R：substr（）在for循环和向量中的结果不同

Question

我正在使用substr来截断字符串列表中的最后三个字母：postal_code0

postal_code0
>[1] "n14BE"   "n14BE"   "n14BE"   "n14BE"   "n14BE"   "N16DD"   "N16DD"   "N16DD"   "N16DD"   "N16DD"  
>[11] "N11TW"   "N11TW"   "N11TW"   "N11TW"   "N11TW"   "n5"      "N160LU"  "N2"      "N200AU"  "N200AU" 
>[21] "london"  "n15"     "N5"      ""    

> outcode <- substr(postal_code0, 1, nchar(postal_code0)-3)
> outcode
[1] "n1"   "n1"   "n1"   "n1"   "n1"   "N1"   "N1"   "N1"   "N1"   "N1"   "N1"   "N1"   "N1"   "N1"   "N1"  
[16] ""     "N16"  ""     "N20"  "N20"  "lon"  ""     ""     ""       

它完全删除了一些我想保留的元素，但是如果我使用for循环来获得相同的函数，它会返回我想要的结果。

> outcode0 <- c()
> for (i in 1: length(postal_code0)){
+ outcode0[i] <- substr(postal_code0[i], 1, nchar(postal_code0)-3)
+ }

> outcode0
[1] "n1" "n1" "n1" "n1" "n1" "N1" "N1" "N1" "N1" "N1" "N1" "N1" "N1" "N1" "N1" "n5" "N1" "N2" "N2" "N2" "lo"
[22] "n1" "N5" "" 

这两个功能有什么区别？以及如何获得

outcode0

不使用for循环？

Answer 1

根据您最后的评论，以下内容应该如下：

len <- nchar(postal_code0)
substring(postal_code0, 1, ifelse(len <= 3, len, len - 3))