从字符串匹配中总结答案

【问题标题】：summarize from string matches从字符串匹配中总结
【发布时间】：2019-02-20 13:38:27
【问题描述】：

我有这个 df 列：

df <- data.frame(Strings = c("ñlas onepojasd", "onenañdsl", "ñelrtwofkld", "asdthreeasp", "asdfetwoasd", "fouroqwke","okasdtwo", "acmofour", "porefour", "okstwo"))
> df
          Strings
1  ñlas onepojasd
2       onenañdsl
3     ñelrtwofkld
4     asdthreeasp
5     asdfetwoasd
6       fouroqwke
7        okasdtwo
8        acmofour
9        porefour
10         okstwo

我知道df$Strings 中的每个值都将与单词one, two, three or four 匹配。而且我也知道它只会与其中一个词匹配。所以要匹配它们：

str_detect(df$Strings,"one")
str_detect(df$Strings,"two")
str_detect(df$Strings,"three")
str_detect(df$Strings,"four")

但是，我被困在这里，因为我正在尝试做这张桌子：

Homes  Quantity Percent
  One         2     0.3
  Two         4     0.4
Three         1     0.1
 Four         3     0.3
Total        10       1

【问题讨论】：

标签： r string summary

【解决方案1】：

使用tidyverse 和janitor，您可以：

df %>%
 mutate(Homes = str_extract(Strings, "one|two|three|four"),
        n = n()) %>%
 group_by(Homes) %>%
 summarise(Quantity = length(Homes),
           Percent = first(length(Homes)/n)) %>%
 adorn_totals("row")

 Homes Quantity Percent
  four        3     0.3
   one        2     0.2
 three        1     0.1
   two        4     0.4
 Total       10     1.0

或者只用tidyverse:

 df %>%
 mutate(Homes = str_extract(Strings, "one|two|three|four"),
        n = n()) %>%
 group_by(Homes) %>%
 summarise(Quantity = length(Homes),
           Percent = first(length(Homes)/n)) %>%
 rbind(., data.frame(Homes = "Total", Quantity = sum(.$Quantity), 
                     Percent = sum(.$Percent)))

在这两种情况下，代码首先提取匹配模式并计算案例数量。其次，它按匹配的单词分组。第三，它计算每个单词的案例数以及给定单词在所有单词中的比例。最后，它添加了一个“总计”行。

【讨论】：

【解决方案2】：

您可以使用str_extract，然后执行table 和prop.table，即

library(stringr)

str_extract(df1$Strings, 'one|two|three|four')
#[1] "one"   "one"   "two"   "three" "two"   "four"  "two"   "four"  "four"  "two"  

table(str_extract(df1$Strings, 'one|two|three|four'))
# four   one three   two 
#    3     2     1     4 

prop.table(table(str_extract(df1$Strings, 'one|two|three|four')))
# four   one three   two 
#  0.3   0.2   0.1   0.4

【讨论】：

【解决方案3】：

base R 选项将是 regmatches/regexpr 和 table

table(regmatches(df$Strings, regexpr('one|two|three|four', df$Strings)))
#  four   one three   two 
#    3     2     1     4

添加addmargins 得到sum 然后除以它

out <- addmargins(table(regmatches(df$Strings, 
     regexpr('one|two|three|four', df$Strings))))
out/out[length(out)]

# four   one three   two   Sum 
#  0.3   0.2   0.1   0.4   1.0

【讨论】：