根据另一个数据帧中的列在一个数据帧中应用正则表达式 [英] apply regexp in one data frame based on the column in another data frame

问题描述

我有两个数据帧---表A是模式表，表B是名称表.我想对表B进行子集化，使其与表a中的模式匹配.

I have two data frames --- table A is the pattern table, and table B is the name table. I want to subset table B, where it matches the pattern in table a.

A <- data.frame(pattern = c("aa", "bb", "cc", "dd"))
B <- data.frame(name = "aa1", "bb1", "abc", "def" ,"ddd")

我试图做一个for循环，看起来像:

I'm trying to do a for loop looks like:

for (i in 1:nrow(A)){
for (j in 1:nrow(B)){
DT <- data.frame(grep(A$pattern[i], B$name[j], ignore.case = T, value = T))
}}

我希望我的结果表DT仅包含aa1，bb1和ddd

And I want my resulting table DTto only contains aa1, bb1, and ddd

但这太慢了.我只是想知道是否还有更有效的方法?多谢！

But it's super slow. I just wondering if there's any more efficient way to do it? Many thans!

推荐答案

在您的示例输入数据中似乎有一个小错误(未正确声明缺少B$name，并且两个data.frame对象都需要包含stringsAsFactors = F ):

it appears there's a slight error in your sample input data (missing B$name is not properly declared and need to include stringsAsFactors = F for both data.frame objects):

> A <- data.frame(pattern = c("aa", "bb", "cc", "dd"), stringsAsFactors = F)
> B <- data.frame(name = c("aa1", "bb1", "abc", "def" ,"ddd"), stringsAsFactors = F)

代码

# using sapply with grepl
> indices <- sapply(1:nrow(A), function(z) grepl(A$pattern[z], B$name[z]))
> indices
[1]  TRUE  TRUE FALSE FALSE

> B[indices, ]
[1] "aa1" "bb1" "ddd"

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