我有两个持有2d向量的numpy数组:
import numpy as np
a = np.array([[ 0.999875, 0.015836],
[ 0.997443, 0.071463],
[ 0.686554, 0.727078],
[ 0.93322 , 0.359305]])
b = np.array([[ 0.7219 , 0.691997],
[ 0.313656, 0.949537],
[ 0.507926, 0.861401],
[ 0.818131, 0.575031],
[ 0.117956, 0.993019]])
如您所见,a.shape
为(4,2),b.shape
为(5,2)。
现在,我可以得到我想要的结果:
In [441]: np.array([np.cross(av, bv) for bv in b for av in a]).reshape(5, 4)
Out[441]:
array([[ 0.680478, 0.638638, -0.049784, 0.386403],
[ 0.944451, 0.924694, 0.423856, 0.773429],
[ 0.85325 , 0.8229 , 0.222097, 0.621377],
[ 0.562003, 0.515094, -0.200055, 0.242672],
[ 0.991027, 0.982051, 0.595998, 0.884323]])
我的问题是:什么是获得上述内容的更“'numpythonic'方式(即没有嵌套列表推导)?我已经尝试了np.cross()
我能想到的每一个组合,我通常会得到这样的结果:
In [438]: np.cross(a, b.T, axisa=1, axisb=0)
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-438-363c0765a7f9> in <module>()
----> 1 np.cross(a, b.T, axisa=1, axisb=0)
D:\users\ae4652t\Python27\lib\site-packages\numpy\core\numeric.p<snipped>
1242 if a.shape[0] == 2:
1243 if (b.shape[0] == 2):
-> 1244 cp = a[0]*b[1] - a[1]*b[0]
1245 if cp.ndim == 0:
1246 return cp
ValueError: operands could not be broadcast together with shapes (4) (5)
答案 0 :(得分:8)
我对此有所了解。
>>> a
array([[ 0.999875, 0.015836],
[ 0.997443, 0.071463],
[ 0.686554, 0.727078],
[ 0.93322 , 0.359305]])
>>> b
array([[ 0.7219 , 0.691997],
[ 0.313656, 0.949537],
[ 0.507926, 0.861401],
[ 0.818131, 0.575031],
[ 0.117956, 0.993019]])
>>> c = np.tile(a, (b.shape[0], 1))
>>> d = np.repeat(b, a.shape[0], axis=0)
>>> np.cross(c, d).reshape(5,4)
array([[ 0.68047849, 0.63863842, -0.0497843 , 0.38640316],
[ 0.94445125, 0.92469424, 0.42385605, 0.77342875],
[ 0.85324981, 0.82290048, 0.22209648, 0.62137629],
[ 0.5620032 , 0.51509455, -0.20005522, 0.24267187],
[ 0.99102692, 0.98205036, 0.59599795, 0.88432301]])
一些时间:
import timeit
s="""
import numpy as np
a=np.random.random(100).reshape(-1, 2)
b=np.random.random(1000).reshape(-1, 2)
"""
ophion="""
np.cross(np.tile(a,(b.shape[0],1)),np.repeat(b,a.shape[0],axis=0))"""
subnivean="""
np.array([np.cross(av, bv) for bv in b for av in a]).reshape(b.shape[0], a.shape[0])"""
DSM="""
np.outer(b[:,1], a[:,0]) - np.outer(b[:,0], a[:,1])"""
Jamie="""
np.cross(a[None], b[:, None, :])"""
h=timeit.timeit(subnivean,setup=s,number=10)
m=timeit.timeit(ophion,setup=s,number=10)
d=timeit.timeit(DSM,setup=s,number=10)
j=timeit.timeit(Jamie,setup=s,number=10)
print "subnivean's method took",h,'seconds.'
print "Ophion's method took",m,'seconds.'
print "DSM's method took",d,'seconds.'
"
subnivean's method took 1.99507117271 seconds.
Ophion's method took 0.0149450302124 seconds.
DSM's method took 0.0040500164032 seconds.
Jamie's method took 0.00390195846558 seconds."
当长度a = 10且b = 100时:
"
subnivean's method took 0.0217308998108 seconds.
Ophion's method took 0.00046181678772 seconds.
DSM's method took 0.000531911849976 seconds.
Jamie's method took 0.000334024429321 seconds."
嗯,你再次改变了交叉产品的顺序,如果你想要(5,4)或(4,5),就会显示两个答案。
答案 1 :(得分:7)
我还没有计算我的代码,但我几乎可以肯定没有更多 numpythonic 这样做的方式而不是简单明了:
>>> np.cross(a[None], b[:, None])
array([[ 0.68047849, 0.63863842, -0.0497843 , 0.38640316],
[ 0.94445125, 0.92469424, 0.42385605, 0.77342875],
[ 0.85324981, 0.82290048, 0.22209648, 0.62137629],
[ 0.5620032 , 0.51509455, -0.20005522, 0.24267187],
[ 0.99102692, 0.98205036, 0.59599795, 0.88432301]])
Broadcasting总是答案......