r来自不同长度和不同键的数据帧列表的多个连接

时间:2018-01-25 21:47:04

标签: r list dataframe left-join

假设我有这个数据框列表:

library(tidyverse)
df_list <- list(data.frame(cheese = c("ex","ok","bd"), 
                          cheese_val = c(3:1), 
                          stringsAsFactors = F),
               data.frame(egg = c("great","good","bad", "eww"), 
                          egg_val = c(4:1),
                          stringsAsFactors = F),
               data.frame(milk = c("good","bad"), 
                          milk_val = c(2:1), 
                          stringsAsFactors = F))

我有这个核心数据集:

core_dat <- data.frame(cheese = c("ex","ok","ok", "bd", "ok"), 
                      egg = c("great", "bad", "bad", "eww", "great"), 
                      milk = c("good", "good", "good", "bad", "good"), 
                      stringsAsFactors = F)

我想core_dat单独加入df_list的每个元素。

然后我尝试了这个:

for(i in 1:length(df_list)) {
  gg<-core_dat %>% 
    left_join(df_list[[i]], by = names(df_list[[i]][1]), copy = T)
}

已运行但仅将联接应用于milk列,因此core_dat中唯一的其他列为milk_val,但我希望看到cheese_val和{{ 1}}。

我怀疑这里有比for循环更合适的选项,我正在寻找建议。请注意,我的实际数据集比这个小例子有更多的df。

我不应该期望结果数据框(在本例中为egg_val)总共包含6列(3个标准名称+ 3,带有“val”后缀),使得它看起来像这样的打印版本:

gg

我在这里看到了许多“多个连接”的答案,但没有一个与我在这里想要完成的事情完全一致(不同的键列,不同的数据长度)。

2 个答案:

答案 0 :(得分:2)

您可以使用map获取已加入数据框的列表,然后使用reduce将它们全部加入。

map(df_list, right_join, rownames_to_column(core_dat)) %>%
  reduce(full_join)
# Joining, by = "cheese"
# Joining, by = "egg"
# Joining, by = "milk"
# Joining, by = c("cheese", "rowname", "egg", "milk")
# Joining, by = c("cheese", "rowname", "egg", "milk")
#   cheese cheese_val rowname   egg milk egg_val milk_val
# 1     ex          3       1 great good       4        2
# 2     ok          2       2   bad good       2        2
# 3     ok          2       3   bad good       2        2
# 4     bd          1       4   eww  bad       1        1
# 5     ok          2       5 great good       4        2

答案 1 :(得分:2)

这应该给出所需的输出:

Reduce(merge,c(df_list,list(core_dat)))
  cheese   egg milk cheese_val egg_val milk_val
1     bd   eww  bad          1       1        1
2     ex great good          3       4        2
3     ok   bad good          2       2        2
4     ok   bad good          2       2        2
5     ok great good          2       4        2