将多个列粘贴在一起

Question

我在数据框中有一堆列,我想粘贴在一起(用" - "分隔),如下所示:

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  

我想成为:

a x  
1 a-d-g  
2 b-e-h  
3 c-f-i  

我通常可以这样做:

within(data, x <- paste(b,c,d,sep='-'))

然后删除旧列,但不幸的是我不知道具体列的名称,只是所有列的集体名称,例如我会知道 cols <- c('b','c','d')

有谁知道这样做的方法？

Answer 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 ) ]

Answer 2

作为baptiste答案的变体,data定义为你所拥有的和你想要放在一起的列cols

cols <- c("b", "c", "d")

您可以使用添加新列data和删除旧列

data$x <- do.call(paste, c(data[cols], sep="-"))
for (co in cols) data[co] <- NULL

这使

> data
  a     x
1 1 a-d-g
2 2 b-e-h
3 3 c-f-i

Answer 3

使用tidyr包,可以在1个函数调用中轻松处理.

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

编辑:排除第一列,其他所有内容都被粘贴.

# 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

Answer 4

我将构建一个新的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 = '-'))))

Answer 5

只是为了添加额外的解决方案Reduce,可能do.call比apply它更慢,但探测性能更好,因为它会避免matrix转换.另外,for我们可以使用循环setdiff来删除不需要的列

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

或者,我们可以data使用data.table包更新(假设新数据)

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

另一种选择是使用.SDcols,而不是mget作为

setDT(data)[, x := Reduce(function(...) paste(..., sep = "-"), .SD), .SDcols = cols]

Answer 6

我在一个小样本上对 Anthony Damico、Brian Diggs 和 data_steve 的答案进行了基准测试，tbl_df并得到了以下结果。

> 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

但是，当我自己评估tbl_df大约 100 万行和 10 列时，结果完全不同。

> 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

Answer 7

在我看来，sprintf-function 也应该在这些答案中占有一席之地。您可以sprintf按如下方式使用：

do.call(sprintf, c(d[cols], '%s-%s-%s'))

这使：

 [1] "a-d-g" "b-e-h" "c-f-i"

并创建所需的数据框：

data.frame(a = d$a, x = do.call(sprintf, c(d[cols], '%s-%s-%s')))

给予：

  a     x
1 1 a-d-g
2 2 b-e-h
3 3 c-f-i

虽然与@BrianDiggssprintf的do.call/paste组合相比没有明显的优势，但当您还想填充所需字符串的某些部分或要指定数字时，它特别有用。有关?sprintf多个选项，请参阅。

另一种变体是使用pmapfrom purrr：

pmap(d[2:4], paste, sep = '-')

注意：此pmap解决方案仅在列不是因子时才有效。

更大数据集的基准：

# 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)

结果是：

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 

使用数据：

d <- data.frame(a = 1:3, b = c('a','b','c'), c = c('d','e','f'), d = c('g','h','i')) 

Answer 8

这是一种非常规（但快速）的方法：使用fwritefromdata.table将列“粘贴”在一起，然后fread将其读回。为方便起见，我将这些步骤编写为名为 的函数fpaste：

fpaste <- function(dt, sep = ",") {
  x <- tempfile()
  fwrite(dt, file = x, sep = sep, col.names = FALSE)
  fread(x, sep = "\n", header = FALSE)
}

下面是一个例子：

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")

fpaste(d[cols], "-")
#       V1
# 1: a-d-g
# 2: b-e-h
# 3: c-f-i

它的表现如何？

d2 <- d[sample(1:3,1e6,TRUE),]
  
library(microbenchmark)
microbenchmark(
  docp = do.call(paste, c(d2[cols], sep="-")),
  tidr = tidyr::unite_(d2, "x", cols, sep="-")$x,
  docs = do.call(sprintf, c(d2[cols], '%s-%s-%s')),
  appl = apply( d2[, cols ] , 1 , paste , collapse = "-" ),
  fpaste = fpaste(d2[cols], "-")$V1,
  dt2 = as.data.table(d2)[, x := Reduce(function(...) paste(..., sep = "-"), .SD), .SDcols = cols][],
  times=10)
# Unit: milliseconds
#    expr        min         lq      mean     median         uq       max neval
#    docp  215.34536  217.22102  220.3603  221.44104  223.27224  225.0906    10
#    tidr  215.19907  215.81210  220.7131  220.09636  225.32717  229.6822    10
#    docs  281.16679  285.49786  289.4514  286.68738  290.17249  312.5484    10
#    appl 2816.61899 3106.19944 3259.3924 3266.45186 3401.80291 3804.7263    10
#  fpaste   88.57108   89.67795  101.1524   90.59217   91.76415  197.1555    10
#     dt2  301.95508  310.79082  384.8247  316.29807  383.94993  874.4472    10

Answer 9

Simple and straightforward code with unite from {tidyr} v1.2.0

Solution with {tidyr v1.2.0}

library(tidyr)

data %>% unite("x", all_of(cols), remove = T, sep = "-")

• "x" is the name of the new column.
• all_of(cols) is a selection of what columns we want to merge. Using <tidy-select> the column names don't need to be hardcoded in.
• remove = T we remove the input columns
• sep = "-" we define the separator between values
• if there is NA, we can also add na.rm = TRUE

Output

#   a     x
# 1 1 a-d-g
# 2 2 b-e-h
# 3 3 c-f-i

Input Data

data <- 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

#   a b c d
# 1 1 a d g
# 2 2 b e h
# 3 3 c f i

*This solution is different from what already posted.