r - 在 R 中对大型、非常稀疏的二元矩阵进行聚类

Question

我有一个大的稀疏二进制矩阵(大约 39,000 x 14,000；大多数行只有一个“1”条目)。我想将相似的行聚集在一起，但我最初的计划需要很长时间才能完成:

d <- dist(inputMatrix, method="binary")
hc <- hclust(d, method="complete")

第一步没有完成，所以我不确定第二步会如何。在 R 中，有哪些方法可以有效地对大型稀疏二元矩阵的相似行进行分组？

Answer 1

我编写了一些 Rcpp 代码和 R 代码，可以计算出二进制矩阵的二进制/杰卡德距离。比 dist(x, method = "binary") 快 80 倍。它将输入矩阵转换为原始矩阵，该矩阵是输入的转置(以便位模式在内部处于正确的顺序)。然后在一些 C++ 代码中使用它，将数据作为 64 位无符号整数处理以提高速度。两个向量 x 和 y 的杰卡德距离等于 x ^ y / (x | y)哪里^是异或运算符。 Hamming Weight计算用于计算设置的位数，如果xor的结果或or非零。

我已将代码放在 github 上 https://github.com/NikNakk/binaryDist/并复制了下面的两个文件。我已经确认结果与 dist(x, method = "binary") 相同对于一些随机数据集。

在 39000 行 x 14000 列、每行 1-5 个数据集上，大约需要 11 分钟。输出距离矩阵为 5.7 GB。

bDist.cpp

#include <Rcpp.h>
using namespace Rcpp;

//countBits function taken from https://en.wikipedia.org/wiki/Hamming_weight#Efficient_implementation

const uint64_t m1  = 0x5555555555555555; //binary: 0101...
const uint64_t m2  = 0x3333333333333333; //binary: 00110011..
const uint64_t m4  = 0x0f0f0f0f0f0f0f0f; //binary:  4 zeros,  4 ones ...
const uint64_t h01 = 0x0101010101010101; //the sum of 256 to the power of 0,1,2,3...

int countBits(uint64_t x) {
  x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits
  x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits 
  x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits 
  return (x * h01)>>56;  //returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24) + ... 
}

// [[Rcpp::export]]
int countBitsFromRaw(RawVector rv) {
  uint64_t* x = (uint64_t*)RAW(rv);
  return(countBits(*x));
}

// [[Rcpp::export]]
NumericVector bDist(RawMatrix mat) {
  int nr(mat.nrow()), nc(mat.ncol());
  int nw = nr / 8;
  NumericVector res(nc * (nc - 1) / 2);
  // Access the raw data as unsigned 64 bit integers
  uint64_t* data = (uint64_t*)RAW(mat);
  uint64_t a(0);
  // Work through each possible combination of columns (rows in the original integer matrix)
  for (int i = 0; i < nc - 1; i++) {
    for (int j = i + 1; j < nc; j++) {
      uint64_t sx = 0;
      uint64_t so = 0;
      // Work through each 64 bit integer and calculate the sum of (x ^ y) and (x | y)
      for (int k = 0; k < nw; k++) {
        uint64_t o = data[nw * i + k] | data[nw * j + k];
        // If (x | y == 0) then (x ^ y) will also be 0
        if (o) {
          // Use Hamming weight method to calculate number of set bits
          so = so + countBits(o);
          uint64_t x = data[nw * i + k] ^ data[nw * j + k];
          if (x) {
            sx = sx + countBits(x);
          }
        }
      }
      res(a++) = (double)sx / so;
    }
  }
  return (res);
}

R 源

library("Rcpp")
library("plyr")
sourceCpp("bDist.cpp")

# Converts a binary integer vector into a packed raw vector,
# padding out at the end to make the input length a multiple of packWidth
packRow <- function(row, packWidth = 64L) {
  packBits(as.raw(c(row, rep(0, (packWidth - length(row)) %% packWidth))))
}

as.PackedMatrix <- function(x, packWidth = 64L) {
  UseMethod("as.PackedMatrix")
}

# Converts a binary integer matrix into a packed raw matrix
# padding out at the end to make the input length a multiple of packWidth
as.PackedMatrix.matrix <- function(x, packWidth = 64L) {
  stopifnot(packWidth %% 8 == 0, class(x) %in% c("matrix", "Matrix"))
  storage.mode(x) <- "raw"
  if (ncol(x) %% packWidth != 0) {
    x <- cbind(x, matrix(0L, nrow = nrow(x), ncol = packWidth - (ncol(x) %% packWidth)))
  }
  out <- packBits(t(x))
  dim(out) <- c(ncol(x) %/% 8, nrow(x))
  class(out) <- "PackedMatrix"
  out
}

# Converts back to an integer matrix
as.matrix.PackedMatrix <- function(x) {
  out <- rawToBits(x)
  dim(out) <- c(nrow(x) * 8L, ncol(x))
  storage.mode(out) <- "integer"
  t(out)
}

# Generates random sparse data for testing the main function
makeRandomData <- function(nObs, nVariables, maxBits, packed = FALSE) {
  x <- replicate(nObs, {
    y <- integer(nVariables)
    y[sample(nVariables, sample(maxBits, 1))] <- 1L
    if (packed) {
      packRow(y, 64L)
    } else {
      y
    }
  })
  if (packed) {
    class(x) <- "PackedMatrix"
    x
  } else {
    t(x)
  }
}

# Reads a binary matrix from file or character vector
# Borrows the first bit of code from read.table
readPackedMatrix <- function(file = NULL, text = NULL, packWidth = 64L) {
  if (missing(file) && !missing(text)) {
    file <- textConnection(text)
    on.exit(close(file))
  }
  if (is.character(file)) {
    file <- file(file, "rt")
    on.exit(close(file))
  }
  if (!inherits(file, "connection")) 
    stop("'file' must be a character string or connection")
  if (!isOpen(file, "rt")) {
    open(file, "rt")
    on.exit(close(file))
  }
  lst <- list()
  i <- 1
  while(length(line <- readLines(file, n = 1)) > 0) {
    lst[[i]] <- packRow(as.integer(strsplit(line, "", fixed = TRUE)[[1]]), packWidth = packWidth)
    i <- i + 1
  }
  out <- do.call("cbind", lst)
  class(out) <- "PackedMatrix"
  out
}

# Wrapper for the C++ code which 
binaryDist <- function(x) {
  if (class(x) != "PackedMatrix") {
    x <- as.PackedMatrix(x)
  }
  dst <- bDist(x)
  attr(dst, "Size") <- ncol(x)
  attr(dst, "Diag") <- attr(dst, "Upper") <- FALSE
  attr(dst, "method") <- "binary"
  attr(dst, "call") <- match.call()
  class(dst) <- "dist"
  dst
}

x <- makeRandomData(2000, 400, maxBits = 5, packed = TRUE)

system.time(bd <- binaryDist(x))

来自原始答案:

其他需要考虑的事情是对两行与单行之间的比较进行一些预过滤，因为距离对于重复项为 0，对于任何其他可能性为 1。

一些相对简单的选项可能会更快，而不需要太多代码，这些选项是 vegdist vegan 包和 Dist 中的函数amap 包中的函数。如果您有多个核心并利用它支持并行化的事实，后者可能只会更快。