在R中的数据框内移动NAs

Question

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楚..

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

看起来像一个简单的任务,但我坚持从哪里开始..你能帮忙吗？

Answer 1

使用的另一种解决方案lapply(不根据您的意见对数据进行排序/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

或者使用data.table以便df通过引用进行更新,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

一些基准测试表明,基本解决方案是迄今为止最快的:

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

Answer 2

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固定底部的NAs.