通过拆分模式的字符列在这里匹配

         Province     ElecDistName        Candidate Votes Majority Vper MajPer 
              <chr>       <chr>         <chr> <int> <int> <dbl> <dbl> 
1 Newfoundland and Labrador/Terre-Neuve-et-Labrador St. John's East/St. John's-Est      Nick Whalen Liberal 20974  646 46.7 1.4 
2 Newfoundland and Labrador/Terre-Neuve-et-Labrador St. John's East/St. John's-Est Jack Harris ** NDP-New Democratic Party 20328  NA 45.3  NA 
3 Newfoundland and Labrador/Terre-Neuve-et-Labrador St. John's East/St. John's-Est   Deanne Stapleton Conservative 2938  NA 6.5  NA 
4 Newfoundland and Labrador/Terre-Neuve-et-Labrador St. John's East/St. John's-Est  David Anthony Peters Green Party 500  NA 1.1  NA 
5 Newfoundland and Labrador/Terre-Neuve-et-Labrador St. John's East/St. John's-Est     Sean Burton Communist 140  NA 0.3  NA 
6     New Brunswick/Nouveau-Brunswick     Fundy Royal     Alaina Lockhart Liberal 19136  1775 40.9 3.8

Top of Dataset 通过拆分模式的字符列在这里匹配

业余的问题，我想在候选列一分为二，一个包含名称，另含的一方。我尝试了一些在这里发布的单独功能：

separate(ElecResults, Candidate, into = c("Name", "Party"), sep = " (?=[^ ]+$)")

但是，这似乎错过了很多意见。对于有三个名字的候选人来说，这个问题是显而易见的，但也有一些人似乎完全错过了（对于一个人来说，这个候选人带着莫名其妙的双星号）。

我试图想到如果函数与grepl结合在一起，它会识别自由派，保守派，NDP和绿色等最常见的派对名称，并创建一个名为派对的新列，每次尝试时不断收到错误消息。

如果有人对我如何拆分这个专栏有一个想法，这将是一个巨大的帮助。

谢谢！

下面是使用dput代码：

structure(list(Province = c("Newfoundland and Labrador/Terre-Neuve-et-Labrador", 
"Newfoundland and Labrador/Terre-Neuve-et-Labrador", "Newfoundland and Labrador/Terre-Neuve-et-Labrador", 
"Newfoundland and Labrador/Terre-Neuve-et-Labrador", "Newfoundland and Labrador/Terre-Neuve-et-Labrador", 
"New Brunswick/Nouveau-Brunswick"), ElecDistName = c("St. John's East/St. John's-Est", 
"St. John's East/St. John's-Est", "St. John's East/St. John's-Est", 
"St. John's East/St. John's-Est", "St. John's East/St. John's-Est", 
"Fundy Royal"), Candidate = c("Nick Whalen Liberal", "Jack Harris ** NDP-New Democratic Party", 
"Deanne Stapleton Conservative", "David Anthony Peters Green Party", 
"Sean Burton Communist", "Alaina Lockhart Liberal"), Votes = c(20974L, 
20328L, 2938L, 500L, 140L, 19136L), Majority = c(646L, NA, NA, 
NA, NA, 1775L), Vper = c(46.7, 45.3, 6.5, 1.1, 0.3, 40.9), MajPer = c(1.4, 
NA, NA, NA, NA, 3.8)), .Names = c("Province", "ElecDistName", 
"Candidate", "Votes", "Majority", "Vper", "MajPer"), row.names = c(NA, 
-6L), class = c("tbl_df", "tbl", "data.frame"))

来源

2017-06-09 lazslo93

难以帮助，而无需查看数据对象的外观。 – ssp3nc3r

对不起，我已经发布了代码 – lazslo93

你有多少个不同的派对？并且它们是否统一命名（例如，保守性总是保守的） – Val

下面是一个使用fuzzyjoin包

library(tidyverse) 
library(fuzzyjoin) 

parties <- data_frame(party = c("Liberal", "NDP-New Democratic Party", "Conservative", "Green Party", "Communist")) 

df %>% 
    regex_left_join(parties, by = c(Candidate = "party")) %>% 
    replace_na(list(party = "minor")) %>% 
    mutate(Candidate = str_replace(Candidate, party, "")) %>% 
    select(Candidate, party) 
#> # A tibble: 6 x 2 
#>    Candidate     party 
#>     <chr>     <chr> 
#> 1   Nick Whalen     Liberal 
#> 2  Jack Harris ** NDP-New Democratic Party 
#> 3  Deanne Stapleton    Conservative 
#> 4 David Anthony Peters    Green Party 
#> 5   Sean Burton     Communist 
#> 6  Alaina Lockhart     Liberal

注意，最后选择只加到表明，该方法适用的另一种方法。我特别喜欢这种方法，因为可能在数据框中显示的其他方可以很好地处理使用replace_na

来源

2017-06-09 22:20:58 yeedle

谢谢，这工作得很好！ – lazslo93

下面是一些基本的代码，你需要国防部。将每个聚会名称放在由|分隔的引号内

require(dplyr) 
require(stringr) 

df <- data.frame(Candidate = "Nick Whalen Liberal", Majority = 1) 
parties <- c("Liberal|Conservative") 
df %>% mutate(Name = str_sub(Candidate, 1, str_locate(Candidate, parties)[1] - 1))

来源

2017-06-09 19:13:34 ssp3nc3r

谢谢！我会玩这个 – lazslo93

通过拆分模式的字符列在这里匹配

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