成对组合以及data.table计数

Question

我有一个data.framed如下。

d <- structure(list(sno = 1:7, list = c("SD1, SD44, SD384, SD32", 
"SD23, SD1, SD567", "SD42, SD345, SD183", "SD345, SD340, SD387", 
"SD455, SD86, SD39", "SD12, SD315, SD387", "SD32, SD1, SD40")), .Names = c("sno", 
"list"), row.names = c(NA, -7L), class = "data.frame")

d
  sno                   list
1   1 SD1, SD44, SD384, SD32
2   2       SD23, SD1, SD567
3   3     SD42, SD345, SD183
4   4    SD345, SD340, SD387
5   5      SD455, SD86, SD39
6   6     SD12, SD315, SD387
7   7        SD32, SD1, SD40

我想得到d$list中由“，”分隔的所有字符串的成对组合。

我可以使用lappy获得它，如下所示。

d2 <- strsplit(d$list, split = ", ")
d2 <- lapply(d2, function(x) as.data.frame(t(combn(x, m=2))))
library(data.table)
d2 <- rbindlist(d2)

我不想在d$list中包含每个组的计数以及组合列表d2作为新列。如何使用data.table？

library(stringi)
stri_count_fixed(d$list,", ")

所需输出如下

out <- structure(list(V1 = structure(c(1L, 1L, 1L, 3L, 3L, 2L, 4L, 4L, 
1L, 6L, 6L, 5L, 5L, 5L, 7L, 8L, 8L, 9L, 10L, 10L, 11L, 12L, 12L, 
1L), .Label = c("SD1", "SD384", "SD44", "SD23", "SD345", "SD42", 
"SD340", "SD455", "SD86", "SD12", "SD315", "SD32"), class = "factor"), 
    V2 = structure(c(3L, 2L, 1L, 2L, 1L, 1L, 4L, 5L, 5L, 7L, 
    6L, 6L, 8L, 9L, 9L, 11L, 10L, 10L, 12L, 9L, 9L, 4L, 13L, 
    13L), .Label = c("SD32", "SD384", "SD44", "SD1", "SD567", 
    "SD183", "SD345", "SD340", "SD387", "SD39", "SD86", "SD315", 
    "SD40"), class = "factor"), count = c(4, 4, 4, 4, 4, 4, 3, 
    3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3)), .Names = c("V1", 
"V2", "count"), row.names = c(NA, -24L), class = "data.frame")


out 
      V1    V2 count
1    SD1  SD44     4
2    SD1 SD384     4
3    SD1  SD32     4
4   SD44 SD384     4
5   SD44  SD32     4
6  SD384  SD32     4
7   SD23   SD1     3
8   SD23 SD567     3
9    SD1 SD567     3
10  SD42 SD345     3
11  SD42 SD183     3
12 SD345 SD183     3
13 SD345 SD340     3
14 SD345 SD387     3
15 SD340 SD387     3
16 SD455  SD86     3
17 SD455  SD39     3
18  SD86  SD39     3
19  SD12 SD315     3
20  SD12 SD387     3
21 SD315 SD387     3
22  SD32   SD1     3
23  SD32  SD40     3
24   SD1  SD40     3

Answer 1

使用gsub可以删除除分隔符（，）之外的所有字符，使用nchar计算字符数，添加1以获得字数，并使用transform创建一个新列“count”。使用来自splitstackshape的cSplit，我们可以按方向将“列表”列拆分，通过将方向指定为长，我们重新格式化数据集。加载splitstackshape也将加载数据。表，因此我们可以使用数据。表聚合方法。按“sno”和“Count”（（sno，Count））分组，我们得到“list”的combn，根据combn输出的交替值创建两列（“V1”，“V2”），并将“sno”列指定为NULL（如果不需要）

library(splitstackshape)
d1 <-  transform(d, Count=nchar(gsub('[^,]', '', list))+1L)
cSplit(d1, 'list', ', ', 'long')[, {
       tmp <- combn(as.character(list), 2)
       list(V1=tmp[c(TRUE, FALSE)], V2= tmp[c(FALSE, TRUE)])
         }, .(sno, Count)][, 
         sno:= NULL]
 #   Count    V1    V2
 #1:     4   SD1  SD44
 #2:     4   SD1 SD384
 #3:     4   SD1  SD32
 #4:     4  SD44 SD384
 #5:     4  SD44  SD32
 #6:     4 SD384  SD32
 #7:     3  SD23   SD1
 #8:     3  SD23 SD567
 #9:     3   SD1 SD567
#10:     3  SD42 SD345
#11:     3  SD42 SD183
#12:     3 SD345 SD183
#13:     3 SD345 SD340
#14:     3 SD345 SD387
#15:     3 SD340 SD387
#16:     3 SD455  SD86
#17:     3 SD455  SD39
#18:     3  SD86  SD39
#19:     3  SD12 SD315
#20:     3  SD12 SD387
#21:     3 SD315 SD387
#22:     3  SD32   SD1
#23:     3  SD32  SD40
#24:     3   SD1  SD40

或者修改您的代码，我们使用Map/cbind在“d2”中创建“Count”列，如本文所述，执行rbindlist以将列表折叠为单个数据。“表”对象。

library(stringi)
library(data.table)
Count <- stri_count_fixed(d$list,", ")+1
d2 <- strsplit(d$list, split = ", ")
d2 <- lapply(d2, function(x) as.data.frame(t(combn(x, m=2))))
rbindlist(Map(cbind, d2, Count=Count))
#      V1    V2 Count
# 1:   SD1  SD44     4
# 2:   SD1 SD384     4
# 3:   SD1  SD32     4
# 4:  SD44 SD384     4
# 5:  SD44  SD32     4
# 6: SD384  SD32     4
# 7:  SD23   SD1     3
# 8:  SD23 SD567     3
# 9:   SD1 SD567     3
#10:  SD42 SD345     3
#11:  SD42 SD183     3
#12: SD345 SD183     3
#13: SD345 SD340     3
#14: SD345 SD387     3
#15: SD340 SD387     3
#16: SD455  SD86     3
#17: SD455  SD39     3
#18:  SD86  SD39     3
#19:  SD12 SD315     3
#20:  SD12 SD387     3
#21: SD315 SD387     3
#22:  SD32   SD1     3
#23:  SD32  SD40     3
#24:   SD1  SD40     3