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使用 R dplyr 将 NA 替换为组平均值,但在计算平均值之前省略组中的一些值

[英]using R dplyr replace NA with group mean but omitting some values from group before mean calculation

This seems like it should be a simple one but I can't see it.这似乎应该是一个简单的,但我看不到它。

Say I have a dataframe like:假设我有一个 dataframe 像:

df <- data.frame(type=c(rep("A", 5), rep("B",5)),
                 stage=rep(c("1","2", "3", "4", "5"),2),
                 val=c(rnorm(n=5, mean=1000, sd=300),rnorm(n=4, mean=1000, sd=100), NA)
                 )

I want to be able to replace the NA in group type=="B" by the mean of group B, but omit some "vals" from B (eg where "stage" equals 1 OR stage equals 2 - or any other condition).我希望能够通过 B 组的平均值替换组type=="B"中的 NA,但从 B 中省略一些“vals”(例如,“stage”等于 1 或 stage 等于 2 - 或任何其他条件) . Using dplyr and zoo its easy to fill in with the group mean:使用dplyrzoo很容易用组平均值填充:

df %>% dplyr::group_by(type) %>% dplyr::mutate_at("val", zoo::na.aggregate) 

but I can't work out how to exclude vals from the group based on condition in "stage".但我不知道如何根据“阶段”中的条件从组中排除 val。 Ideally, a dplyr solution but including zoo would also be good.理想情况下,dplyr 解决方案但包括动物园也很好。

This is how you could do it with the condition stage != 2 :这就是你可以如何使用条件stage != 2

library(tidyverse)

set.seed(12345)
df <- data.frame(type=c(rep("A", 5), rep("B",5)),
                 stage=rep(c("1","2", "3", "4", "5"),2),
                 val=c(rnorm(n=5, mean=1000, sd=300),rnorm(n=4, mean=1000, sd=100), NA)
)



df %>% 
  group_by(type) %>% 
  mutate(val = replace_na(val, mean(val[stage != 2], na.rm = TRUE)))
#> # A tibble: 10 x 3
#> # Groups:   type [2]
#>    type  stage   val
#>    <fct> <fct> <dbl>
#>  1 A     1     1176.
#>  2 A     2     1213.
#>  3 A     3      967.
#>  4 A     4      864.
#>  5 A     5     1182.
#>  6 B     1      818.
#>  7 B     2     1063.
#>  8 B     3      972.
#>  9 B     4      972.
#> 10 B     5      921.

Created on 2020-05-08 by the reprex package (v0.3.0)代表 package (v0.3.0) 于 2020 年 5 月 8 日创建

I set the seed to a number so everybody gets the same numbers.我将种子设置为一个数字,这样每个人都会得到相同的数字。

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