如何在R中使用dplyr计算离开平均数? [英] How to compute a leave one out average using dplyr in R?

问题描述

我正在尝试使用dplyr在所有行中找到一个变量的留一平均值.由于dplyr提供了一个称为row_number()的便捷函数，因此我认为我可以像这样使用它:

Hi I am trying to find a leave one out average of a variable in all rows using dplyr. Since dplyr provides a convenient function called row_number(), I thought I could use it like this:

library(dplyr)

iris %>% 
  tbl_df %>% 
  select(Sepal.Length) %>%
  mutate(loo_avg=mean(Sepal.Length[-row_number()]))  # leave one out average

但这会返回如下结果:

Source: local data frame [150 x 2]

   Sepal.Length loo_avg
          (dbl)   (dbl)
1           5.1     NaN
2           4.9     NaN
3           4.7     NaN
4           4.6     NaN
5           5.0     NaN
6           5.4     NaN
7           4.6     NaN
8           5.0     NaN
9           4.4     NaN
10          4.9     NaN
..          ...     ...

您如何解决此问题?

推荐答案

我特别喜欢data.table方法:

library(data.table)

DT <- as.data.table(iris)

DT[ , loo_avg := DT[-.BY$left_out, mean(Sepal.Length)], 
    by = .(left_out = 1:nrow(DT))
    ][,.(Sepal.Length, loo_avg)]
#      Sepal.Length  loo_avg
#   1:          5.1 5.848322
#   2:          4.9 5.849664
#   3:          4.7 5.851007
#   4:          4.6 5.851678
#   5:          5.0 5.848993
#  ---                      
# 146:          6.7 5.837584
# 147:          6.3 5.840268
# 148:          6.5 5.838926
# 149:          6.2 5.840940
# 150:          5.9 5.842953

请注意，除了j中的mean之外，这种方法还使您执行所需的操作变得异常简单.

Note that this approach also makes it incredibly easy to do whatever you want besides mean in j.

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