Question

我有一个包含两列的数据集：

Quantity    SKU
1,1         2494008,2493953
1,1,1       2167550,1336380,2365409
3,2,1,6,1   1428608,1137956,2401393,2679310,2579183

结束状态是一个如下所示的数据集：

Quantity    SKU
1           2494008
1           2493953
1           2167550
1           1336380
1           2365409
3           1428608
2           1137956
1           2401393

6 2679310 1 2579183

cplit和strsplit如果要分割单个变量here，则可以工作，但我需要拆分两个变量（上面的数量和SKU）。

Answer 1

dat <- read.table(text="Quantity    SKU
1,1         2494008,2493953
1,1,1       2167550,1336380,2365409
3,2,1,6,1   1428608,1137956,2401393,2679310,2579183", header=TRUE, stringsAsFactors=FALSE)
dat2<-data.frame(Quantity = unlist(strsplit(dat$Quantity, split=",")), 
             SKU=unlist(strsplit(dat$SKU, split=",")), row.names = NULL)

dat3 <- as.data.frame(do.call(cbind, lapply(dat, function(x) unlist(strsplit(x, ",")))))

# Quantity     SKU
# 1         1 2494008
# 2         1 2493953
# 3         1 2167550
# 4         1 1336380
# 5         1 2365409
# 6         3 1428608
# 7         2 1137956
# 8         1 2401393
# 9         6 2679310
# 10        1 2579183

Answer 2

不出所料，data.table解决方案非常类似于提议的基础R解决方案by lmo：

library(data.table)
data.table(dat)[, lapply(.SD, function(x) unlist(strsplit(x, ",")))]

    Quantity     SKU
 1:        1 2494008
 2:        1 2493953
 3:        1 2167550
 4:        1 1336380
 5:        1 2365409
 6:        3 1428608
 7:        2 1137956
 8:        1 2401393
 9:        6 2679310
10:        1 2579183

如果需要，可以保留行号：

data.table(dat)[, rn := .I][, lapply(.SD, function(x) unlist(strsplit(x, ","))), rn]

    rn Quantity     SKU
 1:  1        1 2494008
 2:  1        1 2493953
 3:  2        1 2167550
 4:  2        1 1336380
 5:  2        1 2365409
 6:  3        3 1428608
 7:  3        2 1137956
 8:  3        1 2401393
 9:  3        6 2679310
10:  3        1 2579183

数据

dat <- structure(list(Quantity = c("1,1", "1,1,1", "3,2,1,6,1"), SKU = c("2494008,2493953", 
"2167550,1336380,2365409", "1428608,1137956,2401393,2679310,2579183"
)), .Names = c("Quantity", "SKU"), class = "data.frame", row.names = c(NA, -3L))

Answer 3

来自separate_rows的{{1}}：

tidyr

将多个逗号分隔列拆分为单独的行

3 个答案:

数据