我有一个带有“冒号”列的data.frame。每一项罪行都包括一篇文章(Art),一段(Abs)和一段(Ziff):
df<-data.frame(offence=c("Art. 110 Abs. 3 StGB","Art. 10 Abs. 1 StGB", "Art. 122 SVG", "Art. 1 Ziff. 2 UWG"))
> df
offence
1 Art. 110 Abs. 3 StGB
2 Art. 10 Abs. 1 StGB
3 Art. 122 SVG
4 Art. 1 Ziff. 2 UWG
但我需要以这种形式拥有它:
Art Ziff Abs Law
1 110 NA 3 StGB
2 10 NA 1 StGB
3 122 NA NA SVG
4 1 2 NA UWG
获得此结果的最佳方法是什么? lapply?
谢谢!
答案 0 :(得分:1)
您可以使用str_extract
中的stringr
:
library(stringr)
library(dplyr)
df$offence %>%
{data.frame(Art = str_extract(., "(?<=Art[.]\\s)\\d+"),
Ziff = str_extract(., "(?<=Ziff[.]\\s)\\d+"),
Abs = str_extract(., "(?<=Abs[.]\\s)\\d+"),
Law = str_extract(., "\\w+$"))}
<强>结果:强>
Art Ziff Abs Law
1 110 <NA> 3 StGB
2 10 <NA> 1 StGB
3 122 <NA> <NA> SVG
4 1 2 <NA> UWG
答案 1 :(得分:1)
使用gsub
将其转换为dcf格式(即关键字:值),然后使用read.dcf
将其读取。最后将read.dcf
生成的矩阵转换为数据框,并将任意数字列转换为数字。没有包使用。
s <- gsub("(\\S+)[.] (\\d+)", "\\1: \\2\n", df[[1]]) # convert to keyword: value
s <- sub(" (\\D+)$", "Law: \\1\n\n", s) # handle Law column
us <- trimws(unlist(strsplit(s, "\n"))) # split into separate components
DF <- as.data.frame(read.dcf(textConnection(us)), stringsAsFactors = FALSE)
DF[] <- lapply(DF, type.convert)
,并提供:
Art Abs Law Ziff
1 110 3 StGB NA
2 10 1 StGB NA
3 122 NA SVG NA
4 1 NA UWG 2