for循环中使用不同的列名称的left_join

Question

我有一个名为a的数据框架，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")
               )

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Features中的所有b添加到集合a中，以便X1，X2和X3在集合a中具有其对应的功能列。换句话说，集合a中的列变为：-

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

我该如何在for循环中使用left_join？

Answer 1

在基数R中，我们可以将unlist a数据帧和match与b$Xn一起获得相应的Feature值。我们可以将此数据帧cbind转换为原始数据帧以获得最终答案。

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，将数据合并并恢复为宽格式。

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, .)

Answer 2

我们可以使用map生成具有适当名称的数据帧列表以进行连接，然后使用reduce进行连续的连接。例如：

library(tidyverse)

c(list(a), 
  map(1:3, ~b %>% set_names(paste0(c("X","Feature"), .x)))) %>% 
  reduce(left_join)

  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

为了解压，map语句使用b创建我们要加入原始数据帧a的三个数据帧的列表。

map(1:3, ~b %>% set_names(paste0(c("X","Feature"), .x)))

[[1]]
  X1 Feature1
1  A     some
2  B     more
3  C     what
4  D      why
5  A      zzz

[[2]]
  X2 Feature2
1  A     some
2  B     more
3  C     what
4  D      why
5  A      zzz

[[3]]
  X3 Feature3
1  A     some
2  B     more
3  C     what
4  D      why
5  A      zzz

[我们使用c函数将原始数据帧a放在我们刚刚创建的列表的前面，从而为我们提供了四个数据帧的列表。

c(list(a), 
  map(1:3, ~b %>% set_names(paste0(c("X","Feature"), .x))))

现在我们要连续地将map创建的三个数据帧连接到a。这就是reduce(left_join)所做的。