R中不精确字符串的连接

我想把两张桌子合在一起。。但是，我希望加入的数据并不完全匹配。。加入NFL球员名单

数据集如下

> dput(att75a)
structure(list(rusher_player_name = c("A.Ekeler", "A.Jones", 
"A.Kamara", "A.Mattison", "A.Peterson", "B.Hill"), mean_epa = c(-0.110459963350783, 
0.0334332018597805, -0.119488111742492, -0.155261835310445, -0.123485646124451, 
-0.0689611296359916), success_rate = c(0.357664233576642, 0.40495867768595, 
0.401129943502825, 0.283018867924528, 0.322727272727273, 0.35
), plays = c(137L, 242L, 177L, 106L, 220L, 80L)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -6L))

> dput(rb2019capa)
structure(list(rusher_player_name = c("Aaron Jones", "Adrian Peterson", 
"Alexander Mattison", "Alvin Kamara", "Austin Ekeler", "Brian Hill"
), Team = c("Packers", "Redskins", "Vikings", "Saints", "Chargers", 
"Falcons"), `Salary Cap Value` = c(695487, 1780000, 700545, 1050693, 
646668, 645000), `Cash Spent` = c(645000, 2530000, 1317180, 807500, 
645000, 645000)), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA, 
-6L))

例如，我正试图加入A.Mattison关于Alexander Mattison的节目。。等等

我尝试了stringdist和fuzzyjoin，但无法解决我的问题

请考虑…将每个数据集的head（）浓缩为每个问题提问指南。。原始数据集的长度为51 OB。和168个obs。。。这会影响连接的执行方式吗

清理这些名字最好的方法是什么

---

感谢您的时间。

将点替换为%生成SQL模式，并根据与之匹配的内容加入

library(sqldf)

sqldf("select * 
  from att75a a 
  left join rb2019capa r 
    on r.rusher_player_name like replace(a.rusher_player_name, '.', '%')")

给予：

  rusher_player_name    mean_epa success_rate plays rusher_player_name..5
1           A.Ekeler -0.11045996    0.3576642   137         Austin Ekeler
2            A.Jones  0.03343320    0.4049587   242           Aaron Jones
3           A.Kamara -0.11948811    0.4011299   177          Alvin Kamara
4         A.Mattison -0.15526184    0.2830189   106    Alexander Mattison
5         A.Peterson -0.12348565    0.3227273   220       Adrian Peterson
6             B.Hill -0.06896113    0.3500000    80            Brian Hill
      Team Salary Cap Value Cash Spent
1 Chargers           646668     645000
2  Packers           695487     645000
3   Saints          1050693     807500
4  Vikings           700545    1317180
5 Redskins          1780000    2530000
6  Falcons           645000     645000

---

使用

sub

将名字替换为首字母

library(dplyr)

rb2019capa %>%
  mutate(rusher_player_name=
         sub("^([A-Z])\\S+\\s([A-Za-z].*)$", "\\1.\\2", rusher_player_name)) %>%
  inner_join(att75a, by="rusher_player_name") # or left_join (up to you)

---

#一个tible:6 x 7
rusher\u球员\u姓名团队`工资上限值'`现金支出`平均epa成功率\u比赛
1 A.琼斯封隔器公司695487 645000 0.0334 0.405 242
2 A.彼得森红人队1780000 2530000-0.123 0.323 220
3 A.马蒂森维京人700545 1317180-0.155 0.283 106
4 A.卡马拉圣徒1050693 807500-0.119 0.401177
5 A.Ekeler充电器646668 645000-0.110 0.358 137
6 B.山鹰645000 645000-0.0690 0.35 80

将

Alexander Mattison

转换为

a.Mattison

的代码是否可以执行此任务？通过将名字替换为首字母来修改第二个df。只要没有重复项，就可以将其与第一个df精确地连接起来。谢谢，我将使用字符串来尝试此操作。那么您尝试了什么？在另一篇帖子（不应重复发布）中，有人建议我使用sub<代码>子（（。\\\.（.*），“\\1\\2”，rb2019capa$rusher\U player\U name）。。匹配名称后加入成功。。。当我稍后回到我的环境中时，我将尝试下面的答案。。。

# A tibble: 6 x 7
  rusher_player_name Team     `Salary Cap Value` `Cash Spent` mean_epa success_rate plays
  <chr>              <chr>                 <dbl>        <dbl>    <dbl>        <dbl> <int>
1 A.Jones            Packers              695487       645000   0.0334        0.405   242
2 A.Peterson         Redskins            1780000      2530000  -0.123         0.323   220
3 A.Mattison         Vikings              700545      1317180  -0.155         0.283   106
4 A.Kamara           Saints              1050693       807500  -0.119         0.401   177
5 A.Ekeler           Chargers             646668       645000  -0.110         0.358   137
6 B.Hill             Falcons              645000       645000  -0.0690        0.35     80