Mat*_*ave 6 performance r cluster-analysis matrix sparse-matrix
我有一个大的,稀疏的二进制矩阵(大约39,000 x 14,000;大多数行只有一个"1"条目).我想将类似的行聚集在一起,但我的初始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将如何进行.有哪些方法可以有效地对R中大型稀疏二进制矩阵的相似行进行分组？
Nic*_*edy 16
我写了一些Rcpp代码和R代码,它们可以计算出二进制矩阵的二进制/ Jaccard距离.比80倍快dist(x, method = "binary").它将输入矩阵转换为原始矩阵,该矩阵是输入的转置(以便位模式在内部以正确的顺序).然后将其用于某些C++代码中,该代码将数据处理为64位无符号整数以提高速度.两个矢量x和y的杰卡德距离等于x ^ y / (x | y),其中^是XOR运算符.的汉明权重的计算用于计算设定的比特数,如果的结果xor或or为非零.
我已经在https://github.com/NikNakk/binaryDist/上将代码放在github上,并重现了下面的两个文件.我已经确认结果与dist(x, method = "binary")一些随机数据集相同.
在39000行乘14000列的数据集上,每行1-5个,大约需要11分钟.输出距离矩阵为5.7 GB.
#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);
}
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:
其他要考虑的事情是对两行与单个行进行一些预过滤比较,因为重复距离为0或任何其他可能性为1.
一对夫妇的相对简单的选项可能会更快,而不需要太多的代码是vegdist从功能的素食包和Dist从功能AMAP包.如果你有多个内核并利用它支持并行化的事实,后者可能会更快.
这需要花费很长时间来计算的原因是,调用dist是计算和存储超过7.6亿的成对距离.如果您的数据存储稀疏,则需要很长时间和大量存储空间.如果您的数据没有稀疏存储,那么每个距离计算至少需要14,000次操作,总操作次数超过1千万亿!
一种快得多的方法是k均值聚类,因为它不需要预先计算距离矩阵; 在每次迭代中,您只需要39000*k个距离计算,其中k是簇的数量.要获得与Jaccard索引类似的成对距离(如果相同,则为0,如果没有索引重合,则为1,如果某些但不是所有索引重合,则在两者之间),您可以将每行x除以sqrt(2*sum(x^2)).例如,如果您有以下输入矩阵:
(mat <- rbind(c(1, 0, 0, 0, 0), c(0, 0, 0, 1, 1)))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 0 0 0 0
# [2,] 0 0 0 1 1
标准化版本将是(仅在矩阵中假设二进制值;如果不是这种情况,则使用rowSums(mat^2)):
(mat.norm <- mat / sqrt(2*rowSums(mat)))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0.7071068 0 0 0.0 0.0
# [2,] 0.0000000 0 0 0.5 0.5
这两个观测值(没有共同的指数)具有欧几里德距离1,与此情况下的Jaccard距离一致.
dist(mat.norm, "euclidean")
# 1
# 2 1
另外,相同的观察将清楚地具有欧几里德距离0,再次对应于Jaccard距离.