Question

我有一组数据框，每列有4列和1,000,000行。对于每一行，我想运行超几何测试，将这些列的4个值作为输入并返回p值（使用超几何分布的累积概率密度函数）。

我已经尝试了两种基于SciPy的实现（下面），但两者都严重缩放。有没有其他方法可以提高我的效率？我有一个用R编写的工作解决方案（在底部）但不幸的是代码必须用Python编写，因为它将在Airflow任务中使用，该任务从Postgres数据库加载数据，此时R没有Postgres挂钩。

慢速SciPy实施

这样创建的样本数据（使用10,000行而不是完整的52 * 1,000,000行）：

import numpy as np
import pandas as pd
from scipy.stats import hypergeom
from timeit import default_timer as timer

n_rows = 10000
n_total = 1000
max_good = 400
max_sample = 200
s = 100

df = pd.DataFrame({
  'ngood': np.random.hypergeometric(ngood=max_good, nbad=n_total - max_good,
                                    nsample=s, size=n_rows),
  'nsamp': np.random.hypergeometric(ngood=max_sample, nbad=n_total - max_sample,
                                    nsample=s, size=n_rows)
})

df = df.assign(kgood=np.array([
    np.random.hypergeometric(ngood=ngood, nbad=n_total - ngood,
                             nsample=nsamp, size=1)
    for ngood, nsamp
    in zip(df.ngood, df.nsamp)
]))

基于理解的缓慢实现：

start = timer()
res = [
    hypergeom.cdf(k=ngood_found, M=n_total, n=ngood, N=nsamp)
    for ngood_found, ngood, nsamp
    in zip(df.kgood, df.ngood, df.nsamp)
]
end = timer()
print(res[0:10])
print("Elapsed time: %fs" % (end - start))

[0.44247900002512713, 0.71587318053768023, 0.97215178135616498]
Elapsed time: 2.405838s

基于numpy矢量化的缓慢实现：

vectorized_test = np.vectorize(hypergeom.cdf, otypes=[np.float], excluded='M')
start = timer()
res = vectorized_test(k=df.kgood.values, M=n_total,
                      n=df.ngood.values, N=df.nsamp.values)
end = timer()
print(res[0:10])
print("Elapsed time: %fs" % (end - start))

[ 0.442479    0.71587318  0.97215178]
Elapsed time: 2.518952s

快速实施

这表明上述计算可以在几毫秒内完成。诀窍是phyper在C级上被矢量化，而不是基本上是python循环AFAIK的numpy矢量化。

library(microbenchmark)

n_rows <- 10000
n_total <- 1000
max_good <- 400
max_sample <- 200
s <- 100

df <- data.frame(
  ngood = rhyper(nn=n_rows, m=max_good, n=n_total - max_good, k=s),
  nsamp = rhyper(nn=n_rows, m=max_sample, n=n_total - max_sample, k=s)
)

df$kgood <- rhyper(nn=n_rows, m=df$ngood, n=n_total - df$ngood, k=df$nsamp)

microbenchmark(
  res <- phyper(q = df$k, m = df$ngood, n = n_total - df$ngood, k=df$nsamp)
)

Unit: milliseconds
                                                                 expr      
 phyper(q = df$k, m = df$ngood, n = n_total - df$ngood, k = df$nsamp) 

     min      lq     mean   median       uq      max neval
2.984852 3.00838 3.350509 3.134745 3.439138 5.462694   100

Answer 1

通过将hypergeom.cdf的结果缓存为：

，可以获得小的改进

from functools import lru_cache

#@lru_cache(maxsize = 16*1024)
#def fn(k, n, N):
#    return hypergeom.cdf(k = k, M=n_total, n = n, N = N)

data = {}
def fn(k, n, N):
    key = (k, n, N)
    if not key in data:
        val = hypergeom.cdf(k = k, M=n_total, n = n, N = N)
        data[key] = val
    else:
        val = data[key]
    return val

start = timer()
res = [
    fn(ngood_found, ngood, nsamp)
    for ngood_found, ngood, nsamp
    in zip(df.kgood, df.ngood, df.nsamp)
]

end = timer()
print(res[0:10])
print("Elapsed time: %fs" % (end - start))

这在我的机器上显示：Elapsed time: 0.279891s（0.315840s与lru_cache）

修改：

事实上，似乎瓶颈似乎是超几何CDF本身的计算（而不是for循环的开销）。为了测试这一点，我从GSL包中为函数gsl_cdf_hypergeometric_P创建了一个SWIG文件_cdf.i。

%module cdf %{ #include "gsl/gsl_cdf.h" %} double gsl_cdf_hypergeometric_P(int, int, int, int);

然后转换此文件＆＃34;进入包裹：

swig -c++ -python _cdf.i g++ -fPIC -c _cdf_wrap.c -I${HOME}/venvs/p3/include/python3.5m g++ -shared _cdf_wrap.o -o _cdf.so -lgsl

然后可以在原始示例中直接使用它：

import numpy as np import pandas as pd from scipy.stats import hypergeom from timeit import default_timer as timer from cdf import gsl_cdf_hypergeometric_P n_rows = 10000 n_total = 1000 max_good = 400 max_sample = 200 s = 100 df = pd.DataFrame({ 'ngood': np.random.hypergeometric(ngood=max_good, nbad=n_total - max_good, nsample=s, size=n_rows), 'nsamp': np.random.hypergeometric(ngood=max_sample, nbad=n_total - max_sample, nsample=s, size=n_rows) }) df = df.assign(kgood=np.array([ np.random.hypergeometric(ngood=ngood, nbad=n_total - ngood, nsample=nsamp, size=1) for ngood, nsamp in zip(df.ngood, df.nsamp) ])) start = timer() res = [ hypergeom.cdf(k=ngood_found, M=n_total, n=ngood, N=nsamp) for ngood_found, ngood, nsamp in zip(df.kgood, df.ngood, df.nsamp) ] end = timer() print(res[0:10]) print("Elapsed time: %fs" % (end - start)) def cdf(k, M, n, N): return gsl_cdf_hypergeometric_P(int(k), int(n), int(M-n), int(N)) start = timer() res = [ cdf(k=ngood_found, M=n_total, n=ngood, N=nsamp) for ngood_found, ngood, nsamp in zip(df.kgood, df.ngood, df.nsamp) ] end = timer() print(res[0:10]) print("Elapsed time: %fs" % (end - start))

这会产生：

[0.58605423287644209, 0.38055520197355552, 0.70597920363472055, 0.99728041338849138, 0.79797439957395955, 0.42245057292366844, 0.58627644982763727, 0.74819471224742817, 0.75121042470714849, 0.48561471798885397] Elapsed time: 2.069916s [0.5860542328771666, 0.38055520197384757, 0.7059792036350717, 0.997280413389543, 0.7979743995750694, 0.4224505729249291, 0.5862764498272103, 0.7481947122472634, 0.7512104247082603, 0.4856147179890127] Elapsed time: 0.018253s

因此，即使使用普通的for循环，速度也非常快。

如何有效地矢量化超几何CDF计算？

慢速SciPy实施

快速实施

1 个答案: