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r - 聚合随机采样的列以迭代更大的 bin 大小

问题描述

我有一个像这样的矩阵：

mat <- matrix(c(1,0,0,0,0,0,1,0,0,0,0,0,0,0,2,0,
                2,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,
                0,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,
                0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
                0,0,0,0,1,0,0,1,0,1,1,0,0,1,0,1,
                1,1,0,0,0,0,0,0,1,0,1,2,1,0,0,0), nrow=16, ncol=6)
dimnames(mat) <- list(c("a", "c", "f", "h", "i", "j", "l", "m",
                        "p", "q", "s", "t", "u", "v","x", "z"), 
                      c("1", "2", "3", "4", "5", "6"))

我想对列进行分组或分箱，然后为每个组聚合数据。对大小为 x 的 bin 重复采样 n 次。对于 x+1 的 bin 大小，将重复此过程。

对于第一次迭代，两个随机列被分箱。我想在不替换的情况下进行采样，这样两列的组合不会被采样两次（但是，如果一列与不同的列配对，则可以使用两次）。聚合被定义为计算合并列的行总和。聚合结果将作为该 bin 大小的结果矩阵中的新列添加。结果矩阵中的列数将限制为随机采样的 bin 数。

垃圾箱大小继续变得越来越大。对于下一次迭代，bin 大小增加到 3，以便聚合 3 个随机选择的列。聚合数据将被放入不同的结果矩阵中。这个过程将一直持续到 bin 达到数据帧的大小为止，在这种情况下，重新采样是不可能的。所有结果矩阵将被放入矩阵列表中。

下面是resultList给定上述矩阵的前两个 bin 大小的预期结果。

# Bin size =2 
# The randomly sampled columns are columns 1&2, 2&3, 3&4, 4&5, 5&6. 
mat1 <- matrix(c(3,0,0,0,1,0,1,0,0,0,0,0,0,0,2,0,
                 2,0,1,1,2,0,0,0,0,0,0,0,0,0,1,0,
                 0,1,1,1,1,1,0,0,0,0,0,0,0,0,0,0,
                 0,1,0,0,1,1,0,1,0,1,1,0,0,1,0,1,
                 1,1,0,0,1,0,0,1,1,1,2,2,1,1,0,1), nrow=16)
dimnames(mat1) <- list(c("a", "c", "f", "h", "i", "j", "l", "m", 
                         "p", "q", "s", "t", "u", "v","x", "z"), 
                       c("1_2", "2_3", "3_4", "4_5", "5_6"))

# Bin size= 3
# The randomly selected columns to be joined are columns 1,2&3, 
# 2,3&4, 3,4&5, 4,5&6. 
mat2 <- matrix(c(3,0,1,1,2,0,1,0,0,0,0,0,0,0,3,0,
                 2,1,1,1,2,1,0,0,0,0,0,0,0,0,1,0,
                 0,1,1,1,2,1,0,1,0,1,1,0,0,1,0,1,
                 1,2,0,0,1,1,0,1,1,1,2,2,1,1,0,1), nrow=16)
dimnames(mat2) <- list(c("a", "c", "f", "h", "i", "j", "l", "m",
                         "p", "q", "s", "t", "u", "v","x", "z"), 
                       c("1_2_3", "2_3_4", "3_4_5", "4_5_6"))
resultList <- list(mat1, mat2)

我在这里发布了一个关于替代分箱技术的类似问题：Bin columns and aggregate data via random sample with replacement for iteratively large bin size

这是我尝试对随机选择的列进行分箱并将每个箱大小的结果放入矩阵列表中。我尝试使用、do和删除那些选定的配对列来选择j随机列，以便在下一次迭代中不会重复这些列：samplerowSumsj

lapply(seq_len(ncol(mat) - 1), function(j) 
  do.call(cbind, 
          lapply(sample(ncol(mat) - j, size= ), function(i) 
            rowSums(mat[, i:(i - j)]))))

标签： rlapplysample

解决方案

根据您在最终输出中需要多少列，我们可以修改该方法，但目前这提供了所有可能的组合。

#Get column names of the matrices
all_cols <- colnames(mat)

#Select bin value from 2:ncol(mat)
total_out <- lapply(seq_len(ncol(mat))[-1], function(j) {
      #Create all combinations taking j items at a time
      temp <- combn(all_cols, j, function(x) 
              #Take rowSums for the current combination
              #Also paste column names to assign column names later
              list(rowSums(mat[, x]), paste0(x, collapse = "_")), simplify = FALSE)
      #Combine rowSums matrix 
      new_mat <- sapply(temp, `[[`, 1)
      #Assign column names
      colnames(new_mat) <- sapply(temp, `[[`, 2)
      #Return new matrix
      new_mat
})

当前输出看起来像

total_out
#[[1]]
#  1_2 1_3 1_4 1_5 1_6 2_3 2_4 2_5 2_6 3_4 3_5 3_6 4_5 4_6 5_6
#a   3   1   1   1   2   2   2   2   3   0   0   1   0   1   1
#c   0   0   1   0   1   0   1   0   1   1   0   1   1   2   1
#f   0   1   0   0   0   1   0   0   0   1   1   1   0   0   0
#h   0   1   0   0   0   1   0   0   0   1   1   1   0   0   0
#i   1   1   0   1   0   2   1   2   1   1   2   1   1   0   1
#j   0   0   1   0   0   0   1   0   0   1   0   0   1   1   0
#l   1   1   1   1   1   0   0   0   0   0   0   0   0   0   0
#m   0   0   0   1   0   0   0   1   0   0   1   0   1   0   1
#p   0   0   0   0   1   0   0   0   1   0   0   1   0   1   1
#q   0   0   0   1   0   0   0   1   0   0   1   0   1   0   1
#s   0   0   0   1   1   0   0   1   1   0   1   1   1   1   2
#t   0   0   0   0   2   0   0   0   2   0   0   2   0   2   2
#u   0   0   0   0   1   0   0   0   1   0   0   1   0   1   1
#v   0   0   0   1   0   0   0   1   0   0   1   0   1   0   1
#x   3   2   2   2   2   1   1   1   1   0   0   0   0   0   0
#z   0   0   0   1   0   0   0   1   0   0   1   0   1   0   1
#...
#....
#....
#[[5]]
#  1_2_3_4_5_6
#a           4
#c           2
#f           1
#h           1
#i           3
#j           1
#l           1
#m           1
#p           1
#q           1
#s           2
#t           2
#u           1
#v           1
#x           3
#z           1

请注意，共有 5 ( ncol - 1) 个矩阵total_out，列数为

length(total_out)
#[1] 5

sapply(total_out, ncol)
#[1] 15 20 15  6  1

因为，我们知道列表中的最后一个元素将是一个单列矩阵，我们可以删除它们并nc/2从剩余矩阵中选择随机列。

total_out <- total_out[-length(total_out)]

lapply(total_out, function(x) {
     nc <- ncol(x)
     x[, sample(nc, ceiling(nc/2))]
})

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