dplyr 使用 purrr::map 计算 tibbles 列表中的单个观察值

dplyr count single observations in list of tibbles using purrr::map

我正在尝试计算包含由“;”分隔的单个观察结果的小标题列表中出现的频率。当我在 purrr::map() 中使用 purrr::map() 时,我 运行 遇到了一个错误。我怀疑我遗漏了一些简单的东西,因此不胜感激。

以不同客户购买水果为例,同时购买的水果用“;”分隔

# Fruit purchases across days with different number of customers.
day_1 <- as_data_frame(setNames(list(c("oranges;peaches;apples", "pears;apples", "bananas", "oranges;apples", "apples")), "fruits"))
day_2 <- as_data_frame(setNames(list(c("oranges;apples", "peaches","apples;bananas;", "pears", "apples;peaches", "oranges")), "fruits"))
day_3 <- as_data_frame(setNames(list(c("peaches;pears","apples","bananas")), "fruits"))

# Create list of fruit purchases.
fruit_list <- list(day_1, day_2, day_3)

这个 returns 三个小标题的列表,是我的数据的一般格式。我可以使用 dplyr/purrr:

来计算每个水果每天的总观察次数
fruit_list %>% 
  map(function(x) strsplit(x$fruits, ";")) %>% 
  map(unlist) %>% 
  map(table)

但是,当我尝试使用 map() 中的 map() 来隔离和统计 tibbles 列表中的单一水果购买时,我 运行 进入错误

"Error: .x is not a vector (closure)"

fruit_list %>% 
  map(mutate(fruit_count = map(function(x) strsplit(x$fruits, ";"), length))) %>% 
  filter(fruit_count==1) %>% 
  count(solo_fruits = fruits) 

我可以在单个 tibble/df 上执行此功能,但不能跨 tibbles 列表。我是否遗漏了 map() 函数的某些内容或更明显的内容?谢谢!

第一个小标题所需的结果格式:

# A tibble: 2 x 2
  solo_fruits     n
  <chr>       <int>
1 apples          1
2 bananas         1

我如何得出上述单个样本的答案:

day_1_df <- as.data.frame(fruit_list[[1]]) 
day_1_df %>% 
  mutate(fruit_count = map(strsplit(day_1_df$fruits, ";"), length)) %>% 
  filter(fruit_count==1) %>% 
  count(solo_fruits = fruits) 

不完全符合您的要求,但它可能会以不同的方式解决您的问题:

library(tidyverse)

day_1 <- as_data_frame(setNames(list(c("oranges;peaches;apples", "pears;apples", "bananas", "oranges;apples", "apples")), "fruits"))
day_2 <- as_data_frame(setNames(list(c("oranges;apples", "peaches","apples;bananas;", "pears", "apples;peaches", "oranges")), "fruits"))
day_3 <- as_data_frame(setNames(list(c("peaches;pears","apples","bananas")), "fruits"))

df <- tibble(day = 1:3, fruits = c(day_1, day_2, day_3)) %>% 
  unnest() %>% 
  mutate(fruits = strsplit(fruits, ";"), customer = row_number()) %>% 
  unnest()

df %>% 
  group_by(customer) %>% 
  filter(n() == 1) %>% 
  group_by(customer, day, fruits) %>% 
  summarise(n = n())

# # A tibble: 7 x 4
# # Groups:   customer, day [?]
#   customer   day fruits      n
#      <int> <int> <chr>   <int>
# 1        3     1 bananas     1
# 2        5     1 apples      1
# 3        7     2 peaches     1
# 4        9     2 pears       1
# 5       11     2 oranges     1
# 6       13     3 apples      1
# 7       14     3 bananas     1

编辑:误会后修改

您可以使用 str_detect 来捕获没有 ; 的行。或者你可以使用 str_count 来计算 ; 然后加 1.

fruit_list%>%
     map(~filter(.x,!str_detect(fruits,";"))%>%
             mutate(solo_fruits = fruits,count = 1,fruits=NULL))
[[1]]
# A tibble: 2 x 2
  solo_fruits count
  <chr>       <dbl>
1 bananas         1
2 apples          1

[[2]]
# A tibble: 3 x 2
  solo_fruits count
  <chr>       <dbl>
1 peaches         1
2 pears           1
3 oranges         1

[[3]]
# A tibble: 2 x 2
  solo_fruits count
  <chr>       <dbl>
1 apples          1
2 bananas         1

我的意思是使用 str_count:这将为您提供每行的水果总数。而不是拆分然后使用 length

fruit_list%>%
    map(~mutate(.x,count = str_count(fruits,";") + 1))