找到R中匹配元素的数量

Question

sorry for the noob question ! I am trying to calculate the number of elements that matched in basket x with basket y. I have the following data:

user_id basket.x basket.y
1         1,2,3    2,3,4
2         5,6,7    1,2,7

I have tried the following loop, but it doesnot work

df["total"] <- 0
df["TP"] <- 0
for(i in 1:nrow(df)){
 for(j in 1:nrow(df)){
  if(all(df$basket.x[i] %in% df$basket.y[j])){
     df$total <- total + 1
     df$TP <- TP + 1
  }
 }
}

And returns this:

user_id basket.x basket.y   total TP
1         1,2,3    2,3,4     0    0
2         5,6,7    1,2,7     0    0

However, the desired result is:

user_id basket.x basket.y   total TP
1         1,2,3    2,3,4     3    2
2         5,6,7    1,2,7     3    1

Could anyone point me please where i am made mistake ? Thank you

Running the dput() :

structure(list(user_id = c(2957L, 7306L, 10219L, 11290L, 13222L, 
13554L), basket.x = c("13870,22963,1158,18362"),basket.y = 
c("24852,432,47626,33647,6015,1158,24852,24852,24852")
), row.names = c(NA, 
6L), class = "data.frame")

Answer 1

As noted by @JohnColeman, there is something wrong with your dput so I am using a combination of that and your original example.

df = structure(list(user_id = c(2957L, 7306L, 10219L), 
basket.x = c("13870,22963,1158,18362", "1,2,3", "5,6,7"),
basket.y = c("24852,432,47626,33647,6015,1158,24852,24852,24852",
"2,3,4", "1,2,7")
), row.names = c(1L,2L,3L), class = "data.frame")
df
  user_id               basket.x
1    2957 13870,22963,1158,18362
2    7306                  1,2,3
3   10219                  5,6,7
                                           basket.y
1 24852,432,47626,33647,6015,1158,24852,24852,24852
2                                             2,3,4
3                                             1,2,7

Using this data, we can get the individual elements of the lists using strsplit . Once we have the elements, we can use intersect to find the elements that are in both basket.x and basket.y . To get how many elements the two baskets share, we can just take the length of the intersection. Of course, we need to apply this across all of the rows of df . Putting this together, we get

sapply(1:nrow(df), function(i) 
    length(intersect(strsplit(df$basket.x, ",")[[i]],
            strsplit(df$basket.y, ",")[[i]])))
[1] 1 2 1

Edit Thanks to @thelatemail for noticing that the way I wrote this is very inefficient. Better is:

sapply(1:nrow(df), function(i) 
    length(intersect(unlist(strsplit(df$basket.x[[i]], ",")),
            unlist(strsplit(df$basket.y[[i]], ",")))))

Answer 2

A variation on @G5W's answer would be possible through Map to replace (well, hide) the loop over each row index:

spl <- unname(lapply(df[-1], strsplit, ","))
lengths(do.call(Map, c(intersect, spl)))
#[1] 1 2 1

Although you have to save the intermediate spl , this should be significantly faster if you're dealing with larger datasets.