r中的自设计矩阵

以下是我的数据：

sub <- paste ("s", 1:6, sep = "")
mark1a <- c("A", "A", "B", "d1", "A", 2)
mark1b <- c("A", "B", "d1", 2, "d1", "A")
myd <- data.frame (sub, mark1a, mark1b)
myd 
     sub mark1a mark1b
1  s1      A      A
2  s2      A      B
3  s3      B     d1
4  s4     d1      2
5  s5      A     d1
6  s6      2      A

---

sub您可以尝试以下方法：

cbind(myd, t(apply(myd, 1, function(x) sapply(unique(unlist(myd[, 2:3])), function(y) sum(x==y)))))
1  s1      A      A 2 0  0 0
2  s2      A      B 1 1  0 0
3  s3      B     d1 0 1  1 0
4  s4     d1      2 0 0  1 1
5  s5      A     d1 1 0  1 0
6  s6      2      A 1 0  0 1

我想说@jmsigner的解决方案是单行程序的解决方案，但我通常会被那些嵌套的
apply
（及其相关）解决方案弄糊涂

这里有一个类似的解决方案：

# Identify all the levels in `mark1a` and `mark1b`
mydLevels = unique(c(levels(myd$mark1a), levels(myd$mark1b)))
# Use these levels and an anonymous function with `lapply`
temp = data.frame(lapply(mydLevels, 
                         function(x) rowSums(myd[-1] == x)+0))
colnames(temp) = mydLevels
# This gives you the correct output, but not in the order
# that you have in your original question.
cbind(myd, temp)
#   sub mark1a mark1b 2 A B d1
# 1  s1      A      A 0 2 0  0
# 2  s2      A      B 0 1 1  0
# 3  s3      B     d1 0 0 1  1
# 4  s4     d1      2 1 0 0  1
# 5  s5      A     d1 0 1 0  1
# 6  s6      2      A 1 1 0  0

首先，确保
mark1a
和
mark1b
列共享相同的级别：

all.levels <- levels(myd["mark1a", "mark1b"])
levels(myd$mark1a) <- all.levels
levels(myd$mark1b) <- all.levels

此外，您可能对
paste0
函数感兴趣。使用“表格”结果进行矩阵加法。我喜欢。我怎么知道级别的顺序（这里没有显示，是按字母顺序排列的吗
library(plyr)
cbind(myd, ddply(myd, "sub", function(x)table(x$mark1a) + table(x$mark1b))[,-1])
#   sub mark1a mark1b 2 A B d1
# 1  s1      A      A 0 2 0  0
# 2  s2      A      B 0 1 1  0
# 3  s3      B     d1 0 0 1  1
# 4  s4     d1      2 1 0 0  1
# 5  s5      A     d1 0 1 0  1
# 6  s6      2      A 1 1 0  0