R 通过切换1和0局部搜索获得矩阵

让我们从下面的矩阵开始

M <- matrix(c(0,0,0,1,0,0,1,1,
          0,0,1,0,0,1,1,0,
          0,0,0,0,0,1,1,1,
          0,0,0,1,1,0,1,0,
          0,0,0,1,1,1,0,0,
          0,0,1,0,1,0,0,1),nrow = 8,ncol = 6)


      [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    0    0    0    0    0    0
[2,]    0    0    0    0    0    0
[3,]    0    1    0    0    0    1
[4,]    1    0    0    1    1    0
[5,]    0    0    0    1    1    1
[6,]    0    1    1    0    1    0
[7,]    1    1    1    1    0    0
[8,]    1    0    1    0    0    1

---

M这里有一个解决方案。你可以把整个事情包装成一个函数。它生成一个矩阵列表，
results
，其中
results[[i]]
是一个矩阵列表，其第
i
列已切换

    column_switcher = function(x) {
        ones = which(x == 1)
        zeros = which(x == 0)
        results = matrix(rep(x, length(ones) * length(zeros)), nrow = length(x))
        counter = 1
        for (one in ones) {
            for (zero in zeros) {
                results[one, counter] = 0
                results[zero, counter] = 1
                counter = counter + 1
            }
        }
        return(results)
    }

    switched = lapply(1:ncol(M), function(col) column_switcher(M[, col]))
    results = lapply(seq_along(switched), function(m_col) {
        lapply(1:ncol(switched[[m_col]]), function(i) {
            M[, m_col] = switched[[m_col]][, i]
            return(M)
            })
        })

    results[[1]]
    # [[1]]
    #      [,1] [,2] [,3] [,4] [,5] [,6]
    # [1,]    1    0    0    0    0    0
    # [2,]    0    0    0    0    0    0
    # [3,]    0    1    0    0    0    1
    # [4,]    0    0    0    1    1    0
    # [5,]    0    0    0    1    1    1
    # [6,]    0    1    1    0    1    0
    # [7,]    1    1    1    1    0    0
    # [8,]    1    0    1    0    0    1
    # 
    # [[2]]
    #      [,1] [,2] [,3] [,4] [,5] [,6]
    # [1,]    0    0    0    0    0    0
    # [2,]    1    0    0    0    0    0
    # [3,]    0    1    0    0    0    1
    # [4,]    0    0    0    1    1    0
    # [5,]    0    0    0    1    1    1
    # [6,]    0    1    1    0    1    0
    # [7,]    1    1    1    1    0    0
    # [8,]    1    0    1    0    0    1
    # 
    # ...

检查列表的长度和子列表的长度，它们都在那里

    length(results)
    # [1] 6
    lengths(results)
    # [1] 15 15 15 15 15 15

仔细检查，是否要交换第1列（第1-6行）中的1和0，然后将同一列交换给后面的行，然后分别交换给每一列？看起来您的最终目标是同时交换所有1和0。同时交换所有1和0的一种快速方法是：M@MatthewBorg，我想对单个列执行。我只需为

循环编写一个（嵌套的）

。。。逻辑似乎很简单
for（其中一个（x==1））{for（其中零个（x==0））{…}
。把它放在一个对单个列进行操作的函数中，然后你可以
将该函数应用到每一列。@Gregor我不知道该怎么做。