Question

我有一个项目列表和一个搜索词列表，我正在尝试做两件事：

搜索与任何搜索字词匹配的项目，然后返回true iff找到匹配。
对于返回true的所有项目（即匹配），我想还会返回在步骤1中匹配的原始搜索字词。

因此，给出以下数据框：

             items
1             alex
2 alex is a person
3   this is a test
4            false
5    this is cathy

以及以下搜索字词列表：

"alex"      "bob"       "cathy"     "derrick"   "erica"     "ferdinand"

我想创建以下输出：

             items matches original
1             alex    TRUE     alex
2 alex is a person    TRUE     alex
3   this is a test   FALSE     <NA>
4            false   FALSE     <NA>
5    this is cathy    TRUE     cathy

第1步非常简单，但我遇到了第（2）步的问题。要创建“匹配”＃39;如果grepl()中的行位于搜索字词列表中，则使用TRUE创建d$items变量，否则FALSE。grep()。

对于第2步，我的想法是我应该能够在指定value = T时使用items matches original 1 alex TRUE alex 2 alex is a person TRUE alex is a person 3 this is a test FALSE <NA> 4 false FALSE <NA> 5 this is cathy TRUE this is cathy，如下面的代码所示。但是，这会返回错误的值：而不是返回与grep匹配的原始搜索词，而是返回匹配项的值。所以我得到以下输出：

# Dummy data and search terms
d = data.frame(items = c("alex", "alex is a person", "this is a test", "false", "this is cathy"))
searchTerms = c("alex", "bob", "cathy", "derrick", "erica", "ferdinand")

# Return true iff search term is found in items column, not between letters
d$matches = grepl(paste("(^| |[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVQXYZ])", 
    searchTerms, "($| |[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVQXYZ])", sep = "", 
    collapse = "|"), d[,1], ignore.case = TRUE
)

# Subset data
dMatched = d[d$matches==T,]   

# This is where the problem is: return the value that was originally matched with grepl above
dMatched$original = grep(paste("(^| |[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVQXYZ])", 
    searchTerms, "($| |[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVQXYZ])", sep = "", 
    collapse = "|"), dMatched[,1], ignore.case = TRUE, value = TRUE
)


d$original[d$matches==T] = dMatched$original

这是我现在正在使用的代码。任何想法都会非常感激！

{{1}}

Answer 1

感谢Dason的帮助提示！我能够使用regmatches()来解决我的问题。这是我的代码，从最初的问题开始：

# This is where the problem is: return the value that was originally matched with grepl above
m = regexpr(paste("(^| |[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVQXYZ])", 
    searchTerms, "($| |[^abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVQXYZ])", sep = "", 
    collapse = "|"), dMatched[,1], ignore.case = TRUE 
)

dMatched$original = regmatches(dMatched[,1], m)

d$original[d$matches==T] = dMatched$original

这将返回以下输出，这正是我想要的：

             items matches original
1             alex    TRUE     alex
2 alex is a person    TRUE    alex 
3   this is a test   FALSE     <NA>
4            false   FALSE     <NA>
5    this is cathy    TRUE    cathy

Answer 2

不完全符合您的要求，但您可以使用qdap的{{1}}功能来执行此操作。如果您在同一个句子中有两个名字，这将有所帮助：

termco

要获得qdap所需的内容，您可以使用：

library(qdap)
termco(d$items, 1:nrow(d), searchTerms)

## > termco(d$items, 1:nrow(d), searchTerms)
##   nrow(d word.count       alex bob     cathy derrick erica ferdinand
## 1      1          1 1(100.00%)   0         0       0     0         0
## 2      2          4  1(25.00%)   0         0       0     0         0
## 3      3          4          0   0         0       0     0         0
## 4      4          1          0   0         0       0     0         0
## 5      5          3          0   0 1(33.33%)       0     0         0

返回R中grep的原始搜索词

2 个答案: