Question

我正试图摆脱一些角色：

list(structure(list(X1 = c("Station Name: Roches_Point", "Station Height: 40 M", 
"Latitude:51.789", "Longitude: -8.240")), .Names = "X1", row.names = c(NA, 
-4L), class = c("tbl_df", "tbl", "data.frame")), structure(list(
    X1 = c("Station Name: Newport", "Station Height: 22 M", "Latitude:53.920", 
    "Longitude: -9.570")), .Names = "X1", row.names = c(NA, -4L
), class = c("tbl_df", "tbl", "data.frame")))

此功能可以实现我想要的功能，但我无法删除车站高度中的字母M.

metadata_list <- lapply(metadata_list, function(x) 
  sapply(x, function(y){
    str_replace(y,"Station Name: |Station Height: | M|Latitude:|Longitude: ", "" )
  }))

如果我只使用这个表达式再次运行它，它可以工作：

metadata_list <- lapply(metadata_list, function(x) 
  sapply(x, function(y){
    str_replace(y," M", "" )
  }))

只是好奇是什么原因？

Answer 1

原因是您提供的模式将与该字符串的"Station Height: "匹配，并替换一次并停在那里。您应该使用str_replace_all来替换每个字符串中的所有匹配项。只需稍加改动：

metadata_list <- lapply(metadata_list, function(x) 
  sapply(x, function(y){
    str_replace_all(y,"Station Name: |Station Height: | M|Latitude:|Longitude: ", "" )
  }))
print(metadata_list)
[[1]]
     X1            
[1,] "Roches_Point"
[2,] "40"          
[3,] "51.789"      
[4,] "-8.240"      

[[2]]
     X1       
[1,] "Newport"
[2,] "22"     
[3,] "53.920" 
[4,] "-9.570"

删除特定字符不起作用

1 个答案: