使用map()将变量列表作为参数传递给自定义函数

Question

我有一个自定义函数：

1. 获取数据帧列表并传递字符串
2. 进行一些数据转换并且
3. 返回包含转换后数据的数据框

可重现的示例：

# I have 4 dataframes in a listy with a few observations that repeat themselves
df_1 <- data.frame(col1 = c(1, 2, 3, 4), col2 = c('apple', 'pineapple', 'orange', 'grape'))
df_2 <- data.frame(col1 = c(2, 3, 4, 5, 6, 7), col2 = c('watermelon', 'orange', 'halibut', 'apple', 'iron', 'grape'))
df_3 <- data.frame(col1 = c(2, 3, 4, 5, 6, 7, 9, 0), col2 = c('rock', 'pineapple', 'apple', 'tire', 'bomb', 'star', 'coconut', 'grape'))
df_4 <- data.frame(col1 = c(1, 4, 9), col2 = c('grape', 'apple', 'rock'))

# All inside a another list
df_list <- list(df_1, df_2, df_3, df_4)

# now we use a function where
toy_function <- function(df_list, var1) {
  map(df_list, ~.x %>% filter (col2 == var1) %>% mutate(result = col1 * 2)) %>% 
    bind_rows() %>% 
    select(result)
}

# Solution from toy_function()
toy_function(df_list = df_list, var1 = 'apple')

现在，我想做的是将字符串向量传递给 toy_function()，如下所示：

# List of strings to pass to toy_function()
list_of_fruits <- c('apple', 'grape')

# This is where it all goes wrong
map2(.x = df_list, .y = list_of_fruits, .f = toy_function)

# Error
Error in `map2()`:
! Can't recycle `.x` (size 4) to match `.y` (size 2).
Run `rlang::last_trace()` to see where the error occurred.

我想从函数中得到的结果是：

map2(.x = df_list, .y = list_of_fruits, .f = toy_function)

# Expected results

[[1]]
  result
1 2
2 10
3 8
4 8

[[2]]
  result
1 8
2 14
3 0
4 2

编辑

正如评论中指出的，应该修改

toy_function()

以捕获所有变量：

toy_function <- function(df_list, var1) {
  map(df_list, ~ { 
        filtered_df <- .x %>% filter(col2 %in% var1) 
        filtered_df %>% mutate(result = col1 * 2) %>% select(result)
}) %>% 
  bind_rows()
}

但还是出现这个错误

> map2(.x = df_list, .y = list_of_fruits, .f = toy_function)
Error in `map2()`:
! Can't recycle `.x` (size 4) to match `.y` (size 2).
Run `rlang::last_trace()` to see where the error occurred.
````

Answer 1

map2

期望参数具有相同的长度，并且它将“并行”迭代它们，它的工作原理如下：

## this map2 call
map2(.x = df_list, .y = list_of_fruits, .f = toy_function)
## is equivalent to this:
list(
  toy_function(df_list[[1]], list_of_fruits[[1]]), 
  toy_function(df_list[[2]], list_of_fruits[[2]]), 
  toy_function(df_list[[3]], list_of_fruits[[3]]),
  ...
)

注意

df_list

和

list_of_fruits

是如何同时迭代的。

你不希望这样。您已经编写了

toy_function

，因此它已经期望

list

作为它的第一个参数，并且它在内部使用

map

。您不需要另一个包装器来迭代

df_list

。您只需要迭代 1 个对象，即水果列表。

map(list_of_fruits, \(fruit) toy_function(df_list, fruit))
# [[1]]
#   result
# 1      2
# 2     10
# 3      8
# 4      8
# 
# [[2]]
#   result
# 1      8
# 2     14
# 3      0
# 4      2