R 如果一列包含与另一列匹配的单词

假设

A

和

B

是数据集中的列，我想开发一个模糊匹配逻辑-如果A列中至少有一个单词与B列中的一个单词匹配，除了单词“bank”和“of”，我们在新列中分配

1

，如果有0个匹配，我们分配

0

。我想在R做这个

A                          B
BANK OF AMERICA         CHASE BANK
BANK OF AMERICA         BANK OF AMERICA, N.A.
BANK OF HOPE            HOPE BANK
T.D BANK                CHASE BANK

预期产量

A                         B                        C
BANK OF AMERICA         CHASE BANK                 0
BANK OF AMERICA         BANK OF AMERICA, N.A       1
BANK OF HOPE            HOPE BANK                  1
T.D. BANK               CHASE BANK                 0

---

我相信regex和apply的结合在这里很有效

> df <- data.frame(A = c('BANK OF AMERICA', 'BANK OF AMERICA', 'BANK OF HOPE', 'T.D BANK'),
                 B = c('CHASE BANK', 'BANK OF AMERICA, N.A.', 'HOPE BANK', 'CHASE BANK'),
                 stringsAsFactors = FALSE)

> f <- function(x) {
  left <- strsplit(x[1], "(BANK OF\\s|\\s|,|\\sBANK)")[[1]]
  right <- strsplit(x[2],  "(BANK OF\\s|\\s|,|\\sBANK)")[[1]]
  ans <- left %in% right
  as.integer(all(ans[!(left %in% "")]))
}

> df$C <- apply(df, 1, f)
> df
                A                     B C
1 BANK OF AMERICA            CHASE BANK 0
2 BANK OF AMERICA BANK OF AMERICA, N.A. 1
3    BANK OF HOPE             HOPE BANK 1
4        T.D BANK            CHASE BANK 0

---

>df f以下基本R选项可能会有所帮助

df$C <- +do.call(
  function(...) mapply(function(...) any(!intersect(...) %in% c("BANK","OF")),...),
  Map(function(x) strsplit(x,"[[:punct:][:blank:]]",perl = TRUE), df, USE.NAMES = FALSE)
)

这里是另一个选项-使用
dplyr
和
stringr

df <- data.frame(A = c(rep("BANK OF AMERICA", 2), "BANK OF HOPE", "T.D BANK"),
                 B = c("CHASE BANK", "BANK OF AMERICA, N.A.", "HOPE BANK", "CHASE BANK"),
                 stringsAsFactors = FALSE) 


df <- df %>% 
  mutate(C = str_remove_all(B, c("BANK|OF|,")), #remove stopwords
         C = str_trim(C), #remove whitespace from start/end
         C = str_replace_all(C, "  ", ""), #remove double whitespaces
         C = str_replace_all(C, " ", "|")) %>% #replace whitespace with |
  mutate(D = as.numeric(str_detect(A, C))) %>% 
  select(A, B, D)

                A                     B D
1 BANK OF AMERICA            CHASE BANK 0
2 BANK OF AMERICA BANK OF AMERICA, N.A. 1
3    BANK OF HOPE             HOPE BANK 1
4        T.D BANK            CHASE BANK 0

df%#将空格替换为|
变异（D=as.numeric（str_detect（A，C）））%>%
选择（A、B、D）
A、B、D
1美国银行大通银行0
2美国银行美国银行，N.A.1
3希望银行1号银行
4T.D银行大通银行0
请在您的问题中始终以R代码的形式提供您的数据（以帮助我们快速得出可能的答案）+显示示例的预期结果（我想这在您的情况下可能微不足道-只有第二行获得
1
-如果我也忽略停止词“of”），是的，在我的情况下，只有第二行获得1。我将编辑我的帖子。你想处理多少行（粗略估计）？如果您想处理数百万行，并给出迄今为止的答案，那么性能可能至关重要（尽管很难想象您正在处理这么多不同的法律实体）…实际上并没有那么多。大约10k rowsOK，因此请用绿色勾选最佳工作答案（如果有；-）。谢谢！
df <- data.frame(A = c(rep("BANK OF AMERICA", 2), "BANK OF HOPE", "T.D BANK"),
                 B = c("CHASE BANK", "BANK OF AMERICA, N.A.", "HOPE BANK", "CHASE BANK"),
                 stringsAsFactors = FALSE) 


df <- df %>% 
  mutate(C = str_remove_all(B, c("BANK|OF|,")), #remove stopwords
         C = str_trim(C), #remove whitespace from start/end
         C = str_replace_all(C, "  ", ""), #remove double whitespaces
         C = str_replace_all(C, " ", "|")) %>% #replace whitespace with |
  mutate(D = as.numeric(str_detect(A, C))) %>% 
  select(A, B, D)

                A                     B D
1 BANK OF AMERICA            CHASE BANK 0
2 BANK OF AMERICA BANK OF AMERICA, N.A. 1
3    BANK OF HOPE             HOPE BANK 1
4        T.D BANK            CHASE BANK 0