在dplyr - R中的summarize（）中使用索引来引用列 [英] Using index to reference column in summarise() in dplyr - R

问题描述

我想在dplyr中使用其索引而不是名称来引用summary（）内的列。例如：

I would like to reference a column inside the summarise() in dplyr with its index rather than with its name. For example:

        > a

           id visit timepoint bedroom  den
            1   0     0        62      NA 
            2   1     0        53    6.00  
            3   2     0        56    2.75   
            4   0     1        55      NA 
            5   1     2        61      NA 
            6   2     0        54      NA 
            7   0     1        58    2.75   
            8   1     2        59      NA 
            9   2     2        60      NA 
            10  0     1        57      NA 

           # E.g. 
           a %>% group_by(visit) %>% summarise(avg.bedroom = mean(bedroom, na.rm   =T)
           # Returns
        visit avg.dedroom
        <dbl>       <dbl>
     1     0       4.375
     2     1       2.750
     3     2         NaN

我如何使用卧室栏的索引而不是总结条款中的名称？我试过：

How could I use the index of column "bedroom" rather its name in the summarise clause? I tried:

     a %>% group_by(visit) %>% summarise("4" = mean(.[[4]], na.rm = T))

但是这返回了错误结果：

but this returned false results:

       visit      `4`
        <dbl>    <dbl>
      1     0 3.833333
      2     1 3.833333
      3     2 3.833333

我的目标是否可以实现，如果是，如何？谢谢。

Is my objective achievable and if yes how? Thank you.

推荐答案

也许不完全是你想要的，但一种选择是使用 purrr 而不是 dplyr 。类似

Perhaps not exactly what you're looking for, but one option would be to use purrr rather than dplyr. Something like

# Read in data
d <- read.table(textConnection(" id visit timepoint bedroom  den
        1  12     0        62      NA 
        2  14     0        53    6.00  
        3  14     0        56    2.75   
        4  14     1        55      NA 
        5  14     2        61      NA 
        6  15     0        54      NA 
        7  15     1        58    2.75   
        8  16     2        59      NA 
        9  16     2        60      NA 
        10 17     1        57      NA "), 
    header = TRUE)


library(purrr)

d %>% 
    split(.$timepoint) %>% 
    map_dbl(function(x) mean(x[ ,5], na.rm = TRUE))

#     0     1     2 
# 4.375 2.750   NaN 

或者，基数

aggregate(d[ ,5] ~ timepoint, data = d, mean)

#   timepoint d[, 5]
# 1         0  4.375
# 2         1  2.750

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