用中位数替换df中的所有数字

Question

我想用df行的中位数替换我的所有数字,保持NA值.这是我的意见:

df <- 'pr_id  sample1  sample2 sample3 median
            AX-1   NA       120     130  125    
            AX-2   NA       NA     NA  NA
            AX-3   NA       NA     196  196'
df <- read.table(text=df, header=T)

这是我的预期输出:

df <- 'pr_id  sample1  sample2 sample3
            AX-1   NA       125     125    
            AX-2   NA       NA     NA
            AX-3   NA       NA     196'
df <- read.table(text=df, header=T)

有什么想法实现这一目标？

Answer 1

可能的基础解决方案

indx <- which(!is.na(df[-1]), arr.ind = TRUE) # find non-NA incidents
df[-1][indx] <- df$median[indx[, "row"]] # replace them while subsetting accordingly from df$median
df
#   pr_id sample1 sample2 sample3 median
# 1  AX-1      NA     125     125    125
# 2  AX-2      NA      NA      NA     NA
# 3  AX-3      NA      NA     196    196

还有一个好处,如果你还没有中位数,这里有一种可能的方法来计算它们

df[-1][indx] <- matrixStats::rowMedians(as.matrix(df[-1]), na.rm = TRUE)[indx[, "row"]]

Answer 2

纯数学:

cbind(df[1],NA^(is.na(df[,2:4]))*df$median)  

#  pr_id sample1 sample2 sample3
#1  AX-1      NA     125     125
#2  AX-2      NA      NA      NA
#3  AX-3      NA      NA     196

如果您需要计算中位数,只需替换df$median为apply(df[,2:4],1,median,na.rm=TRUE).

Answer 3

使用dplyr,您可以执行以下操作:

library(dplyr)
mutate_each(df, funs(ifelse(is.na(.), ., median)), sample1:sample3)
#  pr_id sample1 sample2 sample3 median
#1  AX-1      NA     125     125    125
#2  AX-2      NA      NA      NA     NA
#3  AX-3      NA      NA     196    196

如果要包含中值计算,可以选择转换为长格式,计算中值并重新转换为宽格式:

library(tidyr)
gather(df, sample, value, sample1:sample3) %>% 
  group_by(pr_id) %>% 
  mutate(value = as.numeric(ifelse(is.na(value), value, median(value, na.rm = TRUE)))) %>% 
  spread(sample, value)