left_join 在具有不同列名的 for 循环中答案

【问题标题】：left_join in a for loop with different columns namesleft_join 在具有不同列名的 for 循环中
【发布时间】：2020-03-24 05:04:37
【问题描述】：

我有一个名为a 的data.frame，其结构类似于：-

a <- data.frame(X1=c("A", "B", "C", "A", "C", "D"),
                X2=c("B", "C", "D", "A", "B", "A"),
                X3=c("C", "D", "A", "B", "A", "B")
               )

我还有另一套是：-

b <- data.frame(Xn=c("A", "B", "C", "D"),
                Feature=c("some", "more", "what", "why"))

我想将集合b中的所有Features添加到集合a，这样X1、X2和X3在集合a中都有对应的特征列。换句话说，集合a 中的列变为：-

colnames(a) <- c("X1", "X2", "X3", "Features1", "Features2", "Features3")

如何在 for 循环中使用 left_join 来做到这一点？

【问题讨论】：

标签： r for-loop left-join cran

【解决方案1】：

在base R中，我们可以unlista数据帧和match它与b$Xn得到对应的Feature值。我们可以cbind这个数据框到原始数据框得到最终答案。

temp <- a
temp[] <- b$Feature[match(unlist(temp), b$Xn)]
names(temp) <- paste0('Feature', seq_along(temp))
cbind(a, temp)

#  X1 X2 X3 Feature1 Feature2 Feature3
#1  A  B  C     some     more     what
#2  B  C  D     more     what      why
#3  C  D  A     what      why     some
#4  A  A  B     some     some     more
#5  C  B  A     what     more     some
#6  D  A  B      why     some     more

在tidyverse 中，我们可以获取长格式的数据，将数据连接起来并恢复为宽格式。

library(dplyr)
library(tidyr)

a %>%
  mutate(row = row_number()) %>%
  pivot_longer(cols = -row) %>%
  left_join(b, by = c('value' = 'Xn'))  %>%
  select(-value) %>%
  pivot_wider(names_from = name, values_from = Feature) %>%
  select(-row) %>%
  rename_all(~paste0('Feature', seq_along(.))) %>%
  bind_cols(a, .)

【讨论】：

【解决方案2】：

这可以通过使用mutate_all 到recode a 中的所有列来完成：

library(tidyverse)

a %>% 
  mutate_all(funs(feat=recode(., !!!set_names(as.character(b$Feature), b$Xn))))

  X1    X2    X3    X1_feat X2_feat X3_feat
1 A     B     C     some    more    what   
2 B     C     D     more    what    why    
3 C     D     A     what    why     some   
4 A     A     B     some    some    more   
5 C     B     A     what    more    some   
6 D     A     B     why     some    more

您可以添加rename_at 以获得所需的名称：

a %>% 
  mutate_all(funs(f=recode(., !!!set_names(as.character(b$Feature), b$Xn)))) %>% 
  rename_at(vars(matches("f")), ~gsub(".([0-9]).*", "Feature\\1", .))

  X1 X2 X3 Feature1 Feature2 Feature3
1  A  B  C     some     more     what
2  B  C  D     more     what      why
3  C  D  A     what      why     some
4  A  A  B     some     some     more
5  C  B  A     what     more     some
6  D  A  B      why     some     more

【讨论】：