the three columns collapsed together
data$x <- apply( data[ , cols ] , 1 , paste , collapse = "-" )
# remove the unnecessary columns
data <- data[ , !( names( data ) %in% cols ) ]
回答2:
As a variant on baptiste's answer, with data defined as you have and the columns that you want to put together defined in cols
cols <- c("b", "c", "d")
You can add the new column to data and delete the old ones with
data$x <- do.call(paste, c(data[cols], sep="-"))
for (co in cols) data[co] <- NULL
which gives
> data
a x
1 1 a-d-g
2 2 b-e-h
3 3 c-f-i
回答3:
Using tidyr package, this can be easily handled in 1 function call.
data <- data.frame('a' = 1:3,
'b' = c('a','b','c'),
'c' = c('d', 'e', 'f'),
'd' = c('g', 'h', 'i'))
tidyr::unite_(data, paste(colnames(data)[-1], collapse="_"), colnames(data)[-1])
a b_c_d
1 1 a_d_g
2 2 b_e_h
3 3 c_f_i
Edit: Exclude first column, everything else gets pasted.
# tidyr_0.6.3
unite(data, newCol, -a)
# or by column index unite(data, newCol, -1)
# a newCol
# 1 1 a_d_g
# 2 2 b_e_h
# 3 3 c_f_i
回答4:
I'd construct a new data.frame:
d <- data.frame('a' = 1:3, 'b' = c('a','b','c'), 'c' = c('d', 'e', 'f'), 'd' = c('g', 'h', 'i'))
cols <- c( 'b' , 'c' , 'd' )
data.frame(a = d[, 'a'], x = do.call(paste, c(d[ , cols], list(sep = '-'))))
回答5:
Just to add additional solution with Reduce which probably is slower than do.call but probebly better than apply because it will avoid the matrix conversion. Also, instead a for loop we could just use setdiff in order to remove unwanted columns
cols <- c('b','c','d')
data$x <- Reduce(function(...) paste(..., sep = "-"), data[cols])
data[setdiff(names(data), cols)]
# a x
# 1 1 a-d-g
# 2 2 b-e-h
# 3 3 c-f-i
Alternatively we could update data in place using the data.table package (assuming fresh data)
library(data.table)
setDT(data)[, x := Reduce(function(...) paste(..., sep = "-"), .SD[, mget(cols)])]
data[, (cols) := NULL]
data
# a x
# 1: 1 a-d-g
# 2: 2 b-e-h
# 3: 3 c-f-i
Another option is to use .SDcols instead of mget as in
setDT(data)[, x := Reduce(function(...) paste(..., sep = "-"), .SD), .SDcols = cols]
回答6:
I benchmarked the answers of Anthony Damico, Brian Diggs and data_steve on a small sample tbl_df and got the following results.
> data <- data.frame('a' = 1:3,
+ 'b' = c('a','b','c'),
+ 'c' = c('d', 'e', 'f'),
+ 'd' = c('g', 'h', 'i'))
> data <- tbl_df(data)
> cols <- c("b", "c", "d")
> microbenchmark(
+ do.call(paste, c(data[cols], sep="-")),
+ apply( data[ , cols ] , 1 , paste , collapse = "-" ),
+ tidyr::unite_(data, "x", cols, sep="-")$x,
+ times=1000
+ )
Unit: microseconds
expr min lq mean median uq max neval
do.call(paste, c(data[cols], sep = "-")) 65.248 78.380 93.90888 86.177 99.3090 436.220 1000
apply(data[, cols], 1, paste, collapse = "-") 223.239 263.044 313.11977 289.514 338.5520 743.583 1000
tidyr::unite_(data, "x", cols, sep = "-")$x 376.716 448.120 556.65424 501.877 606.9315 11537.846 1000
However, when I evaluated on my own tbl_df with ~1 million rows and 10 columns the results were quite different.
> microbenchmark(
+ do.call(paste, c(data[c("a", "b")], sep="-")),
+ apply( data[ , c("a", "b") ] , 1 , paste , collapse = "-" ),
+ tidyr::unite_(data, "c", c("a", "b"), sep="-")$c,
+ times=25
+ )
Unit: milliseconds
expr min lq mean median uq max neval
do.call(paste, c(data[c("a", "b")], sep="-")) 930.7208 951.3048 1129.334 997.2744 1066.084 2169.147 25
apply( data[ , c("a", "b") ] , 1 , paste , collapse = "-" ) 9368.2800 10948.0124 11678.393 11136.3756 11878.308 17587.617 25
tidyr::unite_(data, "c", c("a", "b"), sep="-")$c 968.5861 1008.4716 1095.886 1035.8348 1082.726 1759.349 25
回答7:
In my opinion the sprintf-function deserves a place among these answers as well. You can use sprintf as follows:
do.call(sprintf, c(d[cols], '%s-%s-%s'))
which gives:
[1] "a-d-g" "b-e-h" "c-f-i"
And to create the required dataframe:
data.frame(a = d$a, x = do.call(sprintf, c(d[cols], '%s-%s-%s')))
giving:
a x
1 1 a-d-g
2 2 b-e-h
3 3 c-f-i
Although sprintf doesn't have a clear advantage over the do.call/paste combination of @BrianDiggs, it is especially usefull when you also want to pad certain parts of desired string or when you want to specify the number of digit. See ?sprintf for the several options.
Another variant would be to use pmap from purrr:
pmap(d[2:4], paste, sep = '-')
Note: this pmap solution only works when the columns aren't factors.
A benchmark on a larger dataset:
# create a larger dataset
d2 <- d[sample(1:3,1e6,TRUE),]
# benchmark
library(microbenchmark)
microbenchmark(
docp = do.call(paste, c(d2[cols], sep="-")),
appl = apply( d2[, cols ] , 1 , paste , collapse = "-" ),
tidr = tidyr::unite_(d2, "x", cols, sep="-")$x,
docs = do.call(sprintf, c(d2[cols], '%s-%s-%s')),
times=10)
results in:
Unit: milliseconds
expr min lq mean median uq max neval cld
docp 214.1786 226.2835 297.1487 241.6150 409.2495 493.5036 10 a
appl 3832.3252 4048.9320 4131.6906 4072.4235 4255.1347 4486.9787 10 c
tidr 206.9326 216.8619 275.4556 252.1381 318.4249 407.9816 10 a
docs 413.9073 443.1550 490.6520 453.1635 530.1318 659.8400 10 b
Used data:
d <- data.frame(a = 1:3, b = c('a','b','c'), c = c('d','e','f'), d = c('g','h','i'))
回答8:
library(plyr)
ldply(apply(data, 1, function(x) data.frame(
x = paste(x[2:4],sep="",collapse="-"))))
# x
#1 a-d-g
#2 b-e-h
#3 c-f-i
# and with just the vector of names you have:
ldply(apply(data, 1, function(x) data.frame(
x = paste(x[c('b','c','d')],sep="",collapse="-"))))
# or equally:
mynames <-c('b','c','d')
ldply(apply(data, 1, function(x) data.frame(
x = paste(x[mynames],sep="",collapse="-"))))
来源：https://stackoverflow.com/questions/14568662/paste-multiple-columns-together