R语言的原生相关性函数对于每个变量的200个观测值和1MM行数据大约需要2分钟的时间。虽然我没有试图计算400个变量每列的观测值数,但我的猜测是这将需要近8分钟时间。而我只有不到30秒的时间来完成它。
因此,我想要做以下几件事情:
1- 用C语言编写一个简单的相关性函数,并在块中并行应用它(如下所示)。
2- 块 - 将相关矩阵分为三块(大小为K*K的左上方正方形区域,大小为(p-K)*(p-K)的右下方正方形区域,以及大小为K*(p-K)的右上方矩形区域)。这样就可以覆盖相关矩阵corr的所有单元格,因为我只需要上三角部分。
3- 通过snowfall并行运行C语言函数的.C调用。
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()
这里是我的问题:
对于方法1,它可以工作,但不是我想要的方式。我相信当我传递相关矩阵时,我正在传递一个地址,C将在源代码中进行更改。
# 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
但是原始的
corr
矩阵仍然保持不变:> 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函数在源代码中更改corr
的值?我仍然可以将它们合并以创建一个上三角形相关矩阵,但我想知道是否可能在源代码中进行更改。注意:这不能帮助我实现快速相关性,因为我只是在运行循环。
问题2:对于METHOD 2,如何在init步骤中将共享对象加载到每个核心以进行并行作业(而不是我所做的方式)?
问题3:这个错误是什么意思?我需要一些指针,并且我很乐意自己调试。
问题4:是否有一种快速的方法来计算1MM x 400矩阵的相关性,少于30秒?
当我运行METHOD 2时,我会得到以下错误:
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
以下是我用纯C编写的相关代码:
#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));
}