Question

我有两个角色列表，第一个是A

> mode(A)
[1] "character"
> length(A)
[1] 300
> dim(A)
[1] 150   2

A就是这样的

> head(A)
     [,1]     [,2]    
[1,] "000001" "601166"
[2,] "000027" "600582"
[3,] "000027" "600783"
[4,] "000027" "601088"
[5,] "000166" "000728"
[6,] "000333" "600519"

第二个列表是B，其中包含1000个元素

> length(B)
[1] 1000
> dim(B)
NULL

B的每个元素也具有以下特征

> mode(B[[1]])
[1] "character"
> length(B[[1]])
[1] 300
> dim(B[[1]])
[1] 150   2

B[[1]]是

> head(B[[1]])
     [,1]     [,2]    
[1,] "000001" "601818"
[2,] "000027" "600362"
[3,] "000027" "600827"
[4,] "000166" "601099"
[5,] "000333" "002304"
[6,] "000333" "601318"

我想计算一对A行的频率，例如

"000001" "601166"

在<{1}}的每个元素中

，例如

我尝试了这段代码，但无法获得结果

B[[1]]

Answer 1

我找到了解决方案，我只需要在代码中添加stringsAsFactors = FALSE。正确的代码是

M<-list()
for (i in 1:1000) {M[[i]]<-match(as.data.frame(t(A), stringsAsFactors = FALSE),as.data.frame(t(B[[i]]), stringsAsFactors = FALSE), nomatch=0)}

任何其他改善它的建议将不胜感激。

匹配R中的两个聊天列表

1 个答案: