矢量化 data.table，如 grepl 或类似内容，用于大数据字符串比较

问：

我需要检查一列中的字符串是否包含来自另一列同一行的相应（数字）值，对于所有行。

如果我只检查单个模式的字符串，这将很简单，使用 data.table 的 or .但是，我的模式值对于每一行都是不同的。likegrepl

这里有一个有点相关的问题，但与这个问题不同的是，我需要创建一个逻辑标志来指示模式是否存在。

假设这是我的数据集;

DT <- structure(list(category = c("administration", "nurse practitioner", 
                                  "trucking", "administration", "warehousing", "warehousing", "trucking", 
                                  "nurse practitioner", "nurse practitioner"), industry = c("admin", 
                                                                                            "truck", "truck", "admin", "nurse", "admin", "truck", "nurse", 
                                                                                            "truck")), .Names = c("category", "industry"), class = "data.frame", row.names = c(NA, 
                                                                                                                                                                               -9L))
setDT(DT)
> DT
             category industry
1:     administration    admin
2: nurse practitioner    truck
3:           trucking    truck
4:     administration    admin
5:        warehousing    nurse
6:        warehousing    admin
7:           trucking    truck
8: nurse practitioner    nurse
9: nurse practitioner    truck

我想要的结果将是这样的向量：

> DT
   matches
1: TRUE
2: FALSE
3: TRUE
4: TRUE
5: FALSE
6: FALSE
7: TRUE
8: TRUE
9: FALSE

当然，1 和 0 与 TRUE 和 FALSE 一样好。

以下是我尝试过的一些不起作用的方法：

apply(DT,1,grepl, pattern = DT[,2], x = DT[,1])
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE

> apply(DT,1,grepl, pattern = DT[,1], x = DT[,2])
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE

> grepl(DT[,2], DT[,1])
[1] FALSE

> DT[Vectorize(grepl)(industry, category, fixed = TRUE)]
             category industry
1:     administration    admin
2:           trucking    truck
3:     administration    admin
4:           trucking    truck
5: nurse practitioner    nurse

> DT[stringi::stri_detect_fixed(category, industry)]
             category industry
1:     administration    admin
2:           trucking    truck
3:     administration    admin
4:           trucking    truck
5: nurse practitioner    nurse

> for(i in 1:nrow(DT)){print(grepl(DT[i,2], DT[i,1]))}
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE

> for(i in 1:nrow(DT)){print(grepl(DT[i,2], DT[i,1], fixed = T))}
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE
[1] FALSE

> DT[category %like% industry]
         category industry
1: administration    admin
2: administration    admin
Warning message:
In grepl(pattern, vector) :
  argument 'pattern' has length > 1 and only the first element will be used