R中的数据转换 - 傻瓜

Question

我想与 4 支国家足球队（英格兰、比利时、德国和法国）和 n 个日期合作

Date        Matches
16.03       England X Brazil
16.03       Belgium X Argentina
16.03       Chile X Japan
16.03       Uruguay X Germany
16.03       Italy x France
17.03       South Korea X India
17.03       Germany X France
17.03       Poland  X Belgium
17.03       Colombia X Russia
18.03       South Africa X Mexico
18.03       China X Japon
18.03       Brazil X Venezuela
...          ...

带有假人的期望数据帧。当 dummy = 1 时，队伍进行比赛。当 dummy = 0 球队没有比赛（当天）。重要提示：每行只有一个日期。

Date   Dummy_england   Dummy_belgium    Dummy_germany    Dummy_France
16.03  1               1                1                1
17.03  0               1                1                1
18.03  0               0                0                0

非常感谢！！

Answer 1

我们可以使用tidyverse 方法。

1. 使用str_extract从“匹配”的每一行中提取选定的团队
2. 只保留匹配的行，即用filter 删除NA 行
3. 使用pivot_wider 在selecting 感兴趣的列之后将values_fn 重新整形为“宽” - 将values_fn 指定为length 和values_fill 为0 以将默认NA 更改为0

library(dplyr)
library(tidyr)
library(stringr)
df1 %>%
   mutate(team = str_c('Dummy_', str_extract(Matches, 
       regex('England|Belgium|Germany|France', ignore_case = TRUE)))) %>%
   filter(complete.cases(team)) %>%
   select(-Matches) %>%
   pivot_wider(names_from = team, values_from = team,  
          values_fn = length, values_fill = 0)

-输出

# A tibble: 2 x 5
   Date Dummy_England Dummy_Belgium Dummy_Germany Dummy_France
  <dbl>         <int>         <int>         <int>        <int>
1  16.0             1             1             1            1
2  17.0             0             1             1            0

---

如果我们想在没有匹配的地方保留“日期”，请使用complete

df1 %>%
   mutate(team = str_c('Dummy_', str_extract(Matches, 
    regex('England|Belgium|Germany|France', ignore_case = TRUE)))) %>%
   filter(complete.cases(team)) %>%
   select(-Matches) %>%
   pivot_wider(names_from = team, values_from = team, 
        values_fn = length, values_fill = 0) %>% 
   complete(Date = unique(df1$Date), fill = list(Dummy_England = 0,
         Dummy_Belgium = 0, Dummy_Germany = 0, Dummy_France = 0))

-输出

# A tibble: 3 x 5
   Date Dummy_England Dummy_Belgium Dummy_Germany Dummy_France
  <dbl>         <dbl>         <dbl>         <dbl>        <dbl>
1  16.0             1             1             1            1
2  17.0             0             1             1            0
3  18.0             0             0             0            0

数据

df1 <- structure(list(Date = c(16.03, 16.03, 16.03, 16.03, 16.03, 17.03, 
17.03, 17.03, 17.03, 18.03, 18.03, 18.03), Matches = c("England X Brazil", 
"Belgium X Argentina", "Chile X Japan", "Uruguay X Germany", 
"Italy x France", "South Korea X India", "Germany X France", 
"Poland  X Belgium", "Colombia X Russia", "South Africa X Mexico", 
"China X Japon", "Brazil X Venezuela")), class = "data.frame", row.names = c(NA, 
-12L))

Answer 2

Dummy_england <- tapply(Matches, Date, function(x) {
                      grepl("England", paste(x, collapse = ''))*1
                })