我搜索了这个问题的答案，发现了类似的答案（计算每个组中的行数，计算R数据框列中变量值的唯一组合，按组计算元素的出现次数），但没有一个解决了我的特定问题。

我有一个包含变量年、ID和code的数据框。每个人都有一个ID，并且在（潜在的）多个年的过程中可以有多个code值。

df = data.frame(ID   = c(1,1,1,1, 2,2,2, 3, 4,4,4,4,4,4,4,4, 5,5,5),
                year = c(2018, 2018, 2020, 2020,
                         2020, 2020, 2020,
                         2011,
                         2019, 2019, 2019, 2019, 2020, 2020, 2020, 2020,
                         2018, 2019, 2020),
                code = c("A", "B", "C", "D",
                         "A", "B", "Q",
                         "G",
                         "A", "B", "Q", "G", "C", "D", "T", "S",
                         "S", "Z", "F")

)

df
   ID year code
1   1 2018    A
2   1 2018    B
3   1 2020    C
4   1 2020    D
5   2 2020    A
6   2 2020    B
7   2 2020    Q
8   3 2011    G
9   4 2019    A
10  4 2019    B
11  4 2019    Q
12  4 2019    G
13  4 2020    C
14  4 2020    D
15  4 2020    T
16  4 2020    S
17  5 2018    S
18  5 2019    Z
19  5 2020    F

我想要的是另一个数据帧，给出两个不同值的code在ID和年的分组中同时出现的次数（在这个例子中，A和B共出现3次，A和C共出现0次），然后我将用于网络分析。

到目前为止，我有这个语法：

1：制作数据的宽版本

library(tidyverse)
wide = df %>% 
        group_by(year, ID) %>% 
        mutate(row = row_number()) %>% 
        ungroup() %>% 
        pivot_wider(
            id_cols = c(ID, year),
            names_from = row, 
            names_prefix = "code_", 
            values_from = code
        )

2：制作节点列表

nodes = distinct(df, code) %>% rowid_to_column("id")

3：制作边缘列表

#edge list needs to be three vars: source, dest, and weight
# source and dest are simply code names that (potentially) co-occur in the same year for an ID
# weight is the number of times the codes co-occurred in the same year for each ID.

#all combinations of two codes
edges = combn(x = nodes$code, m = 2 ) %>% 
    t() %>% 
    as.data.frame()

colnames(edges) = c("source", "dest")
edges$weight = NA_integer_


#oh, no! a for() loop! a coder's last ditch effort to make something work
for(i in 1:nrow(edges)){
    
    source = edges$source[i]
    dest = edges$dest[i]
    

    #get the cases with the first code of interest
    temp = df %>% 
        filter( code == source ) %>% 
        select(ID, year)
    
    #get the other codes that occurred for that ID in that year
    temp = left_join(temp, 
                     wide, 
                     by = c("ID", "year"))
    
    
    #convert to a logical showing if the other codes are the one I want
    temp = temp %>% mutate_at(vars(starts_with("code_")),
                            function(x){ x == dest }
    ) 
    
    #sum the number of times `source` and `dest` co-occurred
    temp$dest = temp %>% select(starts_with("code_")) %>% rowSums(., na.rm=TRUE)
    edges$weight[i] = sum(temp$dest, na.rm = TRUE)
    
}

编辑以添加结果：

结果：

edges
   source dest weight
1       A    B      3
2       A    C      0
3       A    D      0
4       A    Q      2
5       A    G      1
6       A    T      0
7       A    S      0
8       A    Z      0
9       A    F      0
10      B    C      0
11      B    D      0
12      B    Q      2
13      B    G      1
14      B    T      0
15      B    S      0
16      B    Z      0
17      B    F      0
18      C    D      2
19      C    Q      0
20      C    G      0
21      C    T      1
22      C    S      1
23      C    Z      0
24      C    F      0
25      D    Q      0
26      D    G      0
27      D    T      1
28      D    S      1
29      D    Z      0
30      D    F      0
31      Q    G      1
32      Q    T      0
33      Q    S      0
34      Q    Z      0
35      Q    F      0
36      G    T      0
37      G    S      0
38      G    Z      0
39      G    F      0
40      T    S      1
41      T    Z      0
42      T    F      0
43      S    Z      0
44      S    F      0
45      Z    F      0

这给了我我想要的（一个数据帧显示A和B共出现3次，A和C共出现0次，A和D共出现0次，A和G共出现1次，A和Q共出现2次，等等…）。所以这是有效的，但即使对于这个小例子也需要一两秒钟。我真正的数据集是~3,000,000次观察。我让它运行了一段时间，但停止了它，却发现它完成了~1%。

有没有更好/更快的方法来做到这一点？