假设我有一个很长的字符串:
"XOVEWVJIEWNIGOIWENVOIWEWVWEW"
我该怎样拆分这个字符串,才能每5个字符后面跟一个空格?
"XOVEW VJIEW NIGOI WENVO IWEWV WEW"
注意最后一个字符串更短。
我可以使用循环进行计数并逐个构建新字符串,但肯定有更好的方法,对吗?使用正则表达式:
gsub("(.{5})", "\\1 ", "XOVEWVJIEWNIGOIWENVOIWEWVWEW")
# [1] "XOVEW VJIEW NIGOI WENVO IWEWV WEW"
"XOVEWVJI EWNIGOIWENVOIW EWVWEW"
。 - Lasarus9使用 sapply
> string <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW"
> sapply(seq(from=1, to=nchar(string), by=5), function(i) substr(string, i, i+4))
[1] "XOVEW" "VJIEW" "NIGOI" "WENVO" "IWEWV" "WEW"
没有*apply的stringi
解决方案:
x <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW"
stri_sub(x, seq(1, stri_length(x),by=5), length=5)
[1] "XOVEW" "VJIEW" "NIGOI" "WENVO" "IWEWV" "WEW"
stri_sub
函数是向量化的,所以我们不需要在这里使用 * apply。s <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW" # Original string
l <- seq(from=5, to=nchar(s), by=5) # Calculate the location where to chop
# Add sentinels 0 (beginning of string) and nchar(s) (end of string)
# and take substrings. (Thanks to @flodel for the condense expression)
mapply(substr, list(s), c(0, l) + 1, c(l, nchar(s)))
输出:
[1] "XOVEW" "VJIEW" "NIGOI" "WENVO" "IWEWV" "WEW"
collapse=' '
将结果向量粘贴,以获得带有空格的单个字符串。substring
是矢量化的 substr
。x <- "XOVEWVJIEWNIGOIWENVOIWEWVWEW"
n <- seq(1, nc <- nchar(x), by = 5)
paste(substring(x, n, c(n[-1]-1, nc)), collapse = " ")
# [1] "XOVEW VJIEW NIGOI WENVO IWEWV WEW"