是否有一个R函数用于制作单选题的列联表？

我只需要为一个有多项选择的问题做一个交叉表

我有一个调查结果。这样做的目的是让了解品牌的人也了解其他品牌

假设我们有一个包含答案的输入表：

Id  Brand1  Brand2  Brand3  Brand4
1   1       1       0       1
2   0       0       1       1
3   1       1       0       0
4   1       0       0       1
5   0       0       1       1
6   0       0       1       0
7   1       1       0       1

我们制作一个交叉表，列出每个品牌的答案总和：

        Brand1  Brand2  Brand3  Brand4
Brand1  4       3       0       3
Brand2  3       3       0       2
Brand3  0       0       3       2
Brand4  3       2       2       5

        Brand1  Brand2  Brand3  Brand4
Brand1  100%    100%    0%      60%
Brand2  75%     100%    0%      40%
Brand3  0%      0%      100%    40%
Brand4  75%     67%     67%     100%

然后从所有了解每个品牌的人中计算列百分比：

        Brand1  Brand2  Brand3  Brand4
Brand1  4       3       0       3
Brand2  3       3       0       2
Brand3  0       0       3       2
Brand4  3       2       2       5

        Brand1  Brand2  Brand3  Brand4
Brand1  100%    100%    0%      60%
Brand2  75%     100%    0%      40%
Brand3  0%      0%      100%    40%
Brand4  75%     67%     67%     100%

---

数据

d = structure(list(Id = 1:7,
                   Brand1 = c(1L, 0L, 1L, 1L, 0L, 0L, 1L),
                   Brand2 = c(1L, 0L, 1L, 0L, 0L, 0L, 1L),
                   Brand3 = c(0L, 1L, 0L, 0L, 1L, 1L, 0L),
                   Brand4 = c(1L, 1L, 0L, 1L, 1L, 0L, 1L)),
              class = "data.frame",
              row.names = c(NA, -7L))

---

数据

d = structure(list(Id = 1:7,
                   Brand1 = c(1L, 0L, 1L, 1L, 0L, 0L, 1L),
                   Brand2 = c(1L, 0L, 1L, 0L, 0L, 0L, 1L),
                   Brand3 = c(0L, 1L, 0L, 0L, 1L, 1L, 0L),
                   Brand4 = c(1L, 1L, 0L, 1L, 1L, 0L, 1L)),
              class = "data.frame",
              row.names = c(NA, -7L))

---

我们可以用矩阵乘法来实现。使用d.b.的数据：

dmat = as.matrix(d[-1])
count = t(dmat) %*% dmat
count
#        Brand1 Brand2 Brand3 Brand4
# Brand1      4      3      0      3
# Brand2      3      3      0      2
# Brand3      0      0      3      2
# Brand4      3      2      2      5
apply(count, 2, function(x) x / max(x))
#        Brand1    Brand2    Brand3 Brand4
# Brand1   1.00 1.0000000 0.0000000    0.6
# Brand2   0.75 1.0000000 0.0000000    0.4
# Brand3   0.00 0.0000000 1.0000000    0.4
# Brand4   0.75 0.6666667 0.6666667    1.0

---

我们可以用矩阵乘法来实现。使用d.b.的数据：

dmat = as.matrix(d[-1])
count = t(dmat) %*% dmat
count
#        Brand1 Brand2 Brand3 Brand4
# Brand1      4      3      0      3
# Brand2      3      3      0      2
# Brand3      0      0      3      2
# Brand4      3      2      2      5
apply(count, 2, function(x) x / max(x))
#        Brand1    Brand2    Brand3 Brand4
# Brand1   1.00 1.0000000 0.0000000    0.6
# Brand2   0.75 1.0000000 0.0000000    0.4
# Brand3   0.00 0.0000000 1.0000000    0.4
# Brand4   0.75 0.6666667 0.6666667    1.0