问题描述
所以我有一个df,它看起来像这样,数字值被分割成逗号而不是点,并且被分类为字符。
var0 <- c("There,are commas","in the text,string","as,well","how,can","i","fix,this","thank you")
var1 <- c("50,0","72,"960,"1.920,"50,0")
var2 <- c("40,"742,"9460,0")
var3<- c("40,"90,"1,30",0")
...
var96 <- c("40,0")
df <- data.frame(cbind(var0,var1,var2,var3))
我知道如何使用gsub手动转换每个变量,但是正如您在下面看到的,我有大约96个。除此之外,我还有其他变量,其中包括不需要转换逗号的文本字符串和因子级别。
关于此的任何提示吗?
谢谢
解决方法
tidyverse软件包非常适合这种事情。
library(tidyverse)
df <- df %>%
# First,remove the points in your numbers b/c otherwise,you'll end up
# with,e.g.,"1.920.0"
mutate_all(.fun = function(x) gsub("\\.","",x)) %>%
# Next,replace all the commas with points and convert to numeric. Only do
# this for the columns that don't contain text,though.
mutate_at(.vars = vars(matches("var[1-3]")),.fun = function(x) as.numeric(gsub(",","\\.",x)))
请注意,在mutate_at
调用中,我假设只有列“ var0”包含您要保留的文本,并且我转换了所有与正则表达式“ var [1-3]相匹配的内容”表示数字数据和使用的点而不是逗号。您需要根据情况调整该正则表达式。
这是一个仅用小数点替换逗号并删除所有其他点的函数,如果出现的所有字符均为数字0-9,点和逗号。
commas2dots <- function(x){
if(any(grepl("[^\\.,[:digit:]]",x))){
x
} else {
y <- gsub("\\.",x)
tc <- textConnection(y)
on.exit(close(tc))
scan(tc,dec = ",quiet = TRUE)
}
}
lapply(df,commas2dots)
#$var0
#[1] "There,are commas" "in the text,string"
#[3] "as,well" "how,can"
#[5] "i" "fix,this"
#[7] "thank you"
#
#$var1
#[1] 50 72 960 1920 50 50 960
#
#$var2
#[1] 40 742 9460 1920 50 50 960
#
#$var3
#[1] 40.0 72.0 90.0 1.3 50.0 50.0 960.0
#
#$var96
#[1] 40 742 9460 1920 50 50 960
要更改data.frame的列,请执行以下操作:
df[] <- lapply(df,commas2dots)
df
# var0 var1 var2 var3 var96
#1 There,are commas 50 40 40.0 40
#2 in the text,string 72 742 72.0 742
#3 as,well 960 9460 90.0 9460
#4 how,can 1920 1920 1.3 1920
#5 i 50 50 50.0 50
#6 fix,this 50 50 50.0 50
#7 thank you 960 960 960.0 960
数据
var0 <- c("There,are commas","in the text,string","as,well","how,can","i","fix,this","thank you")
var1 <- c("50,0","72,"960,"1.920,"50,0")
var2 <- c("40,"742,"9460,0")
var3<- c("40,"90,"1,30",0")
var96 <- c("40,0")
df <- data.frame(var0,var1,var2,var3,var96)