[英]All possible combinations of 2D numpy array
我有四個 numpy arrays,示例如下:
a1=np.array([[-24.4925, 295.77 ],
[-24.4925, 295.77 ],
[-14.3925, 295.77 ],
[-16.4125, 295.77 ],
[-43.6825, 295.77 ],
[-22.4725, 295.77 ]])
a2=np.array([[-26.0075, 309.39 ],
[-24.9975, 309.39 ],
[-14.8975, 309.39 ],
[-17.9275, 309.39 ],
[-46.2075, 309.39 ],
[-23.9875, 309.39 ]])
a3=np.array([[-25.5025, 310.265 ],
[-25.5025, 310.265 ],
[-15.4025, 310.265 ],
[-17.4225, 310.265 ],
[-45.7025, 310.265 ],
[-24.4925, 310.265 ]])
a4=np.array([[-27.0175, 326.895 ],
[-27.0175, 326.895 ],
[-15.9075, 326.895 ],
[-18.9375, 326.895 ],
[-48.2275, 326.895 ],
[-24.9975, 326.895 ]])
我想在 arrays 之間進行所有可能的組合並同時連接,例如:
array[-24.4925, 295.77, -26.0075, 309.39, -25.5025, 310.265, -27.0175, 326.895]
和
array[-24.4925, 295.77, -26.0075, 309.39, -25.5025, 310.265, -27.0175, 326.895]
即[a1[0],a2[0],a3[0],a4[0]]
, [a1[0],a2[0],a3[0],a4[1]]
等等
除了循環四個 arrays 之外,最快的方法是什么?!
好吧,沒有比循環更快的方法了,但是有一種干凈的方法,您不必編寫循環:
import numpy as np
import itertools
a1=np.array([[-24.4925, 295.77 ],
[-24.4925, 295.77 ],
[-14.3925, 295.77 ],
[-16.4125, 295.77 ],
[-43.6825, 295.77 ],
[-22.4725, 295.77 ]])
a2=np.array([[-26.0075, 309.39 ],
[-24.9975, 309.39 ],
[-14.8975, 309.39 ],
[-17.9275, 309.39 ],
[-46.2075, 309.39 ],
[-23.9875, 309.39 ]])
a3=np.array([[-25.5025, 310.265 ],
[-25.5025, 310.265 ],
[-15.4025, 310.265 ],
[-17.4225, 310.265 ],
[-45.7025, 310.265 ],
[-24.4925, 310.265 ]])
a4=np.array([[-27.0175, 326.895 ],
[-27.0175, 326.895 ],
[-15.9075, 326.895 ],
[-18.9375, 326.895 ],
[-48.2275, 326.895 ],
[-24.9975, 326.895 ]])
arrays = [a1, a2, a3, a4]
for pieces in itertools.product(*arrays):
combined = np.concatenate(pieces, axis = 0)
print(combined)
標准庫itertools
模塊( https://docs.python.org/3/library/itertools.html )提供了各種用於生產迭代產品的工具。 由於 numpy 數組恰好是可迭代的(迭代第一個索引),我們可以使用itertools
從每個數組中獲取切片,然后使用 numpy 組合它們。
這是一個numpy
解決方案,基於此處的笛卡爾積實現。
arr = np.stack([a1, a2, a3, a4])
print(arr.shape) # (4, 6, 2)
n, m, k = arr.shape
# from https://stackoverflow.com/questions/11144513/cartesian-product-of-x-and-y-array-points-into-single-array-of-2d-points
def cartesian_product(*arrays):
la = len(arrays)
dtype = np.result_type(*arrays)
arr = np.empty([len(a) for a in arrays] + [la], dtype=dtype)
for i, a in enumerate(np.ix_(*arrays)):
arr[...,i] = a
return arr.reshape(-1, la)
inds = cartesian_product(*([np.arange(m)] * n))
res = np.take_along_axis(arr, inds.T[...,None], 1).swapaxes(0,1).reshape(-1, n*k)
print(res[0])
# [-24.4925 295.77 -26.0075 309.39 -25.5025 310.265 -27.0175 326.895 ]
在此示例中, inds
數組如下所示:
print(inds[:10])
# [[0 0 0 0]
# [0 0 0 1]
# [0 0 0 2]
# [0 0 0 3]
# [0 0 0 4]
# [0 0 0 5]
# [0 0 1 0]
# [0 0 1 1]
# [0 0 1 2]
# [0 0 1 3]]
然后我們可以為每個組合使用np.take_along_axis
到 select 適當的元素。
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