如何在函数中引用列名

Question

我有一个df。如果 'info' 出现在一行中，我想要一个函数来获取列名并将其与单元格中的信息粘贴在一起，它们之间有一个“=”。我已经尝试了下面的函数，它可以工作，除了它没有引用右列中的列名

df <- data.frame('criteria1' = c('info','1', 'info', '', 'info'), "criteria2" = c('y','3', '', 'info', ''), "criteria3" = c('y','7', '', 'info', 'info'))

df[] <- sapply(df, function(x) ifelse(x == 'info', paste(colnames(df)[x], ' = ', x),''))

我的预期输出是这样的（删除列名并不重要，重要的只是来自单元格的信息）

df_exp <- data.frame('criteria1' = c('criteria1= info','', 'criteria1=info', '', 'criteria1 =info'), "criteria2" = c('','', '', 'criteria2 = info', ''), "criteria3" = c('','', '', 'criteria3 = info', 'criteria3 = info'))

Answer 1

我会选择列号，1:ncol(df)（更快）或 seq(df)。我用的是前者。

df <- sapply(1:ncol(df), function(x) 
  ifelse(df[[x]] == 'info', paste(colnames(df)[x], ' = ', df[[x]]),''))
df
#                      [,1]                 [,2]                 [,3]                
# [1,] "criteria1  =  info" ""                   ""                  
# [2,] ""                   ""                   ""                  
# [3,] "criteria1  =  info" ""                   ""                  
# [4,] ""                   "criteria2  =  info" "criteria3  =  info"
# [5,] "criteria1  =  info" ""                   "criteria3  =  info"

另一种使用stack/unstack的好方法：

r <- grep("info", tmp$values)
tmp <- stack(df)
tmp[r, 1] <- apply(tmp[r, 2:1], 1, paste, collapse="=")
tmp[-r, 1] <- ""  ## in case you want non-"info" cells cleared
df <- unstack(tmp)
df
#        criteria1      criteria2      criteria3
# 1 criteria1=info                              
# 2                                             
# 3 criteria1=info                              
# 4                criteria2=info criteria3=info
# 5 criteria1=info                criteria3=info

Answer 2

使用base R（根据需要调整=之前的间距）：

use_names <- names(df)
    data.frame(Map(function(x,y) ifelse(x=="info", paste0(y,"=",x),""), df, use_names))
           criteria1      criteria2      criteria3
    1 criteria1=info                              
    2                                             
    3 criteria1=info                              
    4                criteria2=info criteria3=info
    5 criteria1=info                criteria3=info

purrr:

purrr::map2_df(df, names(df), function(x,y) ifelse(x=="info", paste0(y,"=",x),""))
# A tibble: 5 x 3
  criteria1        criteria2        criteria3       
  <chr>            <chr>            <chr>           
1 "criteria1=info" ""               ""              
2 ""               ""               ""              
3 "criteria1=info" ""               ""              
4 ""               "criteria2=info" "criteria3=info"
5 "criteria1=info" ""               "criteria3=info"

数据：

df <- structure(list(criteria1 = c("info", "1", "info", "", "info"), 
    criteria2 = c("y", "3", "", "info", ""), criteria3 = c("y", 
    "7", "", "info", "info")), class = "data.frame", row.names = c(NA, 
-5L))

如何在函数中引用列名

How to refer to a column name in a function

r

paste

dataframe

sapply