Question

> dc1
  V1                V2
1 20140211-0100     |Box
2 20140211-1782     |Office|Ball
3 20140211-1783     |Office
4 20140211-1784     |Office
5 20140221-0756     |Box
6 20140203-0418     |Box
> strsplit(as.character(dc1[,2]),"^\\|")
[[1]]
[1] ""    "Box"


[[2]]
[1] ""             "Office" "Ball"


[[3]]
[1] ""             "Office"


[[4]]
[1] ""             "Office"


[[5]]
[1] ""    "Box"


[[6]]
[1] ""    "Box"

如何从 strsplit 结果中删除空白（＆＃34;＆＃34;）。结果应如下所示：

[[1]]
[1] "Box"

[[2]]
[1] "Office"    "Ball"

Answer 1

您可以在列表中查看使用lapply。我更改了strsplit的定义以匹配您的预期输出。

dc1 <- read.table(text = 'V1                V2
1 20140211-0100     |Box
2 20140211-1782     |Office|Ball
3 20140211-1783     |Office
4 20140211-1784     |Office
5 20140221-0756     |Box
6 20140203-0418     |Box', header = TRUE)

out <- strsplit(as.character(dc1[,2]),"\\|")

> lapply(out, function(x){x[!x ==""]})
[[1]]
[1] "Box"

[[2]]
[1] "Office" "Ball"  

[[3]]
[1] "Office"

[[4]]
[1] "Office"

[[5]]
[1] "Box"

[[6]]
[1] "Box"

Answer 2

我没有全球解决方案，但对于您的示例，您可以尝试：

strsplit(sub("^\\|", "", as.character(dc1[,2])),"\\|")

在执行拆分之前，它删除了第一个|（这是正则表达式"^\\|"所说的），这是""的原因。

Answer 3

您可以使用：

library(stringr)
str_extract_all(dc1[,2], "[[:alpha:]]+")
[[1]]
 [1] "Box"

[[2]]
 [1] "Office" "Ball"  

[[3]]
 [1] "Office"

[[4]]
 [1] "Office"

[[5]]
 [1] "Box"

[[6]]
 [1] "Box"

Answer 4

在这种情况下，您只需在"["

中调用sapply即可删除每个向量的第一个元素

> sapply(strsplit(as.character(dc1[,2]), "\\|"), "[", -1)
# [[1]]
# [1] "Box"

# [[2]]
# [1] "Office" "Ball"  

# [[3]]
# [1] "Office"

# [[4]]
# [1] "Office"

# [[5]]
# [1] "Box"

# [[6]]
# [1] "Box"

Answer 5

另一种方法在取消列出nzchar()：

的结果后使用strsplit()

out <- unlist(strsplit(as.character(dc1[,2]),"\\|"))

out[nzchar(x=out)] # removes the extraneous "" marks

Answer 6

library("stringr")

lapply(str_split(dc1$V2, "\\|"), function(x) x[-1])

[[1]]
[1] "Box"

[[2]]
[1] "Office" "Ball"  

[[3]]
[1] "Office"

[[4]]
[1] "Office"

[[5]]
[1] "Box"

[[6]]
[1] "Box"

从R中的strsplit删除空白

6 个答案: