我有一个python函数，它接受两个列表，在两个输入中查找两个在同一索引处具有正值的对，并通过附加到这两个正值中的每一个来创建两个输出列表。我有一个工作职能：
def get_pairs_in_first_quadrant(x_in, y_in):
"""If both x_in[i] and y_in[i] are > 0 then both will appended to the output list. If either are negative
then the pair of them will be absent from the output list.
:param x_in: A list of positive or negative floats
:param y_in: A list of positive or negative floats
:return: A list of positive floats <= in length to the inputs.
"""
x_filtered, y_filtered = [], []
for x, y in zip(x_in, y_in):
if x > 0 and y > 0:
x_filtered.append(x)
y_filtered.append(y)
return x_filtered, y_filtered
如何使用numpy加快速度？
答案 0 :(得分：3)
你可以通过简单地找到它们都是正面的指数来做到这一点：
import numpy as np
a = np.random.random(10) - .5
b = np.random.random(10) - .5
def get_pairs_in_first_quadrant(x_in, y_in):
i = np.nonzero( (x_in>0) & (y_in>0) ) # main line of interest
return x_in[i], y_in[i]
print a # [-0.18012451 -0.40924713 -0.3788772 0.3186816 0.14811581 -0.04021951 -0.21278312 -0.36762629 -0.45369899 -0.46374929]
print b # [ 0.33005969 -0.03167875 0.11387641 0.22101336 0.38412264 -0.3880842 0.08679424 0.3126209 -0.08760505 -0.40921421]
print get_pairs_in_first_quadrant(a, b) # (array([ 0.3186816 , 0.14811581]), array([ 0.22101336, 0.38412264]))
<小时/> 我对Jaime建议在不调用
nonzero的情况下使用布尔索引感兴趣，所以我运行了一些时序测试。结果有点令人感兴趣，因为它们的优势比与正匹配的数量是非单调的，但基本上，至少对于速度而言，使用哪一个并不重要（尽管nonzero通常有点更快，可以快两倍）：
threshold = .6
a = np.random.random(10000) - threshold
b = np.random.random(10000) - threshold
def f1(x_in, y_in):
i = np.nonzero( (x_in>0) & (y_in>0) ) # main line of interest
return x_in[i], y_in[i]
def f2(x_in, y_in):
i = (x_in>0) & (y_in>0) # main line of interest
return x_in[i], y_in[i]
print threshold, len(f1(a,b)[0]), len(f2(a,b)[0])
print timeit("f1(a, b)", "from __main__ import a, b, f1, f2", number = 1000)
print timeit("f2(a, b)", "from __main__ import a, b, f1, f2", number = 1000)
对于不同的阈值，它给出了：
0.05 9086 9086
0.0815141201019
0.104746818542
0.5 2535 2535
0.0715141296387
0.153401851654
0.95 21 21
0.027126789093
0.0324990749359