[英]How to split a character column into multiple columns in R
我有一个数据框x
:
dput(x)
structure(list(District = structure(c(6L, 6L, 6L, 6L, 6L, 6L), .Label = c("District - Central (06)",
"District - East (04)", "District - New Delhi (05)", "District - North (02)",
"District - North East (03)", "District - North West (01)", "District - South (09)",
"District - South West (08)", "District - West (07)"), class = "factor"),
Age = structure(c(103L, 1L, 2L, 14L, 25L, 36L), .Label = c("0",
"1", "10", "100+", "11", "12", "13", "14", "15", "16", "17",
"18", "19", "2", "20", "21", "22", "23", "24", "25", "26",
"27", "28", "29", "3", "30", "31", "32", "33", "34", "35",
"36", "37", "38", "39", "4", "40", "41", "42", "43", "44",
"45", "46", "47", "48", "49", "5", "50", "51", "52", "53",
"54", "55", "56", "57", "58", "59", "6", "60", "61", "62",
"63", "64", "65", "66", "67", "68", "69", "7", "70", "71",
"72", "73", "74", "75", "76", "77", "78", "79", "8", "80",
"81", "82", "83", "84", "85", "86", "87", "88", "89", "9",
"90", "91", "92", "93", "94", "95", "96", "97", "98", "99",
"Age not stated", "All ages"), class = "factor"), Total = c(3656539L,
56131L, 58644L, 63835L, 63859L, 64945L), Rural = c(213950L,
3589L, 3757L, 4200L, 4102L, 4223L), Urban = c(3442589L, 52542L,
54887L, 59635L, 59757L, 60722L)), .Names = c("District",
"Age", "Total", "Rural", "Urban"), row.names = c(NA, 6L), class = "data.frame")
我想拆分District
列,将地区名称提取到新的Name
列中。 例如,“分区-西北(01)”应拆分为“西北”。 我尝试了str_split_fixed
并得到:
x
District Age Total Rural Urban 1 name
1 District - North West (01) All ages 3656539 213950 3442589 North West (01)
2 District - North West (01) 0 56131 3589 52542 North West (01)
3 District - North West (01) 1 58644 3757 54887 North West (01)
4 District - North West (01) 2 63835 4200 59635 North West (01)
5 District - North West (01) 3 63859 4102 59757 North West (01)
6 District - North West (01) 4 64945 4223 60722 North West (01)
我尝试再次使用相同的功能来拆分name
列,以将区域名称与代码分开,但是它给了我以下错误:
stri_split_regex中的错误(字符串,模式,n = n,简化= TRUE,opts_regex = attr(模式,:在regexp模式中嵌套的括号不正确。)(U_REGEX_MISMATCHED_PAREN)
有没有一种方法可以根据单个函数中的模式将字符列分为多个列?
你可以用
library(stringr)
data.frame(str_split_fixed(df$District, " ", 3))
X1 X2 X3
1 District - North West (01)
2 District - North West (01)
3 District - North West (01)
4 District - North West (01)
5 District - North West (01)
6 District - North West (01)
您可以使用gsub
删除此处的多余内容,
gsub("[[:digit:]]","",df$X3)
gsub("[[:punct:]]","",df$X3)
等等
您可以使用gsub
获得所需的内容:
gsub("^.* +- +([A-Za-z ]+) \\(.*$", "\\1", df$District)
[1] "North West" "North West" "North West" "North West" "North West" "North West"
gsub
(“ ^。* +-+([A-Za-z] +)\\(。* $”)的第一个参数是一个正则表达式,其解释如下:
从字符串“ ^”的开头开始,匹配任何字符“。*”,后跟至少一个空格,一个连字符和至少一个空格“ +-+”。 然后捕获由(至少一个)字母和空格“ [A-Za-z] +”组成的下一个文本“()”。 到达带括号“ \\\\(”的空格时停止捕获,然后匹配所有内容,直到文本“。* $”的末尾。
gsub
的第二个参数“ \\\\ 1”表示用括号捕获的文本替换文本。
要将其分配给变量:
df$name <- gsub("^.* +- +([A-Za-z ]+) \\(.*$", "\\1", df$District)
您还可以匹配并提取:
library(stringi)
library(dplyr)
library(purrr)
mutate(x,
name=map_chr(stri_match_all_regex(District, "- ([[:alpha:]]+ [[:alpha:]]+) "), function(x) x[,2]),
code=map_chr(stri_match_all_regex(District, "\\(([[:digit:]]+)\\)"), function(x) x[,2]))
## District Age Total Rural Urban name code
## 1 District - North West (01) All ages 3656539 213950 3442589 North West 01
## 2 District - North West (01) 0 56131 3589 52542 North West 01
## 3 District - North West (01) 1 58644 3757 54887 North West 01
## 4 District - North West (01) 2 63835 4200 59635 North West 01
## 5 District - North West (01) 3 63859 4102 59757 North West 01
## 6 District - North West (01) 4 64945 4223 60722 North West 01
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