Question

我在使用此代码时遇到问题，该代码尝试编辑dplyr管道中的某些字符串。以下是一些会引发以下错误的数据。有什么想法吗？

data_frame(id = 1:5,
           name = c('this and it pretty long is a',
                    'name is a',
                    'so and so and so and so  and so',
                    'this is a',
                    'this is a variabel name')) 
%>% 

str_trunc(.,
   string = .$name,
   width = 10,
   side='right',
   ellipsis = '')

给我这个错误：Error in str_trunc(., string = .$name, width = 10, side = "right", ellipsis = ". . . ") : unused argument (.)。

感谢。

Answer 1

您需要mutate或mutate_at/if/all来更改列的内容。

data_frame(id = 1:5,
       name = c('this and it pretty long is a',
                'name is a',
                'so and so and so and so  and so',
                'this is a',
                'this is a variabel name')) %>% 
mutate_at("name", str_trunc, width = 10, side='right', ellipsis = '')

# A tibble: 5 x 2
     id name        
  <int> <chr>       
1     1 this and i  
2     2 name is a   
3     3 "so and so "
4     4 this is a   
5     5 "this is a "

我出于个人喜好使用mutate_at。请注意，变异列会自动作为第一个参数传递。如果您想将其放在函数调用中的其他位置，请将其称为.。

Answer 2

data中没有str_trunc参数，因此您需要将其string提供给它。你可以使用

data_frame(id = 1:5,
           name = c('this and it pretty long is a',
                    'name is a',
                    'so and so and so and so  and so',
                    'this is a',
                    'this is a variabel name'))$name %>% 
  str_trunc(width = 10,
            side='right',
            ellipsis = '')

Answer 3

如果要添加/更新现有列，请使用mutate功能。

你不能直接在管道中使用stringr函数。

data_frame(id = 1:5,
           name = c('this and it pretty long is a',
                    'name is a',
                    'so and so and so and so  and so',
                    'this is a',
                    'this is a variabel name'))  %>% 
           mutate(name=str_trunc(name,width=10,side='right',ellipsis=''))
## # A tibble: 5 x 2
##      id name        
##   <int> <chr>       
## 1     1 this and i  
## 2     2 name is a   
## 3     3 "so and so "
## 4     4 this is a   
## 5     5 "this is a "

mutate（blah blah）相当于以下

> df<-data_frame(id = 1:5,
        name = c('this and it pretty long is a',
                 'name is a',
                 'so and so and so and so  and so',
                 'this is a',
                 'this is a variabel name'))

> df
## # A tibble: 5 x 2
##      id name                           
##   <int> <chr>                          
## 1     1 this and it pretty long is a   
## 2     2 name is a                      
## 3     3 so and so and so and so  and so
## 4     4 this is a                      
## 5     5 this is a variabel name        
> df$name<-str_trunc(df$name,width=10,side='right',ellipsis='')
> df  
## # A tibble: 5 x 2
##      id name        
##   <int> <chr>       
## 1     1 this and i  
## 2     2 name is a   
## 3     3 "so and so "
## 4     4 this is a   
## 5     5 "this is a

如何在`dplyr`管道中使用`stringr`

3 个答案: