Replacing NA depending on distribution type of gender in R

问题

When i selected NA value here

data[data=="na"] <- NA
data[!complete.cases(data),]

i must replace it, but depending on type of distribution. If using Shapiro.test the distribution by variables not normal, then missing value must be replace by median, If it's normal, than replace by mean. But distribution for each gender(1 girl, 2 -man)

data=structure(list(sex = c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), emotion = c(20L, 
15L, 49L, NA, 34L, 35L, 54L, 45L), IQ = c(101L, 98L, 105L, NA, 
123L, 120L, 115L, NA)), .Names = c("sex", "emotion", "IQ"), class = "data.frame", row.names = c(NA, 
-8L))

the desired output

sex emotion IQ
1   20  101
1   15  98
1   49  105
1   28  101
2   34  123
2   35  120
2   54  115
2   45  119

回答1:

Following code will replace NA values according to the Shapiro Test:

library(dplyr)

data %>% 
 group_by(sex) %>%
 mutate(
  emotion = ifelse(!is.na(emotion), emotion,
   ifelse(shapiro.test(emotion)$p.value > 0.05,
    mean(emotion, na.rm=TRUE), quantile(emotion, na.rm=TRUE, probs=0.5) ) ),
  IQ = ifelse(!is.na(IQ), IQ,
   ifelse(shapiro.test(IQ)$p.value > 0.05,
    mean(IQ, na.rm=TRUE), quantile(IQ, na.rm=TRUE, probs=0.5) )
  )
 ) 

来源：https://stackoverflow.com/questions/51326684/replacing-na-depending-on-distribution-type-of-gender-in-r