问题描述
我想使用 NA 中的估算值将 df 中的列替换为 df2 以获得 df3。
我可以用 left_join 和 coalesce 做到这一点，但我认为这种方法不能很好地概括。有没有更好的办法？
library(tidyverse)
df <- tibble(c = c("a","a","b","b"),d = c(1,2,3,1,3),x = c(1,NA,4,5,6),y = c(1,z = c(1,7,6))
# I want to replace NA in df by df2
df2 <- tibble(c = c("a","a"),2))
# to get
df3 <- tibble(c = c("a",6))
# is there a better solution than coalesce?
df3 <- df %>% left_join(df2,by = c("c","d")) %>%
mutate(x = coalesce(x.x,x.y),y = coalesce(y.x,y.y)) %>%
select(-x.x,-x.y,-y.x,-y.y)
Created on 2021-06-17 by the reprex package (v2.0.0)
解决方法
这是一个自定义函数，用于合并 all .x 和 .y 列，可选择重命名和删除列。
#' Coalesce all columns duplicated in a previous join.
#'
#' Find all columns resulting from duplicate names after a join
#' operation (e.g.,`dplyr::*_join` or `base::merge`),then coalesce
#' them pairwise.
#'
#' @param x data.frame
#' @param suffix character,length 2,the same string suffixes
#' appended to column names of duplicate columns; should be the same
#' as provided to `dplyr::*_join(.,suffix=)` or `base::merge(.,#' suffixes=)`
#' @param clean logical,whether to remove the suffixes from the LHS
#' columns and remove the columns on the RHS columns
#' @param strict logical,whether to enforce same-classes in the LHS
#' (".x") and RHS (".y") columns; while it is safer to set this to
#' true (default),sometimes the conversion of classes might be
#' acceptable,for instance,if one '.x' column is 'numeric' and its
#' corresponding '.y' column is 'integer',then relaxing the class
#' requirement might be acceptable
#' @return 'x',coalesced,optionally cleaned
#' @export
coalesce_all <- function(x,suffix = c(".x",".y"),clean = FALSE,strict = TRUE) {
nms <- colnames(x)
Xs <- endsWith(nms,suffix[1])
Ys <- endsWith(nms,suffix[2])
# x[Xs] <- Map(dplyr::coalesce,x[Xs],x[Ys])
# x[Xs] <- Map(data.table::fcoalesce,x[Ys])
x[Xs] <- Map(function(dotx,doty) {
if (strict) stopifnot(identical(class(dotx),class(doty)))
isna <- is.na(dotx)
replace(dotx,isna,doty[isna])
},x[Ys])
if (clean) {
names(x)[Xs] <- gsub(glob2rx(paste0("*",suffix[1]),trim.head = TRUE),"",nms[Xs])
x[Ys] <- NULL
}
x
}
在行动：
df %>%
left_join(df2,by = c("c","d")) %>%
coalesce_all()
# # A tibble: 6 x 7
# c d x.x y.x z x.y y.y
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 a 1 1 1 1 1 1
# 2 a 2 2 2 2 2 2
# 3 a 3 3 2 7 3 2
# 4 b 1 4 4 4 NA NA
# 5 b 2 5 5 5 NA NA
# 6 b 3 6 6 6 NA NA
df %>%
left_join(df2,"d")) %>%
coalesce_all(clean = TRUE)
# # A tibble: 6 x 5
# c d x y z
# <chr> <dbl> <dbl> <dbl> <dbl>
# 1 a 1 1 1 1
# 2 a 2 2 2 2
# 3 a 3 3 2 7
# 4 b 1 4 4 4
# 5 b 2 5 5 5
# 6 b 3 6 6 6
我在 Map 中包含了两个合并函数作为 base-R 的替代。一个优点是 strict 参数：dplyr::coalesce 将默默地允许 integer 和 numeric 合并，而 data.table::fcoalesce 不会。如果这是可取的，请使用您喜欢的。 （另一个优点是两个非基础合并函数都接受任意数量的要合并的列，这在本实现中不是必需的。）
您可以通过使用 across 和使用 .names & .keep 参数一次改变所有列，如下所示
library(dplyr,warn.conflicts = F)
df %>% left_join(df2,"d")) %>%
mutate(across(ends_with('.x'),~ coalesce(.,get(gsub('.x','.y',cur_column()))),.names = '{gsub(".x$",.col)}'),.keep = 'unused')
#> # A tibble: 6 x 5
#> c d z x y
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 a 1 1 1 1
#> 2 a 2 2 2 2
#> 3 a 3 7 3 2
#> 4 b 1 4 4 4
#> 5 b 2 5 5 5
#> 6 b 3 6 6 6
由 reprex package (v2.0.0) 于 2021 年 6 月 17 日创建
,我尝试了另一种方法，过滤 c，使用 df 删除 NA 的所有列，使用 df2 加入并将未过滤的 df 的行与df3。
df3 <- df %>% filter(c == "a") %>% select_if(~ !any(is.na(.))) %>%
left_join(df2,"d"))
df3 <- bind_rows(df %>% filter(!c == "a"),df3) %>% arrange(c,d)
df3
#> # A tibble: 6 x 5
#> c d x y z
#> <chr> <dbl> <dbl> <dbl> <dbl>
#> 1 a 1 1 1 1
#> 2 a 2 2 2 2
#> 3 a 3 3 2 7
#> 4 b 1 4 4 4
#> 5 b 2 5 5 5
#> 6 b 3 6 6 6
Created on 2021-06-17 by the reprex package (v2.0.0)