您的代码具有复杂度O(N * N)[genesPerSet
= N]。但使用值的顺序与您无关的事实,我们可以在O(N•log(N))中对它进行排序并计算O(N)中的“分数”。 (有可能会快上千倍)。
而且,我们总共有N * N的比较。然后U_Original + U_Random
= N * N,这意味着我们不需要计算U_Random。 也是你的统计Zn = Umin-N * N/2;当你只有abs(Zn/Zd)在N * N/2附近对称时。我们只需要一个算法。
1.-第一东西可以通过(常数)参考取参数:
int WilcoxinRST::work(const GeneSet &originalGeneSet, const vector<string> &randomGenes)
2.-您填充到载体geneIDs;但不使用它?为什么?
3.-您只能迭代2次。
4.-你保持信号的值(探头intensitat?)一起在一个向量,并使用其它矢量信号是什么每个项目 - 简单的保持两个向量。
5.-你不需要得分矢量,仅苏玛。
6.-为什么要2000万次?我想你正在计算一些“统计”稳定性或BS陷阱。很可能你使用相同的原始基因集很多时间。我想你可以在另一个问题中发布调用这个函数的代码,以便每次都使用值和排序向量。
这里是第一新O(N•日志(N))的代码。
下面是你的代码的清理,但仍O(N * N),速度快,但只能由一个常数因子。
那么同样的代码,但与你的原代码,并与更多的评论混合。
请调试这个,告诉我是怎么回事。
#include<vector>
#include<algorithm>
int WilcoxinRST::work(const GeneSet &originalGeneSet , const vector<string>& randomGenes)
{
size_t genesPerSet = originalGeneSet.geneCount();
std::vector<double> valueOri(genesPerSet), valueRnd(genesPerSet);
/**
* Fill the valueOri vector with original gene set data, and valueRnd with random data
*/
for (size_t i = 0; i < genesPerSet; i++)
{
valueOri[i] = std::fabs(expressionLevels.getValue(originalGeneSet.getMemberGeneAt(i) , statType));
valueRnd[i] = std::fabs(expressionLevels.getValue(randomGenes.at(i) , statType));
}
std::sort(valueOri.begin(),valueOri.end());
std::sort(valueRnd.begin(),valueRnd.end());
/**
* calculate the scores Ua, Ub and U
*/
long U_Original ;
if (statType == Fold_Change || statType == T_Statistic || statType == Z_Statistic
statType == FDR_PValue || statType == PValue)
{
// Higher value is a winner
size_t j=0;
for (size_t i = 0; i < genesPerSet /*totalGenes*/; i++) // i - 2x
{
while(valueOri[i] > valueRnd[j]) ++j ;
U_Original += j;
}
} else { cout << endl << "ERROR. Statistic type not defined." << endl; }
/**
* calculate z
*/
double Zn, Zd, Z;
Zn = U_Original - ((genesPerSet * genesPerSet)/2);
Zd = std::sqrt((double) (((genesPerSet * genesPerSet* (genesPerSet + genesPerSet + 1))))/12.0);
Z = Zn/Zd;
/**
* Return 0/1/2
* 2: p value < 0.01
* 1: 0.01 < p value < 0.05
* 0: p value > 0.05
*/
if (std::fabs(Z) > 2.303) return 2;
else if (std::fabs(Z) > 1.605) return 1;
else return 0;
}
接下来是清理你的代码,但仍然是O(N * N),但速度很快,但只有一个常数因子。
#include<vector>
using namespace std;
class GeneSet ;
class WilcoxinRST;
int WilcoxinRST::work(const GeneSet &originalGeneSet , const vector<string>& randomGenes)
{
size_t genesPerSet = originalGeneSet.geneCount();
vector<double> valueOri(genesPerSet), valueRnd(genesPerSet);
/**
* Fill the valueOri vector with original gene set data, and valueRnd with random data
*/
for (size_t i = 0; i < genesPerSet; i++)
{
valueOri[i] = fabs(expressionLevels.getValue(originalGeneSet.getMemberGeneAt(i) , statType));
valueRnd[i] = fabs(expressionLevels.getValue(randomGenes.at(i) , statType));
}
/**
* calculate the scores Ua, Ub and U
*/
long U_Original = 0, U_Random = 0, U_Final;
if (statType == Fold_Change || statType == T_Statistic || statType == Z_Statistic)
{
// Higher value is a winner
for (size_t i = 0; i < genesPerSet /*totalGenes*/; i++) // i - 2x
{ for (size_t j = 0; j < genesPerSet; j++)
{ U_Random += (valueRnd[i] > valueOri[j]);// i en 2 set=Rnd, j in 1 set=Ori. count how many Ori are less than this Rnd
U_Original+= (valueOri[i] > valueRnd[j]);// i in 1 set=Ori, j in 2 set=Rnd, count how many Rnd are less than this Ori
}
}
} else
if (statType == FDR_PValue || statType == PValue)
{
// Lower value is a winner
for (size_t i = 0; i < genesPerSet; i++)
{
for (size_t j = 0; j < genesPerSet; j++)
{ U_Random += (valueRnd[i] < valueOri[j]);// i en 2 set=Rnd, j in 1 set=Ori. count how many Ori are > than this Rnd
U_Original+= (valueOri[i] < valueRnd[j]);// i in 1 set=Ori, j in 2 set=Rnd, count how many Rnd are > than this Ori
}
}
} else { cout << endl << "ERROR. Statistic type not defined." << endl; }
U_Final = (U_Original < U_Random) ? U_Original : U_Random;
/**
* calculate z
*/
double Zn, Zd, Z;
Zn = U_Final - ((genesPerSet * genesPerSet)/2);
Zd = sqrt(
(double) (((genesPerSet * genesPerSet
* (genesPerSet + genesPerSet + 1))))/12.0);
Z = Zn/Zd;
/**
* Return 0/1/2
* 2: p value < 0.01
* 1: 0.01 < p value < 0.05
* 0: p value > 0.05
*/
if (fabs(Z) > 2.303) return 2;
else if (fabs(Z) > 1.605) return 1;
else return 0;
}
相同的代码,但混有你的原始代码和更多评论。
int WilcoxinRST::work(const GeneSet &originalGeneSet , const vector<string>& randomGenes)
{
size_t genesPerSet = originalGeneSet.geneCount();
unsigned int totalGenes, tempScore;
totalGenes = genesPerSet * 2;
//vector<string> geneIDs;
//vector<bool> isOriginal;
//vector<int> rank;
vector<double> valueOri(genesPerSet), valueRnd(genesPerSet);
//vector<int> score;
/**
* Fill the first half of the vectors with original gene set data
*/
for (size_t i = 0; i < genesPerSet; i++)
{
//geneIDs.push_back(originalGeneSet.getMemberGeneAt(i) );
//isOriginal.push_back(true);
valueOri[i] = fabs(expressionLevels.getValue(originalGeneSet.getMemberGeneAt(i) , statType));
valueRnd[i] = fabs(expressionLevels.getValue(randomGenes.at(i) , statType));
}
/**
* Fill the second half with random data
*/
//for (unsigned int i = genesPerSet; i < totalGenes; i++) {
// geneIDs.push_back(randomGenes.at(i - genesPerSet));
// isOriginal.push_back(false);
// value.push_back(fabs(expressionLevels.getValue(geneIDs[i], statType)));
//}
//totalGenes = geneIDs.size();
/**
* calculate the scores
*/
/**
* calculate Ua, Ub and U
*/
long U_Original = 0, U_Random = 0, U_Final;
//for (int j = 0; j < genesPerSet; j++) // j in 1 set=Ori. count how many Ori are less than this Rnd
//{
// U_Original += score[j];
//}
//for (unsigned int j = genesPerSet; j < totalGenes; j++) // j in 2 set=Rnd, count how many Rnd are less than this Ori
//{
// U_Random += score[j];
//}
if (statType == Fold_Change || statType == T_Statistic || statType == Z_Statistic)
{
// Higher value is a winner
for (size_t i = 0; i < genesPerSet /*totalGenes*/; i++) // i - 2x
{ //tempScore = 0;
//if (!isOriginal[i]) // i en 2 set=Rnd, j in 1 set=Ori. count how many Ori are less than this Rnd
for (size_t j = 0; j < genesPerSet; j++)
{ U_Random += (valueRnd[i] > valueOri[j]);// i en 2 set=Rnd, j in 1 set=Ori. count how many Ori are less than this Rnd
U_Original+= (valueOri[i] > valueRnd[j]);// i in 1 set=Ori, j in 2 set=Rnd, count how many Rnd are less than this Ori
}
//} else
//{
// for (unsigned int j = genesPerSet; j < totalGenes; j++) // i in 1 set=Ori, j in 2 set=Rnd, count how many Rnd are less than this Ori
// { if (value.at(i) > value.at(j)) { tempScore++; }
// }
//}
//score.push_back(tempScore);
}
} else
if (statType == FDR_PValue || statType == PValue)
{
// Lower value is a winner
for (size_t i = 0; i < genesPerSet; i++)
{
for (size_t j = 0; j < genesPerSet; j++)
{ U_Random += (valueRnd[i] < valueOri[j]);// i en 2 set=Rnd, j in 1 set=Ori. count how many Ori are > than this Rnd
U_Original+= (valueOri[i] < valueRnd[j]);// i in 1 set=Ori, j in 2 set=Rnd, count how many Rnd are > than this Ori
}
//} else
//{
// for (unsigned int j = genesPerSet; j < totalGenes; j++) // i in 1 set=Ori, j in 2 set=Rnd, count how many Rnd are less than this Ori
// { if (value.at(i) > value.at(j)) { tempScore++; }
// }
//}
//score.push_back(tempScore);
}
//for (unsigned int i = 0; i < totalGenes; i++)
//{ tempScore = 0;
// if (!isOriginal[i])
// { for (int j = 0; j < genesPerSet; j++) {
// if (value.at(i) < value.at(j)) { // Rnd i < Ori j increm U_Random
// tempScore++;
// }
// }
// } else {
// for (unsigned int j = genesPerSet; j < totalGenes; j++) { // Ori i < Rnd j. Increm U_Original
// if (value.at(i) < value.at(j)) {
// tempScore++;
// }
// }
// }
// score.push_back(tempScore);
//}
} else { cout << endl << "ERROR. Statistic type not defined." << endl; }
U_Final = (U_Original < U_Random) ? U_Original : U_Random;
/**
* calculate z
*/
double Zn, Zd, Z;
Zn = U_Final - ((genesPerSet * genesPerSet)/2);
Zd = sqrt(
(double) (((genesPerSet * genesPerSet
* (genesPerSet + genesPerSet + 1))))/12.0);
Z = Zn/Zd;
/**
* Return 0/1/2
* 2: p value < 0.01
* 1: 0.01 < p value < 0.05
* 0: p value > 0.05
*/
if (fabs(Z) > 2.303)
return 2;
else if (fabs(Z) > 1.605)
return 1;
else
return 0;
}
所以我假设你已经对它进行了基准测试,结果证明它是主要的瓶颈,对吧? – 2013-03-28 07:50:02
不详细看,但一个“显而易见”的观点突出 - 如果事先知道std :: vector的预期大小,先调用resize()或reserve()以避免重复的内存重新分配。 – 2013-03-28 07:50:38
不,我在ubuntu上使用eclipse,我不认为有一种方法可以在此IDE上剖析代码。 – Pranjal 2013-03-28 07:51:40