我有这样一个数据框:
df <- structure(list(a = c(NA, NA, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L), b = c(NA, NA, NA, 1L, 2L, 3L, 4L, 5L, 6L, 7L), d = c(NA, NA, NA, NA, 1L, 2L, 3L, 4L, 5L, 6L)), .Names = c("a", "b", "d"), row.names = c(NA, -10L), class = "data.frame")
> df
a b d
1 NA NA NA
2 NA NA NA
3 1 NA NA
4 2 1 NA
5 3 2 1
6 4 3 2
7 5 4 3
8 6 5 4
9 7 6 5
10 8 7 6
我想向上移动每一列并将NA移动到数据框的底部:
> df.out
a b d
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 NA
8 8 NA NA
9 NA NA NA
10 NA NA NA
更新以使我的问题更清晰..
df <- structure(list(a = c(NA, NA, 1, 5, 34, 7, 3, 5, 8, 4), b = c(NA,
NA, NA, 57, 2, 7, 9, 5, 12, 100), d = c(NA, NA, NA, NA, 5, 7,
2, 8, 2, 5)), .Names = c("a", "b", "d"), row.names = c(NA, -10L
), class = "data.frame")
> df
a b d
1 NA NA NA
2 NA NA NA
3 1 NA NA
4 5 57 NA
5 34 2 5
6 7 7 7
7 3 9 2
8 5 5 8
9 8 12 2
10 4 100 5
应该导致:
a b d
1 1 57 5
2 5 2 7
3 34 7 2
4 7 9 8
5 3 5 2
6 5 12 5
7 8 100 NA
8 4 NA NA
9 NA NA NA
10 NA NA NA
看起来似乎是一件容易的事,但我仍然坚持从哪里开始......你能帮忙吗?
答案 0 :(得分:13)
使用lapply
的另一种解决方案(根据您的意见不对数据进行排序/重新排序)
df[] <- lapply(df, function(x) c(x[!is.na(x)], x[is.na(x)]))
df
# a b d
# 1 1 57 5
# 2 5 2 7
# 3 34 7 2
# 4 7 9 8
# 5 3 5 2
# 6 5 12 5
# 7 8 100 NA
# 8 4 NA NA
# 9 NA NA NA
# 10 NA NA NA
或使用data.table
通过引用更新df
,而不是创建它的副本(该解决方案不会对您的数据进行排序)
library(data.table)
setDT(df)[, names(df) := lapply(.SD, function(x) c(x[!is.na(x)], x[is.na(x)]))]
df
# a b d
# 1: 1 57 5
# 2: 5 2 7
# 3: 34 7 2
# 4: 7 9 8
# 5: 3 5 2
# 6: 5 12 5
# 7: 8 100 NA
# 8: 4 NA NA
# 9: NA NA NA
# 10: NA NA NA
一些基准测试显示,到目前为止,基础解决方案是最快的:
library("microbenchmark")
david <- function() lapply(df, function(x) c(x[!is.na(x)], x[is.na(x)]))
dt <- setDT(df)
david.dt <- function() dt[, names(dt) := lapply(.SD, function(x) c(x[!is.na(x)], x[is.na(x)]))]
microbenchmark(as.data.frame(lapply(df, beetroot)), david(), david.dt())
# Unit: microseconds
# expr min lq median uq max neval
# as.data.frame(lapply(df, beetroot)) 1145.224 1215.253 1274.417 1334.7870 4028.507 100
# david() 116.515 127.382 140.965 149.7185 308.493 100
# david.dt() 3087.335 3247.920 3330.627 3415.1460 6464.447 100
答案 1 :(得分:6)
在完全误解了这个问题之后,这是我的最终答案:
# named after beetroot for being the first to ever need this functionality
beetroot <- function(x) {
# count NA
num.na <- sum(is.na(x))
# remove NA
x <- x[!is.na(x)]
# glue the number of NAs at the end
x <- c(x, rep(NA, num.na))
return(x)
}
# apply beetroot over each column in the dataframe
as.data.frame(lapply(df, beetroot))
它将对NA进行计数,删除NA,并在数据框中为每列的底部粘贴NA。
答案 2 :(得分:4)
为了好玩,您还可以使用length<-
和na.omit
。
以下是该组合的用途:
x <- c(NA, 1, 2, 3)
x
# [1] NA 1 2 3
`length<-`(na.omit(x), length(x))
# [1] 1 2 3 NA
适用于您的问题,解决方案是:
df[] <- lapply(df, function(x) `length<-`(na.omit(x), nrow(df)))
df
# a b d
# 1 1 57 5
# 2 5 2 7
# 3 34 7 2
# 4 7 9 8
# 5 3 5 2
# 6 5 12 5
# 7 8 100 NA
# 8 4 NA NA
# 9 NA NA NA
# 10 NA NA NA
答案 3 :(得分:0)
如果你的列数很少,我建议:
data.frame( a=sort(example$a, na.last=T), b=sort(example$b, na.last=T), d=sort(example$d, na.last=T))
最佳, ADII _