use*_*988 5 c r correlation snowfall
我今天的项目是使用我拥有的基本技能在R中编写一个快速关联例程.我必须找到几乎400个变量之间的相关性,每个变量有近百万个观测值(即大小为p = 1MM行且n = 400个cols的矩阵).
R的原生相关函数对于1MM行需要近2分钟,每个变量需要200个观察值.我没有每列运行400次观察,但我的猜测是需要将近8分钟.我完成它的时间不到30秒.
因此,我想做的事情.
1 - 在C中编写一个简单的相关函数,并将其平行应用于块中(见下文).
2 - 块 - 将相关矩阵分成三个块(大小为K*K的左上方,大小的右下方(pK)(pK)和大小为K(pK)的右上方矩阵).这涵盖了相关矩阵中的所有单元,corr因为我只需要上三角形.
3 - 使用降雪并行地通过.C调用运行C功能.
n = 100
p = 10
X = matrix(rnorm(n*p), nrow=n, ncol=p)
corr = matrix(0, nrow=p, ncol=p)
# calculation of column-wise mean and sd to pass to corr function
mu = colMeans(X)
sd = sapply(1:dim(X)[2], function(x) sd(X[,x]))
# setting up submatrix row and column ranges
K = as.integer(p/2)
RowRange = list()
ColRange = list()
RowRange[[1]] = c(0, K)
ColRange[[1]] = c(0, K)
RowRange[[2]] = c(0, K)
ColRange[[2]] = c(K, p+1)
RowRange[[3]] = c(K, p+1)
ColRange[[3]] = c(K, p+1)
# METHOD 1. NOT PARALLEL
########################
# function to calculate correlation on submatrices
BigCorr <- function(x){
Rows = RowRange[[x]]
Cols = ColRange[[x]]
return(.C("rCorrelationWrapper2", as.matrix(X), as.integer(dim(X)),
as.double(mu), as.double(sd),
as.integer(Rows), as.integer(Cols),
as.matrix(corr)))
}
res = list()
for(i in 1:3){
res[[i]] = BigCorr(i)
}
# METHOD 2
########################
BigCorr <- function(x){
Rows = RowRange[[x]]
Cols = ColRange[[x]]
dyn.load("./rCorrelation.so")
return(.C("rCorrelationWrapper2", as.matrix(X), as.integer(dim(X)),
as.double(mu), as.double(sd),
as.integer(Rows), as.integer(Cols),
as.matrix(corr)))
}
# parallelization setup
NUM_CPU = 4
library('snowfall')
sfSetMaxCPUs() # maximum cpu processing
sfInit(parallel=TRUE,cpus=NUM_CPU) # init parallel procs
sfExport("X", "RowRange", "ColRange", "sd", "mu", "corr")
res = sfLapply(1:3, BigCorr)
sfStop()
这是我的问题:
对于方法1,它可以工作,但不是我想要的方式.我相信,当我通过corr矩阵时,我传递一个地址,C将在源头进行更改.
# Output of METHOD 1
> res[[1]][[7]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 0.1040506 -0.01003125 0.23716384 -0.088246793 0 0 0 0 0
[2,] 0 1.0000000 -0.09795989 0.11274508 0.025754150 0 0 0 0 0
[3,] 0 0.0000000 1.00000000 0.09221441 0.052923520 0 0 0 0 0
[4,] 0 0.0000000 0.00000000 1.00000000 -0.000449975 0 0 0 0 0
[5,] 0 0.0000000 0.00000000 0.00000000 1.000000000 0 0 0 0 0
[6,] 0 0.0000000 0.00000000 0.00000000 0.000000000 0 0 0 0 0
[7,] 0 0.0000000 0.00000000 0.00000000 0.000000000 0 0 0 0 0
[8,] 0 0.0000000 0.00000000 0.00000000 0.000000000 0 0 0 0 0
[9,] 0 0.0000000 0.00000000 0.00000000 0.000000000 0 0 0 0 0
[10,] 0 0.0000000 0.00000000 0.00000000 0.000000000 0 0 0 0 0
> res[[2]][[7]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 -0.02261175 -0.23398448 -0.02382690 -0.1447913 -0.09668318
[2,] 0 0 0 0 0 -0.03439707 0.04580888 0.13229376 0.1354754 -0.03376527
[3,] 0 0 0 0 0 0.10360907 -0.05490361 -0.01237932 -0.1657041 0.08123683
[4,] 0 0 0 0 0 0.18259522 -0.23849323 -0.15928474 0.1648969 -0.05005328
[5,] 0 0 0 0 0 -0.01012952 -0.03482429 0.14680301 -0.1112500 0.02801333
[6,] 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.0000000 0.00000000
[7,] 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.0000000 0.00000000
[8,] 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.0000000 0.00000000
[9,] 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.0000000 0.00000000
[10,] 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.0000000 0.00000000
> res[[3]][[7]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.00000000
[2,] 0 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.00000000
[3,] 0 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.00000000
[4,] 0 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.00000000
[5,] 0 0 0 0 0 0 0.00000000 0.00000000 0.00000000 0.00000000
[6,] 0 0 0 0 0 1 0.03234195 -0.03488812 -0.18570151 0.14064640
[7,] 0 0 0 0 0 0 1.00000000 0.03449697 -0.06765511 -0.15057244
[8,] 0 0 0 0 0 0 0.00000000 1.00000000 -0.03426464 0.10030619
[9,] 0 0 0 0 0 0 0.00000000 0.00000000 1.00000000 -0.08720512
[10,] 0 0 0 0 0 0 0.00000000 0.00000000 0.00000000 1.00000000
但原始corr矩阵保持不变:
> corr
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 0 0 0 0 0
[2,] 0 0 0 0 0 0 0 0 0 0
[3,] 0 0 0 0 0 0 0 0 0 0
[4,] 0 0 0 0 0 0 0 0 0 0
[5,] 0 0 0 0 0 0 0 0 0 0
[6,] 0 0 0 0 0 0 0 0 0 0
[7,] 0 0 0 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 0 0 0 0 0 0
[10,] 0 0 0 0 0 0 0 0 0 0
问题1:有没有办法确保C函数改变corr源的值？我仍然可以合并这三个来创建一个上三角相关矩阵,但我想知道是否可以在源头进行更改.注意:这不能帮助我实现快速关联,因为我只是在运行一个循环.
问题2:对于方法2,如何在init步骤中为每个核心上的并行作业将共享对象加载到每个核心(而不是我如何完成)？
问题3:这个错误是什么意思？我需要一些指针,我很乐意自己调试.
问题#4:在不到30秒的时间内,是否有一种快速计算矩阵1MM乘以400的相关性的方法？
当我运行METHOD 2时,我收到以下错误:
R(6107) malloc: *** error for object 0x100664df8: incorrect checksum for freed object - object was probably modified after being freed.
*** set a breakpoint in malloc_error_break to debug
Error in unserialize(node$con) : error reading from connection
下面是我的普通香草C代码,用于关联:
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <stddef.h>
#include <R.h> // to show errors in R
double calcMean (double *x, int n);
double calcStdev (double *x, double mu, int n);
double calcCov(double *x, double *y, int n, double xmu, double ymu);
void rCorrelationWrapper2 ( double *X, int *dim, double *mu, double *sd, int *RowRange, int *ColRange, double *corr) {
int i, j, n = dim[0], p = dim[1];
int RowStart = RowRange[0], RowEnd = RowRange[1], ColStart = ColRange[0], ColEnd = ColRange[1];
double xyCov;
Rprintf("\n p: %d, %d <= row < %d, %d <= col < %d", p, RowStart, RowEnd, ColStart, ColEnd);
if(RowStart==ColStart && RowEnd==ColEnd){
for(i=RowStart; i<RowEnd; i++){
for(j=i; j<ColEnd; j++){
Rprintf("\n i: %d, j: %d, p: %d", i, j, p);
xyCov = calcCov(X + i*n, X + j*n, n, mu[i], mu[j]);
*(corr + j*p + i) = xyCov/(sd[i]*sd[j]);
}
}
} else {
for(i=RowStart; i<RowEnd; i++){
for (j=ColStart; j<ColEnd; j++){
xyCov = calcCov(X + i*n, X + j*n, n, mu[i], mu[j]);
*(corr + j*p + i) = xyCov/(sd[i]*sd[j]);
}
}
}
}
// function to calculate mean
double calcMean (double *x, int n){
double s = 0;
int i;
for(i=0; i<n; i++){
s = s + *(x+i);
}
return(s/n);
}
// function to calculate standard devation
double calcStdev (double *x, double mu, int n){
double t, sd = 0;
int i;
for (i=0; i<n; i++){
t = *(x + i) - mu;
sd = sd + t*t;
}
return(sqrt(sd/(n-1)));
}
// function to calculate covariance
double calcCov(double *x, double *y, int n, double xmu, double ymu){
double s = 0;
int i;
for(i=0; i<n; i++){
s = s + (*(x+i)-xmu)*(*(y+i)-ymu);
}
return(s/(n-1));
}
And*_*lin 11
使用快速BLAS(通过Revolution R或Goto BLAS),您可以在R中快速计算所有这些相关性,而无需编写任何C代码.在我的第一代Intel i7 PC上需要16秒:
n = 400;
m = 1e6;
# Generate data
mat = matrix(runif(m*n),n,m);
# Start timer
tic = proc.time();
# Center each variable
mat = mat - rowMeans(mat);
# Standardize each variable
mat = mat / sqrt(rowSums(mat^2));
# Calculate correlations
cr = tcrossprod(mat);
# Stop timer
toc = proc.time();
# Show the results and the time
show(cr[1:4,1:4]);
show(toc-tic)
上面的R代码报告了以下时间:
user system elapsed
31.82 1.98 15.74
我在我的MatrixEQTL包中使用这种方法.
http://www.bios.unc.edu/research/genomic_software/Matrix_eQTL/
有关R的各种BLAS选项的更多信息,请访问:http:
//www.bios.unc.edu/research/genomic_software/Matrix_eQTL/runit.html#large