我有一个函数来计算numpy数组中所有行对之间的成对相关性.一切正常,但是后来我想起,我经常不得不处理丢失的数据.我使用蒙版数组尝试解决此问题,但它使计算速度大大降低.关于使用屏蔽函数的任何想法.我认为真正的瓶颈将在np.ma.dot函数中.但是我添加了一些配置文件,并很快使用iPython进行了模拟.我应该说,就这些阵列将要拥有的行数而言,5000在频谱的较低端.有些可能超过30万.带掩码数组的代码看起来比没有掩码的代码慢大约20倍,但是显然由于缺少数据,没有掩码数组的代码经常会产生NaN.

首先,一种快速而肮脏的方式来生成一些样本数据

genotypes = np.empty((5000,200))
for i in xrange(5000):
    genotypes[i] = np.random.binomial(3,.333, size=200)

pValues = np.random.uniform(0,1,5000)

然后测试功能

 def testMask(pValsArray, genotypeArray):
    nSNPs = len(pValsArray)-1
    genotypeArray = np.ma.masked_array(genotypeArray, np.isnan(genotypeArray))
    chisq = np.sum(-2 * np.log(pValsArray))
    ms = genotypeArray.mean(axis=1)[(slice(None,None,None),None)]
    datam = genotypeArray - ms
    datass = np.ma.sqrt(np.ma.sum(datam**2, axis=1))
    runningSum = 0
    for i in xrange(nSNPs):
        temp = np.ma.dot(datam[i:],datam[i].T)
        d = (datass[i:]*datass[i])
        rs = temp / d
        rs = np.absolute(rs)[1:]
        runningSum += 3.25 * np.sum(rs) +  .75 * np.dot(rs, rs)
    sigmaSq = 4*nSNPs+2*runningSum
    E = 2*nSNPs
    df = (2*(E*E))/sigmaSq
    runningSum = sigmaSq/(2*E)
    d = chisq/runningSum
    brownsP = stats.chi2.sf(d, df)
    return brownsP

def testNotMask(pValsArray, genotypeArray):
    nSNPs = len(pValsArray)-1
    chisq = np.sum(-2 * np.log(pValsArray))
    ms = genotypeArray.mean(axis=1)[(slice(None,None,None),None)]
    datam = genotypeArray - ms
    datass = np.sqrt(stats.ss(datam, axis=1))
    runningSum = 0
    for i in xrange(nSNPs):
        temp = np.dot(datam[i:],datam[i].T)
        d = (datass[i:]*datass[i])
        rs = temp / d
        rs = np.absolute(rs)[1:]
        runningSum += 3.25 * np.sum(rs) +  .75 * np.dot(rs, rs)
    sigmaSq = 4*nSNPs+2*runningSum
    E = 2*nSNPs
    df = (2*(E*E))/sigmaSq
    runningSum = sigmaSq/(2*E)
    d = chisq/runningSum
    brownsP = stats.chi2.sf(d, df)
    return brownsP

还有一些时间

%timeit testMask(pValues, genotypes)
1 loops, best of 3: 14.3 s per loop

%timeit testNotMask(pValues, genotypes)
1 loops, best of 3: 678 ms per loop

添加一些丢失的数据,然后再次运行：

randis = np.random.randint(0,5000, 10)
randjs = np.random.randint(0,200, 10)

for i,j in zip(randis, randjs):
    genotypes[i,j] = None



%timeit testMask(pValues, genotypes)
1 loops, best of 3: 14.2 s per loop

%timeit testNotMask(pValues, genotypes)
1 loops, best of 3: 654 ms per loop

和一些分析：

%prun

       2559677 function calls in 15.045 seconds

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
     9791    5.259    0.001    5.259    0.001 {method 'copy' of 'numpy.ndarray' objects}
     4999    2.877    0.001   11.888    0.002 extras.py:949(dot)
     9794    1.566    0.000    1.566    0.000 {numpy.core.multiarray.copyto}
    14997    1.497    0.000    1.564    0.000 {numpy.core._dotblas.dot}
    30007    0.559    0.000    0.559    0.000 {method 'reduce' of 'numpy.ufunc' objects}
    94996    0.450    0.000    0.875    0.000 core.py:2751(_update_from)
   864970    0.347    0.000    0.347    0.000 {getattr}
     5000    0.346    0.000    0.802    0.000 core.py:1065(__call__)
        1    0.240    0.240   15.045   15.045 <ipython-input-115-2aab1c8ea4c5>:1(testMask)
     5000    0.196    0.000    0.196    0.000 core.py:771(__call__)
     5001    0.147    0.000    0.551    0.000 core.py:917(__call__)
    24996    0.143    0.000    0.609    0.000 core.py:2930(__getitem__)
    54998    0.140    0.000    0.775    0.000 core.py:2776(__array_finalize__)
   419985    0.126    0.000    0.126    0.000 {method 'update' of 'dict' objects}
        1    0.093    0.093    0.111    0.111 core.py:5990(power)
   339994    0.077    0.000    0.077    0.000 {isinstance}
    50015    0.072    0.000    0.072    0.000 {numpy.core.multiarray.array}
    60002    0.060    0.000    0.568    0.000 {method 'view' of 'numpy.ndarray' objects}
     5000    0.060    0.000    0.199    0.000 core.py:2626(__new__)
    14999    0.058    0.000    7.412    0.000 core.py:3341(filled)
    25005    0.055    0.000    0.092    0.000 core.py:609(getdata)

编辑：

我尝试了perimosocordiae的答案,但我仍然感到烦恼.看起来像平均值,stats.ss和np.sqrt函数都关心nan值.

def fastNotMask(pValsArray, genotypeArray):
    nSNPs = len(pValsArray)-1
    chisq = np.sum(-2 * np.log(pValsArray))
    ms = genotypeArray.mean(axis=1)[(slice(None,None,None),None)]
    print ms
    datam = genotypeArray - ms
    print datam
    datass = np.sqrt(stats.ss(datam, axis=1))
    print datass
    runningSum = 0
    for i in xrange(nSNPs):
        temp = np.dot(datam[i:],datam[i].T)
        d = (datass[i:]*datass[i])
        rs = temp / d
        rs = np.absolute(rs)[1:]
        runningSum += 3.25 * np.nansum(rs) +  .75 * np.nansum(rs * rs)
    print runningSum
    sigmaSq = 4*nSNPs+2*runningSum
    E = 2*nSNPs
    df = (2*(E*E))/sigmaSq
    runningSum = sigmaSq/(2*E)
    d = chisq/runningSum
    brownsP = stats.chi2.sf(d, df)
    return brownsP

用少量输出进行测试表明,nans没有被正确处理.

pValues = np.random.uniform(0,1,10)

genotypes = np.empty((10,10))
for i in xrange(10):
    genotypes[i] = np.random.binomial(2,.5, size=10)

randis = np.random.randint(0,10, 2)
randjs = np.random.randint(0,10, 2)

for i,j in zip(randis, randjs):
    genotypes[i,j] = None

print testfastMask(pValues, genotypes)



[[ 1.5]
 [ 1.2]
 [ 0.9]
 [ 1.2]
 [ 1.1]
 [ 0.6]
 [ nan]
 [ 1.1]
 [ nan]
 [ 0.8]]
[[-0.5 -0.5  0.5  0.5 -0.5 -0.5  0.5  0.5 -0.5  0.5]
 [-0.2  0.8 -0.2 -0.2 -0.2 -0.2 -0.2  0.8 -0.2 -0.2]
 [-0.9  0.1 -0.9  1.1  0.1 -0.9  1.1  0.1  0.1  0.1]
 [-0.2 -0.2 -0.2 -0.2  0.8 -0.2 -0.2 -1.2  0.8  0.8]
 [-0.1 -0.1 -0.1 -0.1 -0.1 -0.1  0.9 -0.1  0.9 -1.1]
 [-0.6  1.4  0.4 -0.6 -0.6  0.4 -0.6 -0.6  0.4  0.4]
 [ nan  nan  nan  nan  nan  nan  nan  nan  nan  nan]
 [-0.1  0.9 -1.1 -1.1  0.9 -0.1  0.9 -1.1  0.9 -0.1]
 [ nan  nan  nan  nan  nan  nan  nan  nan  nan  nan]
 [ 1.2 -0.8 -0.8  0.2 -0.8  0.2  1.2  0.2  0.2 -0.8]]
[ 1.58113883  1.26491106  2.21359436  1.8973666   1.70293864  2.0976177
         nan  2.62678511         nan  2.36643191]
nan
nan

我在这里想念什么吗？这可能是版本问题.我正在使用python 2.7和numpy 1.7.1？

为帮助加油.

解决方法:

编辑：原始答案不适用于numpy版本< = 1.8,其中np.nansum([NaN,NaN])== 0.0(note the FutureWarning).对于早期版本,您必须手动检查这种情况：

tmp = 3.25 * np.nansum(rs) +  .75 * np.nansum(rs * rs)
if not np.isnan(tmp):
  runningSum += tmp

或者,您可以建立一个tmp值列表/数组,并在其上调用np.nansum.

原始答案：

您需要更改的只是testNotMask中的这一行：

runningSum += 3.25 * np.sum(rs) +  .75 * np.dot(rs, rs)

对此：

runningSum += 3.25 * np.nansum(rs) +  .75 * np.nansum(rs * rs)

所有其他操作都可以使用NaN值正常工作,因此您可以在保持正确结果的同时获得所有非掩码数组的速度.

这是证明.函数fastNotMask只是带有上述更改的testNotMask.

In [63]: genotypes = np.random.binomial(3, .333, size=(5000, 200)).astype(float)

In [64]: pValues = np.random.uniform(0,1,5000)

In [65]: %timeit testMask(pValues, genotypes)
1 loops, best of 3: 11.3 s per loop

In [66]: %timeit testNotMask(pValues, genotypes)
1 loops, best of 3: 3.53 s per loop

In [67]: %timeit fastNotMask(pValues, genotypes)
1 loops, best of 3: 3.96 s per loop

In [68]: randjs = np.random.randint(0,200, 10)

In [69]: randis = np.random.randint(0,5000,10)

In [70]: genotypes[randis,randjs] = None

In [71]: %timeit testMask(pValues, genotypes)
1 loops, best of 3: 33 s per loop

In [72]: %timeit testNotMask(pValues, genotypes)
1 loops, best of 3: 3.6 s per loop

In [73]: %timeit fastNotMask(pValues, genotypes)
1 loops, best of 3: 3.98 s per loop

In [74]: testMask(pValues, genotypes)
Out[74]: 0.47606794747438386

In [75]: testNotMask(pValues, genotypes)
Out[75]: nan

In [76]: fastNotMask(pValues, genotypes)
Out[76]: 0.47613597091679449

请注意,testMask和fastNotMask之间的精度略有不同.我实际上不确定这是哪里来的,但我将假定它并不重要.