Python中成对相关的优化计算

Question

给定一组离散位置(例如"站点"),它们以某些分类方式(例如一般接近度)成对相关并且包含局部级别数据(例如,种群大小),我希望有效地计算本地级别数据之间的平均相关系数.成对位置以相同的关系为特征.

例如,我假设100个站点并使用值1到25随机化它们的成对关系,产生三角矩阵relations:

import numpy as np

sites = 100
categ = 25

relations = np.random.randint(low=1, high=categ+1, size=(sites, sites))
relations = np.triu(relations) # set relation_ij = relation_ji
np.fill_diagonal(relations, 0) # ignore self-relation

我还在每个站点上复制了5000个模拟结果:

sims = 5000
res = np.round(np.random.rand(sites, sims),1)

要计算每个特定关系类别的平均成对相关,我首先计算出每个类别相关i,相关系数rho[j]的模拟结果之间的res每一个独特的网站对j,然后采取所有可能的对平均有关系i:

rho_list = np.ones(categ)*99

for i in range(1, categ+1):
    idr = np.transpose(np.where(relations == i)) # pairwise site indices of the same relation category
    comp = np.vstack([res[idr[:,0]].ravel(), res[idr[:,1]].ravel()]) # pairwise comparisons of simulation results from the same relation category
    comp_uniq = np.reshape(comp.T, (len(idr), res.shape[1], -1)) # reshape above into pairwise comparisons of simulation results between unique site pairs

    rho = np.ones(len(idr))*99 # correlation coefficients of all unique site pairs of current relation category

    for j in range(len(idr)): # loop through unique site pairs
        comp_uniq_s = comp_uniq[j][np.all(comp_uniq!=0, axis=2)[j]].T # shorten comparisons by removing pairs with zero-valued result
        rho[j] = np.corrcoef(comp_uniq_s[0], comp_uniq_s[1])[0,1]

    rho_list[i-1] = np.nanmean(rho)

虽然这个脚本有效,但是一旦我增加sites = 400,那么整个计算可能需要6个多小时才能完成,这让我质疑我对数组函数的使用.这种糟糕表现的原因是什么？我该如何优化算法？

Answer 1

我们可以使用j迭代器对最内层循环进行向量化,masking以处理在该循环的每次迭代中处理的数据的粗糙性.我们也可以慢慢np.corrcoef(受到启发this post).此外,我们可以在外循环开始时优化几个步骤,特别是堆叠步骤,这可能是瓶颈.

因此,完整的代码将减少到这样的东西 -

for i in range(1, categ+1):
    r,c = np.where(relations==i)

    A = res[r]
    B = res[c]

    mask0 = ~((A!=0) & (B!=0))
    A[mask0] = 0
    B[mask0] = 0

    count = mask0.shape[-1] - mask0.sum(-1,keepdims=1)
    A_mA = A - A.sum(-1, keepdims=1)/count
    B_mB = B - B.sum(-1, keepdims=1)/count

    A_mA[mask0] = 0
    B_mB[mask0] = 0

    ssA = np.einsum('ij,ij->i',A_mA, A_mA)
    ssB = np.einsum('ij,ij->i',B_mB, B_mB)
    rho = np.einsum('ij,ij->i',A_mA, B_mB)/np.sqrt(ssA*ssB)

    rho_list[i-1] = np.nanmean(rho)

运行时测试

案例#1:在给定的样本数据上,使用sites = 100

In [381]: %timeit loopy_app()
1 loop, best of 3: 7.45 s per loop

In [382]: %timeit vectorized_app()
1 loop, best of 3: 479 ms per loop

15x+ 加速.

案例#2:有 sites = 200

In [387]: %timeit loopy_app()
1 loop, best of 3: 1min 56s per loop

In [388]: %timeit vectorized_app()
1 loop, best of 3: 1.86 s per loop

In [390]: 116/1.86
Out[390]: 62.36559139784946

62x+ 加速.

案例#3:最后用 sites = 400

In [392]: %timeit vectorized_app()
1 loop, best of 3: 7.64 s per loop

这6hrs+在OP结束时采用了循环方法.

从时间上看,很明显,内部循环的矢量化是获得显着加速的关键sites.