Question

我有以下类型的数据框（这是简化的示例）：

id = c("1", "1", "1", "2", "3", "3", "4", "4")
bank = c("a", "b", "c", "b", "b", "c", "a", "c")
df = data.frame(id, bank)
df

  id bank
1  1    a
2  1    b
3  1    c
4  2    b
5  3    b
6  3    c
7  4    a
8  4    c

在此数据框中，您可以看到对于某些ID，存在多个存储体，即id==1，bank=c(a,b,c)。

我想从此数据框中提取的信息是不同库中的id与计数之间的重叠。

例如，对于银行a：银行a有两个人（唯一的ID）：1和4。对于这些人，我想知道他们还拥有哪些其他银行

对于人1：银行b和c
第4个人：银行c

其他银行的总数：3，其中b = 1，c = 2。

所以我想创建一个如下的重叠表作为输出：

bank overlap amount
a    b       1
a    c       2
b    a       1
b    c       2
c    a       2
c    b       2

Answer 1

花点时间获取结果，所以我将其发布。不像罗纳克·沙赫斯（Ronak Shahs）那样性感，但结果相同。

id = c("1", "1", "1", "2", "3", "3", "4", "4")
    bank = c("a", "b", "c", "b", "b", "c", "a", "c")
    df = data.frame(id, bank)

df$bank <- as.character(df$bank)

resultlist <- list()
dflist <- split(df, df$id)

for(i in 1:length(dflist)) {
    if(nrow(dflist[[i]]) < 2) {
        resultlist[[i]] <- data.frame(matrix(nrow = 0, ncol = 2))
    } else {
        resultlist[[i]] <- as.data.frame(t(combn(dflist[[i]]$bank, 2))) 
    }
}
result <-  setNames(data.table(rbindlist(resultlist)), c("bank", "overlap"))

result %>%
    group_by(bank, overlap) %>%
    summarise(amount = n())

  bank  overlap amount
  <fct> <fct>    <int>
1 a     b            1
2 a     c            2
3 b     c            2

Answer 2

选项为full_join

library(dplyr)
full_join(df, df, by = "id") %>%     
      filter(bank.x != bank.y) %>%
      dplyr::count(bank.x, bank.y) %>%
      select(bank = bank.x, overlap = bank.y, amount = n)
# A tibble: 6 x 3
#  bank  overlap amount
#  <fct> <fct>    <int>
#1 a     b            1
#2 a     c            2
#3 b     a            1
#4 b     c            2
#5 c     a            2
#6 c     b            2

Answer 3

我们可以使用data.table：

df = data.frame(id = c("1", "1", "1", "2", "3", "3", "4", "4"), 
                bank = c("a", "b", "c", "b", "b", "c", "a", "c"))

library(data.table)
setDT(df)[, .(bank = rep(bank, (.N-1L):0L), 
              overlap = bank[(sequence((.N-1L):1L) + rep(1:(.N-1L), (.N-1L):1))]), 
                 by=id][, 
                        .N, by=.(bank, overlap)]
#>    bank overlap N
#> 1:    a       b 1
#> 2:    a       c 2
#> 3:    b       c 2
#> 4: <NA>       b 1

^{由reprex package（v0.3.0）于2019-07-01创建}

请注意，b的{{1}}与其他值不重叠。如果您不希望在最终产品中使用它，只需在输出中应用id==2。

Answer 4

您是否需要同时覆盖两个银行？由于在这种情况下，a-> b与b-> a相同。我们可以使用combn并创建一次取2的唯一bank的组合，找出在该组合中发现的length共同的id。

as.data.frame(t(combn(unique(df$bank), 2, function(x) 
     c(x, with(df, length(intersect(id[bank == x[1]], id[bank == x[2]])))))))

#  V1 V2 V3
#1  a  b  1
#2  a  c  2
#3  b  c  2

数据

id = c("1", "1", "1", "2", "3", "3", "4", "4")
bank = c("a", "b", "c", "b", "b", "c", "a", "c")
df = data.frame(id, bank, stringsAsFactors = FALSE)

计算数据框中每个条目的出现次数

4 个答案: