我有一个data.frame d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 $list中用“,”分隔的所有字符串的成对组合.
我可以使用lapply获得它,如下所示.
d2 <- strsplit(d$list, split = ", ")
d2 <- lapply(d2, function(x) as.data.frame(t(combn(x, m=2))))
library(data.table)
d2 <- rbindlist(d2)
我不希望将d $列表中的每个组的计数与组合列表d2作为新列.如何使用data.table执行此操作？
library(stringi)
stri_count_fixed(d$list,", ")
期望的输出如下
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
最佳答案 使用gsub,我们可以删除除分隔符(,)之外的所有字符,使用nchar计算字符数,添加1以获取字数,并使用transform创建新列“Count”.使用splitstackshape中的cSplit,我们可以将’list’列拆分为,通过指定方向为long,我们重新格式化数据集.加载splitstackshape也会加载data.table,因此我们可以使用data.table聚合方法.通过’sno’和’Count'(.(sno,Count))分组,我们得到’list’的组合,根据combn输出中的交替值创建两列(‘V1′,’V2’),并分配’sno’列为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
或者修改你的代码,我们使用Map / cbind在’d2’中创建’Count’列,并且如帖子中所述,执行rbindlist将列表折叠为单个’data.table’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
