从带有参考 df 的列表中选择 df 和变量以进行变量选择：R，purrr

Question

我有一个 df 列表，其中包含不同的变量。

最后，我想从数据框和变量的子选择中得到一个合成数据框，并将其合并为一个。 我有另一个 df，其中包含感兴趣的变量列表和相应的表。

当然，要在简单的 df 上进行选择，只需进行 select(colnames(.) %in% VarSel$Var) 并对所有 df 进行选择，然后进行连接... 但是，就目前而言，我不能直接在所选 df 列表中的 purrr 中执行此操作。

你知道如何正确地做到这一点吗？

# reprex

## a list of df 

list_df <- list(A = data.frame(ID = letters[1:10],
                               Var1 = rnorm(10),
                               Var2 = rnorm(10),
                                Var3 = rnorm(10)),
                B = data.frame(ID = letters[1:10],
                               Var1 = rnorm(10),
                               VarY = rnorm(10),
                               VarX = rnorm(10)),
                D = data.frame(ID = letters[5:14],
                               Var1 = rnorm(10),
                               VarZ = rnorm(10),
                               Var3 = rnorm(10)),
                E = data.frame(ID = letters[3:12],
                               Var1 = rnorm(10),
                               VarK = rnorm(10),
                               Var3 = rnorm(10)))

## a df for variables & df selection 

VarSel <- data.frame(Var = c("ID", "VarZ", "Var3", "Var3", "VarK"),
                     Df = c("all", "D", "A", "D", "E"))



# the 3 operations that I want to perform (I simply managed the first one)

list_df %>% 
  keep(names(.) %in% unique(VarSel$Df))# keep only df named in VarSel 
                                       # then select matching variable names for thoses DF (not forgetting the ID one) 
                                       # finally produce a synthesis Df joining A & D and variables of interest (ID, VarZ, Var3, Var3)

# desired output 
output <- full_join(list_df[["A"]], list_df[["D"]], by = "ID") %>%
full_join(.,list_df[["E"]], by= "ID") %>% 
select(ID, VarZ, Var3_A = Var3.x, Var3_D = Var3.y, VarK)

Answer 1

你可以这样做：

library(purrr)
map2_dfc(list_df[VarSel$Df], VarSel$Var, ~ .x[.y])

#    ID       VarZ   Var3...3    Var3...4
# 1   a  1.0215807 -0.9307522 -0.79314560
# 2   b -1.3861810 -1.1376876 -3.06921836
# 3   c  0.4661697  0.3690057 -0.08778707
# 4   d  0.1365207  0.5290638  0.86164346
# 5   e  0.5498877 -0.6208638 -2.22677452
# 6   f  0.4925813 -1.3938548  0.21014465
# 7   g  0.3675358  0.8269150 -1.71422099
# 8   h  0.4576200  0.1554632 -1.80290292
# 9   i  0.5560993  0.6597010 -1.43455363
# 10  j  0.2121097 -0.5935487 -1.42085047

---

请注意，我稍微更改了您的 VarSel 以避免无法识别的 "all"。由于它们都是相同的，所以我选择了“A”。

VarSel <- data.frame(Var = c("ID", "VarZ", "Var3", "Var3"),
                     Df = c("A", "D", "A", "D"))

---

要获得完全正确的输出，您可以动态创建新名称向量，然后使用 pmap：

newnames <- ifelse(duplicated(VarSel$Var) | duplicated(VarSel$Var, fromLast = TRUE), 
                   paste0(VarSel$Var, "_", VarSel$Df), VarSel$Var)
pmap_dfc(list(list_df[VarSel$Df], VarSel$Var, newnames), ~ set_names(..1[..2], ..3))

#    ID       VarZ     Var3_A      Var3_D
# 1   a  1.0215807 -0.9307522 -0.79314560
# 2   b -1.3861810 -1.1376876 -3.06921836
# 3   c  0.4661697  0.3690057 -0.08778707
# 4   d  0.1365207  0.5290638  0.86164346
# 5   e  0.5498877 -0.6208638 -2.22677452
# 6   f  0.4925813 -1.3938548  0.21014465
# 7   g  0.3675358  0.8269150 -1.71422099
# 8   h  0.4576200  0.1554632 -1.80290292
# 9   i  0.5560993  0.6597010 -1.43455363
# 10  j  0.2121097 -0.5935487 -1.42085047