9

2d ベクトルを保持する numpy 配列が 2 つあります。

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]])

ご覧のとおり、a.shape(4,2) とb.shape(5,2) です。

これで、次のようにして必要な結果を得ることができます。

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]])

私の質問は次のとおりです:上記を取得するより「numpythonic」な方法は何ですか(つまり、ネストされたリスト内包表記なし)? 考えられるすべての組み合わせを試しましたがnp.cross()、通常は次のような結果が得られます。

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)
4

2 に答える 2

8

これについてもう少し考えました。

>>> 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]])

いくつかのタイミング：

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."

a=10およびb=100の長さの場合：

"
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."

うーん、外積の順序をもう一度切り替えました。（5,4）または（4,5）が必要な場合は、両方の回答が表示されます。

于 2013-01-06T00:40:12.923 に答える
7

私は自分のコードの時間を計っていませんが、これを行うためのnumpythonicの方法は、素晴らしくシンプルな方法よりも少ないとほぼ確信しています。

>>> 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]])

放送は常に答えです...

于 2013-01-06T03:46:24.360 に答える