用未指定列数的组均值替换NA [英] replace NA with groups mean in a non specified number of columns

问题描述

我想用多列中每个单个组collembola和mite的平均值代替NA.这是一个3列的示例，但是我想将其应用于5000列的数据框

I want to replace the NA with mean of each single group collembola and mite in multiple columns. Here it is an example with 3 columns however I want to apply this a data frame with 5000 columns

dat <- read.table(text = 
                  "id    ID        length  width    extra
                  101   collembola  2.1     0.9     1
                  102   mite        NA      0.7     NA
                  103   mite        1.1     0.8     2
                  104   collembola  1       NA      3
                  105   collembola  1.5     0.5     4
                  106   mite        NA      NA      NA
                  106   mite        1.9     NA      4", 
                  header=TRUE)

如果我输入每一列，它将起作用

It works if I enter each column

library(plyr)
impute.mean <- function(x) replace(x, is.na(x), mean(x, na.rm = TRUE))
data2 <- ddply(dat, ~ ID, transform, length = impute.mean(length))

我想应用计算多列中每个组ID collembola和mite的均值的函数，下面是我尝试过的方法(不起作用):

I want to apply the function that calculates the mean of each single group ID collembola and mite across multiple columns, below is what I tried (it does not work):

dat2 <- ddply(dat, ~ ID, transform,  impute.mean(dat[,3:ncol(dat)]))

推荐答案

如果您不介意使用dplyr:

library(dplyr)

dat %>% 
  group_by(ID) %>% 
  mutate_if(is.numeric, function(x) ifelse(is.na(x), mean(x, na.rm = TRUE), x))
#> # A tibble: 7 x 5
#> # Groups:   ID [2]
#>      id         ID length width extra
#>   <int>     <fctr>  <dbl> <dbl> <dbl>
#> 1   101 collembola    2.1  0.90     1
#> 2   102       mite    1.5  0.70     3
#> 3   103       mite    1.1  0.80     2
#> 4   104 collembola    1.0  0.70     3
#> 5   105 collembola    1.5  0.50     4
#> 6   106       mite    1.5  0.75     3
#> 7   106       mite    1.9  0.75     4

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