Grepl 并在 R 中提取匹配项

Question

在 R 中我有：

library(tidyverse)
full_names <- tibble(FIRM = c("APPLE INC.", "MICROSOFT CORPORATION", "GOOGLE", "TESLA INC.", "ABBOTT LABORATORIES"), 
                 TICKER = c("AAPL", "MSFT", "GOOGL", "TSLA", "ABT"),
                 ID = c(111, 222, 333, 444, 555)) # a dataset with full names of firms, including some IDs
abbr_names <- c("Abbott", "Apple", "Coca-Cola", "Pepsi, "Microsoft", "Tesla") # a vector with abbreviated names of firms

我想检查缩写名称是否在全名数据集中，如果为真，则随后将 full_names 行匹配到 abbr_names 向量，例如：

    [1]        [2]                    [3]   [4]
[1] Abbott     ABBOTT LABORATORIES    ABT   555
[2] Apple      APPLE INC.             AAPL  111
[3] Microsoft  MICROSOFT CORPORATION  MSFT  222
[4] Tesla      TESLA INC.             TSLA  444

尝试了几个 str_extract 和 grepl 函数，但仍然无法正常工作。

Answer 1

matches <- unlist(sapply(toupper(abbr_names), grep, x = full_names$FIRM, value = TRUE))

这将为您提供一个向量，其中名称作为缩写，公司作为值

names(matches)
# [1] "ABBOTT"    "APPLE"     "MICROSOFT" "TESLA"  
c(firm_matches, use.names = FALSE)
# [1] "ABBOTT LABORATORIES"   "APPLE INC."            "MICROSOFT CORPORATION" "TESLA INC."

有多种方法可以将其组合在一起...拼凑...

从@Oscar的评论中，我们总共两行代码得到了想要的输出：

matches <- unlist(sapply(toupper(abbr_names), grep, x = full_names$FIRM, value = TRUE))
tibble(ABBR_FIRM = names(matches), FIRM = matches) %>% left_join(., full_names, by = "FIRM")

Answer 2

这个怎么样？

full_names$row_num <- 1:nrow(full_names)

do.call(rbind, 
        lapply(abbr_names, 
               function(x){
                 if(sum(grepl(x, full_names$FIRM, ignore.case = TRUE)) > 0){
                   row <- grepl(x, full_names$FIRM, ignore.case = TRUE) %>% 
                     which()} else {row <- 0}
                 data.frame("name" = x,
                            "row_num" = row)})) %>% 
  right_join(full_names, by = "row_num")

Answer 3

另一个选项可能是例如这个...

map_int(abbr_names, ~ {
  idx <- grep(., full_names$FIRM, ignore.case = TRUE)
  if (length(idx) == 0) return(NA) else return(idx)
}) %>% 
  cbind(ABBR = abbr_names, FIRM = full_names$FIRM[.]) %>% 
  as.tibble() %>% 
  left_join(full_names, by = "FIRM") %>%
  complete(FIRM)

# A tibble: 4 x 5
  FIRM                  .     ABBR      TICKER    ID
  <chr>                 <chr> <chr>     <chr>  <dbl>
1 ABBOTT LABORATORIES   5     Abbott    ABT      555
2 APPLE INC.            1     Apple     AAPL     111
3 MICROSOFT CORPORATION 2     Microsoft MSFT     222
4 TESLA INC.            4     Tesla     TSLA     444

只是想还post而已:)

Answer 4

我的建议是，将所有单词的大写或小写设置为大写或小写。更容易与grepl的功能进行比较。

我的代码：

library(tidyverse)

full_names <- tibble(FIRM = c("APPLE INC.", "MICROSOFT CORPORATION", "GOOGLE", "TESLA INC.", "ABBOTT LABORATORIES"), 
                     TICKER = c("AAPL", "MSFT", "GOOGL", "TSLA", "ABT"),
                     ID = c(111, 222, 333, 444, 555)) # a dataset with full names of firms, including some IDs

abbr_names <- c("Abbott", "Apple", "Coca-Cola", "Microsoft", "Tesla") # a vector with abbreviated names of firms

我在这里创建了一个新列，我们要索引 grepl

的 return 的列

full_names$new_column <- NA

然后，我在我们想要在数据框中索引的名称中做了一个循环

for(i in 1:length(abbr_names)){
  search_test <- grepl(tolower(substr(abbr_names[i], 0,4)), tolower(full_names$FIRM))
  position <- grep("TRUE", search_test)
  full_names$new_column[position] <- abbr_names[i]
}

结果是以下数据框：

   FIRM               TICKER    ID  new_column
1 APPLE INC.            AAPL     111 Apple     
2 MICROSOFT CORPORATION MSFT     222 Microsoft 
3 GOOGLE                GOOGL    333 NA        
4 TESLA INC.            TSLA     444 Tesla     
5 ABBOTT LABORATORIES   ABT      555 Abbott

"GOOG" 不在 abbr_names 向量中，所以 return 是 NA

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