Question

我在变量resHeaders中有一个字符串列表：
每一行的语法是：＆＃34;玩家名称，玩家A分数，玩家B名称，玩家B分数＆＃34;

> resHeaders
[1] "Mackenzie McDonald\n0\n5\n0\nTatsuma Ito\n0\n5\n0"     
[2] "Uladzimir Ignatik\n0\n5\n15\nGleb Sakharov\n0\n3\n30"  
[3] "Evgeny Karlovskiy\n0\n0\n30\nGuillermo Olaso\n1\n0\n15"
[4] "Katherine Sebov\n0\n3\n40\nAmandine Hesse\n0\n2\n40"   
[5] "Karolina Muchova\n1\n1\n15\nElena Bovina\n0\n1\n0"

如何提取＆＃34;玩家名字＆＃34;和＃34;玩家B的名字＆＃34;零件？
对于第一行，结果将是：

球员一个名字：＆＃34; Mackenzie McDonald＆＃34;
玩家B名称：＆＃34; Tatsuma Ito＆＃34;

数据（@ h3rm4n）

vec <- c("Mackenzie McDonald\n0\n5\n0\nTatsuma Ito\n0\n5\n0","Uladzimir Ignatik\n0\n5\n15\nGleb Sakharov\n0\n3\n30",
     "Evgeny Karlovskiy\n0\n0\n30\nGuillermo Olaso\n1\n0\n15","Katherine Sebov\n0\n3\n40\nAmandine Hesse\n0\n2\n40")

Answer 1

使用tidyverse方法，我们可以使用dplyr和purrr中的动词将数据重新整形为 key~value 格式。

library(dplyr)
library(purrr)

chr <- c(
  "Mackenzie McDonald\n0\n5\n0\nTatsuma Ito\n0\n5\n0",
  "Uladzimir Ignatik\n0\n5\n15\nGleb Sakharov\n0\n3\n30",
  "Evgeny Karlovskiy\n0\n0\n30\nGuillermo Olaso\n1\n0\n15",
  "Katherine Sebov\n0\n3\n40\nAmandine Hesse\n0\n2\n40",
  "Karolina Muchova\n1\n1\n15\nElena Bovina\n0\n1\n0"
)

map2_dfr(chr, 1:length(chr), ~{

  df <- as.data.frame(
    matrix(unlist(strsplit(.x, "\n")), ncol = 4, byrow = TRUE),
    stringsAsFactors = FALSE
  )

  df %>%
    transmute(
      match = .y,
      player = c("A", "B"),
      name = V1,
      score = paste(V2, V3, V4, sep = ", ")
    ) %>%
    as_tibble

})

# # A tibble: 10 x 4
#    match player name               score   
#    <int> <chr>  <chr>              <chr>   
#  1     1 A      Mackenzie McDonald 0, 5, 0 
#  2     1 B      Tatsuma Ito        0, 5, 0 
#  3     2 A      Uladzimir Ignatik  0, 5, 15
#  4     2 B      Gleb Sakharov      0, 3, 30
#  5     3 A      Evgeny Karlovskiy  0, 0, 30
#  6     3 B      Guillermo Olaso    1, 0, 15
#  7     4 A      Katherine Sebov    0, 3, 40
#  8     4 B      Amandine Hesse     0, 2, 40
#  9     5 A      Karolina Muchova   1, 1, 15
# 10     5 B      Elena Bovina       0, 1, 0

Answer 2

这可能是一个解决方案：

lapply(sapply(vec, strsplit, split = '\n'), '[', c(1,5))

结果：

$`Mackenzie McDonald\n0\n5\n0\nTatsuma Ito\n0\n5\n0`
[1] "Mackenzie McDonald" "Tatsuma Ito"       

$`Uladzimir Ignatik\n0\n5\n15\nGleb Sakharov\n0\n3\n30`
[1] "Uladzimir Ignatik" "Gleb Sakharov"    

$`Evgeny Karlovskiy\n0\n0\n30\nGuillermo Olaso\n1\n0\n15`
[1] "Evgeny Karlovskiy" "Guillermo Olaso"  

$`Katherine Sebov\n0\n3\n40\nAmandine Hesse\n0\n2\n40`
[1] "Katherine Sebov" "Amandine Hesse"

数据：

vec <- c("Mackenzie McDonald\n0\n5\n0\nTatsuma Ito\n0\n5\n0","Uladzimir Ignatik\n0\n5\n15\nGleb Sakharov\n0\n3\n30",
         "Evgeny Karlovskiy\n0\n0\n30\nGuillermo Olaso\n1\n0\n15","Katherine Sebov\n0\n3\n40\nAmandine Hesse\n0\n2\n40")

Answer 3

您可以使用r egmatches`来提取名称

do.call(rbind,regmatches(x,gregexpr("[A-Z]\\w+\\s[A-Z]\\w+",x,perl = T)))
     [,1]                 [,2]             
[1,] "Mackenzie McDonald" "Tatsuma Ito"    
[2,] "Uladzimir Ignatik"  "Gleb Sakharov"  
[3,] "Evgeny Karlovskiy"  "Guillermo Olaso"
[4,] "Katherine Sebov"    "Amandine Hesse" 
[5,] "Karolina Muchova"   "Elena Bovina"

您也可以使用str_extract_all(x,"[A-Z]\\w+\\s[A-Z]\\w+")

Answer 4

您也可以尝试：

df <- data.frame(strings = c("Mackenzie McDonald\n0\n5\n0\nTatsuma Ito\n0\n5\n0", 
                   "Uladzimir Ignatik\n0\n5\n15\nGleb Sakharov\n0\n3\n30" ), stringsAsFactors = FALSE)

f <- function(x) str_split(x, "\n")[[1]][nchar(str_split(x, "\n")[[1]]) > 3]

df2 <- apply(df, 1, f)
df2[,1] <- paste0("Player A name: ", df2[,1])
df2[,2] <- paste0("Player B name: ", df2[,2])
df2

R提取字符串的字符串部分

4 个答案: