如何将数据表中的单元格向右移动R中每行的NA数量?
示例数据:
data <- data.table(c1=c("a","e","h","j"),
c2=c("b","f","i",NA),
c3=c("c","g",NA,NA),
c4=c("d",NA,NA,NA), stringsAsFactors = F)
c1 c2 c3 c4
1 a b c d
2 e f g <NA>
3 h i <NA> <NA>
4 j <NA> <NA> <NA>
示例中的所需数据:
data.desired <- data.table(
c1=c("a",NA,NA,NA),
c2=c("b","e",NA,NA),
c3=c("c","f","h",NA),
c4=c("d","g","i","j"), stringsAsFactors = F)
c1 c2 c3 c4
1 a b c d
2 <NA> e f g
3 <NA> <NA> h i
4 <NA> <NA> <NA> j
答案 0 :(得分:3)
这是尝试使用矩阵索引和按行计算NA
个值的计数器:
#convert back to a data.frame to take advantage of matrix indexing
setDF(data)
arr <- which(!is.na(data), arr.ind=TRUE)
arr[,"col"] <- arr[,"col"] + rowSums(is.na(data))[arr[,"row"]]
out <- data
out[] <- NA
out[arr] <- data[!is.na(data)]
out
# c1 c2 c3 c4
#1 a b c d
#2 <NA> e f g
#3 <NA> <NA> h i
#4 <NA> <NA> <NA> j
#convert to data.table if necessary
setDT(out)
此选项非常快捷,只需短暂的测试就可以在3-4秒内遍历4列/ 200万行。
答案 1 :(得分:1)
我们可以使用
data.table(t(apply(data, 1, function(x){ c(rep(NA, sum(is.na(x))), x[!is.na(x)])})))
# V1 V2 V3 V4
# 1: a b c d
# 2: <NA> e f g
# 3: <NA> <NA> h i
# 4: <NA> <NA> <NA> j