按行填充缺失值（右/左） [英] Fill missing values rowwise (right / left)

问题描述

我正在寻找一种使用dplyr向右填充（而不是向下/向上） NA 的方法。换句话说，我想将d转换为d2，而不必在mutate调用中显式引用任何列。

I'm looking for a way to "fill" NAs to the right (as opposed to down/up) with dplyr. In other words, I would like to convert d into d2 without having to explicitly reference any columns in a mutate call.

我的实际数据框有几十个字段，这些字段具有交错的块NA跨越可变数目的列。我很好奇是否有一个短方法可以全局继承左侧的第一个非NA值，而不管它出现在哪个字段中。

My real dataframe has several 10s of fields with staggered blocks of NAs spanning variable numbers of columns. I'm curious whether there's a short way to globally inherit the first non-NA value to the left, regardless of what field it occurs in.

d<-data.frame(c1=c("a",1:4), c2=c(NA,2,NA,4,5), c3=c(NA,3,4,NA,6))
d2<-data.frame(c1=c("a",1:4), c2=c("a",2,2,4,5), c3=c("a",3,4,4,6))
d
d2

推荐答案

我们可以将收集转换为'long'格式，进行填充按行号分组，然后传播回到宽格式

We can do a gather into 'long' format, do the fill grouped by the row number and then spread back to 'wide' format

library(tidyverse)
rownames_to_column(d, 'rn') %>% 
    gather(key, val, -rn) %>%
    group_by(rn) %>% 
    fill(val) %>% 
    spread(key, val) %>%
    ungroup %>%
    select(-rn)
# A tibble: 5 x 3
#  c1    c2    c3   
#  <chr> <chr> <chr>
#1 a     a     a    
#2 1     2     3    
#3 2     2     4    
#4 3     4     4    
#5 4     5     6 

---

或不重塑的另一种方法是使用进行行填充na.locf

library(zoo)
d %>% 
    mutate(c1 = as.character(c1)) %>%
    pmap_dfr(., ~ na.locf(c(...)) %>%
                      as.list %>%
                      as_tibble)

---

此外，如果我们使用 na.locf ，它按列运行，因此数据可以转置并直接应用 na.locf

---

Also, if we use na.locf, it run columnwise, so the data can be transposed and apply na.locf directly

d[] <- t(na.locf(t(d)))
d
#  c1 c2 c3
#1  a  a  a
#2  1  2  3
#3  2  2  4
#4  3  4  4
#5  4  5  6

如注释中提到的@ G.Grothendieck那样，为了照顾在行首的NA元素，使用 na.locf0 而不是 na.locf

As @G.Grothendieck mentioned in the comments, inorder to take care of the elements that are NA at the beginning of the row, use na.locf0 instead of na.locf

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