我有一些麻烦的前导和尾随空白在一个数据。框架。

例如,我根据特定条件查看data.frame中的特定行:

> myDummy[myDummy$country == c("Austria"),c(1,2,3:7,19)] 



[1] codeHelper     country        dummyLI    dummyLMI       dummyUMI       

[6] dummyHInonOECD dummyHIOECD    dummyOECD      

<0 rows> (or 0-length row.names)

我想知道为什么我没有得到预期的输出,因为奥地利显然存在于我的数据框架中。在查看了我的代码历史并试图找出错误后,我尝试了:

> myDummy[myDummy$country == c("Austria "),c(1,2,3:7,19)]
   codeHelper  country dummyLI dummyLMI dummyUMI dummyHInonOECD dummyHIOECD
18        AUT Austria        0        0        0              0           1
   dummyOECD
18         1

我所更改的命令只是在奥地利之后增加了一个空白。

显然还会出现更多烦人的问题。例如,当我喜欢根据国家列合并两帧时。一个data.frame使用“Austria”,而另一个frame使用“Austria”。匹配不起作用。

有没有一种很好的方法来“显示”屏幕上的空白,让我意识到这个问题? 我能移除R开头和结尾的空白吗?

到目前为止,我曾经写过一个简单的Perl脚本,它消除了白色的速度,但如果我能以某种方式在R中做到这一点就好了。


当前回答

本线程中主要方法的基准测试。这并没有捕捉到所有奇怪的情况,但到目前为止,我们仍然缺少str_trim删除空格而trimws不删除空格的示例(参见Richard Telford对这个答案的评论)。似乎并不重要- gsub选项似乎是最快的:)

x <- c(" lead", "trail ", " both ", " both and middle ", " _special")
## gsub function from https://stackoverflow.com/a/2261149/7941188 
## this is NOT the function from user Bernhard Kausler, which uses 
## a much less concise regex 
gsub_trim <- function (x) gsub("^\\s+|\\s+$", "", x)

res <- microbenchmark::microbenchmark(
  gsub = gsub_trim(x),
  ## https://stackoverflow.com/a/30210713/7941188
  trimws = trimws(x),
  ## https://stackoverflow.com/a/15007398/7941188
  str_trim = stringr::str_trim(x),
  times = 10^5
)
res
#> Unit: microseconds
#>      expr    min     lq      mean median       uq       max neval cld
#>      gsub 20.201 22.788  31.43943 24.654  28.4115  5303.741 1e+05 a  
#>    trimws 38.204 41.980  61.92218 44.420  51.1810 40363.860 1e+05  b 
#>  str_trim 88.672 92.347 116.59186 94.542 105.2800 13618.673 1e+05   c
ggplot2::autoplot(res)

sessionInfo()
#> R version 4.0.3 (2020-10-10)
#> Platform: x86_64-apple-darwin17.0 (64-bit)
#> Running under: macOS Big Sur 10.16
#> 
#> locale:
#> [1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8
#> 
#> attached base packages:
#> [1] stats     graphics  grDevices utils     datasets  methods   base     
#> 
#> loaded via a namespace (and not attached):
#>  stringr_1.4.0  

其他回答

使用grep或grepl查找带有空格的观测值,并使用sub删除它们。

names<-c("Ganga Din\t", "Shyam Lal", "Bulbul ")
grep("[[:space:]]+$", names)
[1] 1 3
grepl("[[:space:]]+$", names)
[1]  TRUE FALSE  TRUE
sub("[[:space:]]+$", "", names)
[1] "Ganga Din" "Shyam Lal" "Bulbul"

使用dplyr/tidyverse mutate_all和str_trim来修剪整个数据帧:

myDummy %>%
  mutate_all(str_trim)
library(tidyverse)
set.seed(335)
df <- mtcars %>%
        rownames_to_column("car") %>%
        mutate(car = ifelse(runif(nrow(mtcars)) > 0.4, car, paste0(car, " "))) %>%
        select(car, mpg)

print(head(df), quote = T)
#>                    car    mpg
#> 1         "Mazda RX4 " "21.0"
#> 2      "Mazda RX4 Wag" "21.0"
#> 3        "Datsun 710 " "22.8"
#> 4    "Hornet 4 Drive " "21.4"
#> 5 "Hornet Sportabout " "18.7"
#> 6           "Valiant " "18.1"

df_trim <- df %>%
  mutate_all(str_trim)

print(head(df_trim), quote = T)  
#>                   car    mpg
#> 1         "Mazda RX4"   "21"
#> 2     "Mazda RX4 Wag"   "21"
#> 3        "Datsun 710" "22.8"
#> 4    "Hornet 4 Drive" "21.4"
#> 5 "Hornet Sportabout" "18.7"
#> 6           "Valiant" "18.1"

由reprex包于2021-05-07创建(v0.3.0)

本线程中主要方法的基准测试。这并没有捕捉到所有奇怪的情况,但到目前为止,我们仍然缺少str_trim删除空格而trimws不删除空格的示例(参见Richard Telford对这个答案的评论)。似乎并不重要- gsub选项似乎是最快的:)

x <- c(" lead", "trail ", " both ", " both and middle ", " _special")
## gsub function from https://stackoverflow.com/a/2261149/7941188 
## this is NOT the function from user Bernhard Kausler, which uses 
## a much less concise regex 
gsub_trim <- function (x) gsub("^\\s+|\\s+$", "", x)

res <- microbenchmark::microbenchmark(
  gsub = gsub_trim(x),
  ## https://stackoverflow.com/a/30210713/7941188
  trimws = trimws(x),
  ## https://stackoverflow.com/a/15007398/7941188
  str_trim = stringr::str_trim(x),
  times = 10^5
)
res
#> Unit: microseconds
#>      expr    min     lq      mean median       uq       max neval cld
#>      gsub 20.201 22.788  31.43943 24.654  28.4115  5303.741 1e+05 a  
#>    trimws 38.204 41.980  61.92218 44.420  51.1810 40363.860 1e+05  b 
#>  str_trim 88.672 92.347 116.59186 94.542 105.2800 13618.673 1e+05   c
ggplot2::autoplot(res)

sessionInfo()
#> R version 4.0.3 (2020-10-10)
#> Platform: x86_64-apple-darwin17.0 (64-bit)
#> Running under: macOS Big Sur 10.16
#> 
#> locale:
#> [1] en_GB.UTF-8/en_GB.UTF-8/en_GB.UTF-8/C/en_GB.UTF-8/en_GB.UTF-8
#> 
#> attached base packages:
#> [1] stats     graphics  grDevices utils     datasets  methods   base     
#> 
#> loaded via a namespace (and not attached):
#>  stringr_1.4.0  

最好的方法是trimws()。

下面的代码将这个函数应用到整个数据框架。

mydataframe<- data.frame(lapply(mydataframe, trimws),stringsAsFactors = FALSE)

如果输入之间有多个空格,则会出现另一个相关问题:

> a <- "  a string         with lots   of starting, inter   mediate and trailing   whitespace     "

然后,您可以使用split参数的正则表达式轻松地将该字符串拆分为“真实”标记:

> strsplit(a, split=" +")
[[1]]
 [1] ""           "a"          "string"     "with"       "lots"
 [6] "of"         "starting,"  "inter"      "mediate"    "and"
[11] "trailing"   "whitespace"

请注意,如果在(非空)字符串的开头有匹配项,输出的第一个元素是' "" ',但如果在字符串的末尾有匹配项,输出与删除匹配项后相同。