计算 R 数据框列中出现的月份名称答案

【问题标题】：Counting month name occurrences in R dataframe column计算 R 数据框列中出现的月份名称
【发布时间】：2021-05-31 01:33:29
【问题描述】：

我有一个这样的数据框

from <- c('Bob', 'Matt', 'Bob', 'Bob', 'Matt', 'Bob', 'Matt')
subject <- c('See you in June', 'Funny cat video', 'Plans for December', 'Re: Plans for December', 'Re:Re: Plans for December','November sales', 'Out of office in November')

df <- data.frame(from, subject)

  from                   subject
1  Bob           See you in June
2 Matt           Funny cat video
3  Bob        Plans for December
4  Bob    Re: Plans for December
5 Matt Re:Re: Plans for December
6  Bob            November sales
7 Matt Out of office in November

我想根据内置的month.name 向量在subject 中以字符串形式查找月份名称的出现

 [1] "January"   "February"  "March"     "April"     "May"       "June"      "July"     
 [8] "August"    "September" "October"   "November"  "December"

然后将它们按from 值和月份相加。我想得到这样的结果：

from    month counts
1  Bob     June      1
2  Bob November      1
3 Matt November      1
4  Bob December      2
5 Matt December      1

我的猜测是我需要像下面这样使用str_count，但我不知道如何用from 中的名称来计数。

df %>%
  mutate(counts = str_count(subject, month.name)

【问题讨论】：

首先使用 str_extract 创建您的月份列，然后是 groupby，然后是 summarise
同一个字符串中是否有可能出现多个月份？

标签： r string tidyverse

【解决方案1】：

使用regmatches + subset + aggregate 的基本 R 选项

aggregate(
  n ~ from + months,
  subset(
    transform(
      df,
      months = unlist(
        replace(
          u <- regmatches(subject, gregexpr(paste0(month.name, collapse = "|"), subject)),
          lengths(u) == 0,
          NA
        )
      ),
      n = 1
    ),
    !is.na(months)
  ), sum
)

给予

  from   months n
1  Bob December 2
2 Matt December 1
3  Bob     June 1
4  Bob November 1
5 Matt November 1

【讨论】：

【解决方案2】：

另一种解决方案

library(tidyverse)
from <- c('Bob', 'Matt', 'Bob', 'Bob', 'Matt', 'Bob', 'Matt')
subject <- c('See you in June', 'Funny cat video', 'Plans for December', 'Re: Plans for December', 'Re:Re: Plans for December','November sales', 'Out of office in November')

df <- data.frame(from, subject)
df %>% 
  group_by(from) %>% 
  mutate(month = map(strsplit(subject, split = " "), ~intersect(.x, month.name))) %>% 
  unnest(month) %>% 
  count(from, month)
#> # A tibble: 5 x 3
#> # Groups:   from [2]
#>   from  month        n
#>   <chr> <chr>    <int>
#> 1 Bob   December     2
#> 2 Bob   June         1
#> 3 Bob   November     1
#> 4 Matt  December     1
#> 5 Matt  November     1

^{由reprex package (v1.0.0) 于 2021-03-01 创建}

使用 data.table


library(data.table)

setDT(df)[, list(from, month = unlist(tstrsplit(subject, split = " ")))][month %in% month.name][, .N, by = list(from, month)]
#>    from    month N
#> 1:  Bob November 1
#> 2:  Bob December 2
#> 3:  Bob     June 1
#> 4: Matt December 1
#> 5: Matt November 1

^{由reprex package (v1.0.0) 于 2021-03-01 创建}

【讨论】：

【解决方案3】：

这是一个解决方案，使用 str_extract_all，然后在 data.frame 中处理一些列表：

library(tidyverse)

df %>%
    rowwise() %>%
    mutate(months = list(str_extract_all(subject, month.name))) %>%
    unnest(months) %>%
    rowwise() %>%
    filter(length(months) > 0) %>%
    unnest(months) %>%
    count(from, months)

#   from  months       n
# 1 Bob   December     2
# 2 Bob   June         1
# 3 Bob   November     1
# 4 Matt  December     1
# 5 Matt  November     1

【讨论】：

【解决方案4】：

基础 R 解决方案：

subset(
  aggregate(ctr ~ from + months_,
   transform(
     data.frame(do.call(rbind, Map(
       function(y, z) {
         cbind(from = y, months_ = z)
       },
       df$from,
       apply(do.call(rbind, lapply(month.name, function(w) {
         grepl(w, df$subject)
       })), 2, function(x) {
         if (sum(x) == 0) {
           ""
         } else{
           month.name[x]
         }
       })
     )),
     stringsAsFactors = FALSE), ctr = 1
   ), sum), 
  months_ !=  "")

【讨论】：