我想用行的中位数替换df中的所有数字,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)想办法做到这一点吗?
发布于 2016-01-27 11:11:51
一种可能的基解
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还有一个额外的好处,如果你还没有中庸之道,这里有一个可能的方法来计算它们。
df[-1][indx] <- matrixStats::rowMedians(as.matrix(df[-1]), na.rm = TRUE)[indx[, "row"]]发布于 2016-01-27 11:14:46
用纯数学:
cbind(df[1],NA^(is.na(df[,2:4]))*df$median)
# pr_id sample1 sample2 sample3
#1 AX-1 NA 125 125
#2 AX-2 NA NA NA
#3 AX-3 NA NA 196如果需要计算中位数,只需将df$median替换为apply(df[,2:4],1,median,na.rm=TRUE)即可。
发布于 2016-01-27 11:09:33
使用dplyr可以如下所示:
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,一个选项是转换为长格式、计算中间值和重转换为宽格式:
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)https://stackoverflow.com/questions/35035383
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