93

I have a script that reads in data from a CSV file into a data.table and then splits the text in one column into several new columns. I am currently using the lapply and strsplit functions to do this. Here's an example:

library("data.table")
df = data.table(PREFIX = c("A_B","A_C","A_D","B_A","B_C","B_D"),
                VALUE  = 1:6)
dt = as.data.table(df)

# split PREFIX into new columns
dt$PX = as.character(lapply(strsplit(as.character(dt$PREFIX), split="_"), "[", 1))
dt$PY = as.character(lapply(strsplit(as.character(dt$PREFIX), split="_"), "[", 2))

dt 
#    PREFIX VALUE PX PY
# 1:    A_B     1  A  B
# 2:    A_C     2  A  C
# 3:    A_D     3  A  D
# 4:    B_A     4  B  A
# 5:    B_C     5  B  C
# 6:    B_D     6  B  D

In the example above the column PREFIX is split into two new columns PX and PY on the "_" character.

Even though this works just fine, I was wondering if there is a better (more efficient) way to do this using data.table. My real datasets have >=10M+ rows, so time/memory efficiency becomes really important.

UPDATE:

Following @Frank's suggestion I created a larger test case and used the suggested commands, but the stringr::str_split_fixed takes a lot longer than the original method.

library("data.table")
library("stringr")
system.time ({
    df = data.table(PREFIX = rep(c("A_B","A_C","A_D","B_A","B_C","B_D"), 1000000),
                    VALUE  = rep(1:6, 1000000))
    dt = data.table(df)
})
#   user  system elapsed 
#  0.682   0.075   0.758 

system.time({ dt[, c("PX","PY") := data.table(str_split_fixed(PREFIX,"_",2))] })
#    user  system elapsed 
# 738.283   3.103 741.674 

rm(dt)
system.time ( {
    df = data.table(PREFIX = rep(c("A_B","A_C","A_D","B_A","B_C","B_D"), 1000000),
                     VALUE = rep(1:6, 1000000) )
    dt = as.data.table(df)
})
#    user  system elapsed 
#   0.123   0.000   0.123 

# split PREFIX into new columns
system.time ({
    dt$PX = as.character(lapply(strsplit(as.character(dt$PREFIX), split="_"), "[", 1))
    dt$PY = as.character(lapply(strsplit(as.character(dt$PREFIX), split="_"), "[", 2))
})
#    user  system elapsed 
#  33.185   0.000  33.191

So the str_split_fixed method takes about 20X times longer.

4

5 回答 5

141

更新：从版本 1.9.6（截至 15 年 9 月在 CRAN 上），我们可以使用该函数tstrsplit()直接获取结果（并且以更有效的方式）：

require(data.table) ## v1.9.6+
dt[, c("PX", "PY") := tstrsplit(PREFIX, "_", fixed=TRUE)]
#    PREFIX VALUE PX PY
# 1:    A_B     1  A  B
# 2:    A_C     2  A  C
# 3:    A_D     3  A  D
# 4:    B_A     4  B  A
# 5:    B_C     5  B  C
# 6:    B_D     6  B  D

tstrsplit()基本上是 的包装器transpose(strsplit())，其中transpose()最近实现的函数转置了一个列表。请参见?tstrsplit()和?transpose()示例。

请参阅历史以获取旧答案。

于 2013-08-10T01:00:58.647 回答
15

我为不使用data.table v1.9.5并且还想要一个单行解决方案的人添加了答案。

dt[, c('PX','PY') := do.call(Map, c(f = c, strsplit(PREFIX, '-'))) ]
于 2015-04-01T07:48:26.003 回答
7

使用splitstackshape包：

library(splitstackshape)
cSplit(df, splitCols = "PREFIX", sep = "_", direction = "wide", drop = FALSE)
#    PREFIX VALUE PREFIX_1 PREFIX_2
# 1:    A_B     1        A        B
# 2:    A_C     2        A        C
# 3:    A_D     3        A        D
# 4:    B_A     4        B        A
# 5:    B_C     5        B        C
# 6:    B_D     6        B        D
于 2016-05-16T20:07:09.013 回答
6

我们可以尝试：

library(data.table)  
cbind(dt, fread(text = dt$PREFIX, sep = "_", header = FALSE))
    #    PREFIX VALUE V1 V2
    # 1:    A_B     1  A  B
    # 2:    A_C     2  A  C
    # 3:    A_D     3  A  D
    # 4:    B_A     4  B  A
    # 5:    B_C     5  B  C
    # 6:    B_D     6  B  D
于 2018-10-04T16:40:20.293 回答
3

使用 tidyr 的解决方案是：

separate(df,col = "PREFIX",into = c("PX", "PY"), sep = "_")
于 2017-07-30T13:12:32.697 回答