如何在 R 中匹配至少有一个共同单词的字符串？

Question

数据框 1 示例：

NAME;                        CITY; STATE;  SURNAME;
Maria Antonia Sousa          A     X       Antonia Sousa
Josep Oliveira Carlos        A     X       Oliveira Carlos 
Jose Mario Augusto Farias    B     Y       Augusto Farias
Andre Gois Lucas             B     Y       Gois Lucas

我想在第二个数据框中创建一个列familyDummy，以指示与第一个数据框中的姓氏共享至少一个姓氏的人，但前提是他们来自同一城市和州。同一个人可能会出现在两个 df 中，我不想将他们视为家人。 df 的长度不同。

数据框 2 示例：

NAME;                    CITY;  STATE;    SURNAME;          familyDummy;
Maria Antonia Sousa      A      X         Antonia Sousa     0
Angela Oliveira Santos   A      X         Oliveira Santos   1
Fabio Silva Carlos       B      Y         Silva Carlos      0
Luan Gois Lucas          B      Y         Gois Lucas        1

感谢您的帮助。

Answer 1

这里有一个解决您的问题的解决方案。该解决方案首先将 df1 和 df2 的 SURNAME 列分为两个姓氏，以检查单个匹配项（请参阅 df1_bis 和 df2_bis）。然后，它循环遍历 df2 的所有条目以检查在 df1 中是否不找到确切的 NAME，以及是否在 df1 中找到 df2 的每个条目的至少一个姓氏。如果满足这两个条件，它会检查这些条目的CITY 和STATE 是否在df1 和df2 中匹配。如果是这种情况，则将 familyDummy 分配为 1，如果不是，则分配为 0。

library(tidyverse)

# Your data
df1 <-structure(list(NAME = c("Maria Antonia Sousa", "Josep Oliveira Carlos", 
"Jose Mario Augusto Farias", "Andre Gois Lucas"), CITY = c("A", 
"A", "B", "B"), STATE = c("X", "X", "Y", "Y"), SURNAME = c("Antonia Sousa", 
"Oliveira Carlos", "Augusto Farias", "Gois Lucas")), class = "data.frame", row.names = c(NA, 
-4L))

df2 <- structure(list(NAME = c("Maria Antonia Sousa", "Angela Oliveira Santos", 
"Fabio Silva Carlos", "Luan Gois Lucas"), CITY = c("A", "A", 
"B", "B"), STATE = c("X", "X", "Y", "Y"), SURNAME = c("Antonia Sousa", 
"Oliveira Santos", "Silva Carlos", "Gois Lucas"), familyDummy = c(0L, 
1L, 0L, 1L)), class = "data.frame", row.names = c(NA, -4L))

# Divide surnames
df1_bis <- df1 %>%
  # Divide SURNAME into two surnames to check independently for each single surname
  mutate(surname1 = str_extract(SURNAME,"[A-z]+(?=\\s)"),
         surname2 = str_extract(SURNAME,"(?<=\\s)[A-z]+"))

df2_bis <- df2 %>%
  # Divide SURNAME into two surnames to check independently for each single surname
  mutate(surname1 = str_extract(SURNAME,"[A-z]+(?=\\s)"),
         surname2 = str_extract(SURNAME,"(?<=\\s)[A-z]+"))

df2 %>%
# Add the result as another column
  # Use map to cycle over each row in df2
  mutate(familyDummy = map(1:nrow(df2_bis), function(i){
    # Check if the same NAME is in df1 and df2, if it appears assign 0, if not, 1.
    dif_name = str_detect(df2_bis$NAME[i], df1_bis$NAME, negate = T)

    # Check if any of the surnames of df1 is in df2. If it appears, assign 1, if not 0,
    surname_same = ifelse(str_detect(df2_bis$surname1[i], df1_bis$surname1) | str_detect(df2_bis$surname1[i], df1_bis$surname2) | str_detect(df2_bis$surname2[i], df1_bis$surname1) | str_detect(df2_bis$surname2[i], df1_bis$surname2), 1, 0)

    # Get the indices in df1 of the cases that meet the two latter criteria
    temp <- which(dif_name == 1 & surname_same == 1)

    # Check if there are cases where at least one entry matches the two criteria
    if(length(temp) >= 1){
      # Check if city and state in df1 matches that in df2
      # I used %in% instead of == because there might be more than 1 match
      familyDummy = ifelse(df2_bis$CITY[i] %in% df1_bis$CITY[temp] & df2_bis$STATE[i] %in% df1_bis$STATE[temp], 1, 0)
      }else{ # If no case match the previous two criteria return 0
        familyDummy = 0
        }
    return(familyDummy)
    }))

#                    NAME CITY STATE         SURNAME familyDummy
#1    Maria Antonia Sousa    A     X   Antonia Sousa           0
#2 Angela Oliveira Santos    A     X Oliveira Santos           1
#3     Fabio Silva Carlos    B     Y    Silva Carlos           0
#4        Luan Gois Lucas    B     Y      Gois Lucas           1