更改R中字符串的前两个字符

Question

在 R 中，我想查找并替换字符串中的两个字符（只是前两个字符）这是我开始的数据。

time <-  c("153500", "153800", "161400", "161700", "163000", "161800", 
                "201700", "201800")

from <- c("15", "16", "17", "18")
to <- c("10","11", "12", "13" )
repl <- data.frame(from, to)

结果应如下所示：

[1] "103500" "103800" "111400" "111700" "113000" "111800" "201700" "201800"

Answer 1

尝试

v1 <- setNames(to, from)[substr(time, 1, 2)]
as.character(ifelse(!is.na(v1), paste0(v1, sub('^.{2}','', time)), time))
#[1] "103500" "103800" "111400" "111700" "113000" "111800" "201700" "201800"

Answer 2

@grrgrrbla 谢谢，但就我而言，这不合适。我已经从数值转换中逃脱，因为转换为数值变量时省略了 0。另一个例子是 0* hours

> time <-  c("053500", "053800", "061400", "061700", "163000", "161800", "201700", "201800")
> from <- c("05", "06", "16", "20")
> test[substr(time, 1, 2) %in% from] <-
+ as.character(as.numeric(time[substr(time, 1, 2) %in% from]) - 50000)
> cbind(time, test)
     time     test    
[1,] "053500" "3500"  
[2,] "053800" "3800"  
[3,] "061400" "11400" 
[4,] "061700" "11700" 
[5,] "163000" "113000"
[6,] "161800" "111800"
[7,] "201700" "151700"
[8,] "201800" "151800"

这些字符给需要使用时间格式的地方加上冒号(h:m:s)和“0”带来了麻烦。

更改R中字符串的前两个字符

Change first two characters in string in R

r

substr