Question

我有一个从json导入的字符串数组的列表。

data = fromJSON("data.json")

> data
[[1]]
[1] "Finish"  "Accept"  "Request"

[[2]]
[1] "Finish"  "Accept"  "Request"

[[3]]
[1] "Finish"        "Accept"        "UnlockRequest"
[4] "Accept"        "Request"      

[[4]]
[1] "Finish"  "Accept"  "Request"

[[5]]
[1] "Finish"  "Accept"  "Request"

我现在的目标是计算列表对象，类似于数据表的table()函数，但是table(data)返回错误：

table(data)中的错误：所有参数的长度必须相同

这里正确的解决方法是什么？

输出应类似于：

c("Finish", "Accept", "Request") 4
c("Finish", "Accept", "UnlockRequest", "Accept", 
    "Request") 1

数据：

 list(c("Finish", "Accept", "Request"), c("Finish", "Accept", 
"Request"), c("Finish", "Accept", "UnlockRequest", "Accept", 
"Request"), c("Finish", "Accept", "Request"), c("Finish", "Accept", 
"Request"))

Answer 1

请注意这是否“正确”，但这是一种方法。基本上将其变成带有一个列表列的data_frame，操纵列表列（使用as.character），然后对其计数：

library(tidyverse)

data_frame(l1) %>%
    mutate(char = as.character(l1)) %>%
    count(char)

# A tibble: 2 x 2
  char                                                                        n
  <chr>                                                                   <int>
1 "c(\"Finish\", \"Accept\", \"Request\")"                                    4
2 "c(\"Finish\", \"Accept\", \"UnlockRequest\", \"Accept\", \"Request\")"     1

另一种方法是将文本paste在一起，运行table，然后使用strsplit。

table(sapply(l1, paste, collapse = ',')) %>% 
    as_data_frame() %>%
    mutate(list_col = strsplit(Var1, ','))

Source: local data frame [2 x 3]
Groups: <by row>

# A tibble: 2 x 3
  Var1                                           n list_col 
  <chr>                                      <int> <list>   
1 Finish,Accept,Request                          4 <chr [3]>
2 Finish,Accept,UnlockRequest,Accept,Request     1 <chr [5]>

Answer 2

您可以在列表上运行as.character()，然后将结果制成表格。

as.data.frame(table(as.character(x)))
#                                                          Var1 Freq
# 1                            c("Finish", "Accept", "Request")    4
# 2 c("Finish", "Accept", "UnlockRequest", "Accept", "Request")    1

数据：

x <- list(c("Finish", "Accept", "Request"), c("Finish", "Accept", 
"Request"), c("Finish", "Accept", "UnlockRequest", "Accept", 
"Request"), c("Finish", "Accept", "Request"), c("Finish", "Accept", 
"Request"))

从列表中整理序列

2 个答案: