In January I asked how to replace the first N dots of a string: replace the first N dots of a string
在1月份,我问了如何替换一个字符串的第一个N个点:替换字符串的第一个N个点。
DWin's answer was very helpful. Can it be generalized?
德温的回答很有帮助。可以通用吗?
df.1 <- read.table(text = '
my.string other.stuff
1111111111111111 120
..............11 220
11.............. 320
1............... 320
.......1........ 420
................ 820
11111111111111.1 120
', header = TRUE)
nn <- 14
# this works:
df.1$my.string <- sub("^\\.{14}", paste(as.character(rep(0, nn)), collapse = ""),
df.1$my.string)
# this does not work:
df.1$my.string <- sub("^\\.{nn}", paste(as.character(rep(0, nn)), collapse = ""),
df.1$my.string)
2 个解决方案
#1
3
Using sprintf
you can have the desired output
使用sprintf可以得到所需的输出
nn <- 3
sub(sprintf("^\\.{%s}", nn),
paste(rep(0, nn), collapse = ""), df.1$my.string)
## [1] "1111111111111111" "000...........11" "11.............."
## [4] "1..............." "000....1........" "000............."
## [7] "11111111111111.1"
#2
0
pattstr <- paste0("\\.", paste0( rep(".",nn), collapse="") )
pattstr
#[1] "\\..............."
df.1$my.string <- sub(pattstr,
paste0( rep("0", nn), collapse=""),
df.1$my.string)
> df.1
my.string other.stuff
1 1111111111111111 120
2 000000000000001 220
3 11.............. 320
4 100000000000000 320
5 00000000000000. 420
6 00000000000000. 820
7 11111111111111.1 120
#1
3
Using sprintf
you can have the desired output
使用sprintf可以得到所需的输出
nn <- 3
sub(sprintf("^\\.{%s}", nn),
paste(rep(0, nn), collapse = ""), df.1$my.string)
## [1] "1111111111111111" "000...........11" "11.............."
## [4] "1..............." "000....1........" "000............."
## [7] "11111111111111.1"
#2
0
pattstr <- paste0("\\.", paste0( rep(".",nn), collapse="") )
pattstr
#[1] "\\..............."
df.1$my.string <- sub(pattstr,
paste0( rep("0", nn), collapse=""),
df.1$my.string)
> df.1
my.string other.stuff
1 1111111111111111 120
2 000000000000001 220
3 11.............. 320
4 100000000000000 320
5 00000000000000. 420
6 00000000000000. 820
7 11111111111111.1 120