python - 如何有效地合并二维上的两个 3d 数组?
问题描述
假设我有两个形状为 (1.000.000, ???, 50), (??? = 见下文) 的 3 维数组 (a & b)。
如何合并它们,使结果为 (1.000.000, {shape of a's + b's second dimension} , 50)?
以下是示例,如下所示:(np.arrays 也是可能的)
编辑:添加了可用的代码,请滚动^^
[ #a
[
],
[
[1 2 3]
],
[
[0 2 7]
[1 Nan 3]
],
[
[10 0 3]
[NaN 9 9]
[10 NaN 3]
],
[
[8 2 0]
[2 2 3]
[8 1 3]
[1 2 3]
],
[
[0 2 3]
[1 2 9]
[1 2 3]
[1 0 3]
[1 2 3]
]
]
[#b
[
[7 2 3]
[1 2 9]
[1 2 3]
[8 0 3]
[1 7 3]
]
[
[3 9 0]
[2 2 3]
[8 1 3]
[0 2 3]
],
[
[10 0 3]
[0 NaN 9]
[10 NaN 3]
],
[
[0 2 NaN]
[1 Nan 3]
],
[
[1 2 NaN]
],
[
]
]
a = [ [ ],
[ [1, 2, 3] ],
[ [0, 2, 7], [1,np.nan,3] ],
[
[10,0,3], [np.nan,9,9], [10,np.nan,3]
],
[
[8,2,0], [2,2,3], [8,1,3], [1,2,3]
],
[
[0,2,3], [1,2,9], [1,2,3], [1,0,3], [1,2,3]
]
]
b = [
[
[7,2,3], [1,2,9], [1,2,3], [8,0,3], [1,7,3]
],
[
[3,9,0], [2,2,3], [8,1,3], [0,2,3]
],
[
[10,0,3], [0,np.nan,9], [10,np.nan,3]
],
[
[0,2,np.nan], [1,np.nan,3]
],
[
[1,2,np.nan]
],
[
]
]
预期结果:
[
[ [7 2 3]# from b
[1 2 9]# from b
[1 2 3]# from b
[8 0 3]# from b
[1 7 3]# from b
],
[
[1 2 3]
[3 9 0]# from b
[2 2 3]# from b
[8 1 3]# from b
[0 2 3]# from b
],
[
[0 2 7]
[1 Nan 3]
[10 0 3]# from b
[0 NaN 9]# from b
[10 NaN 3]# from b
],
[
[10 0 3]
[NaN 9 9]
[10 NaN 3]
[0 2 NaN]# from b
[1 Nan 3]# from b
],
[
[8 2 0]
[2 2 3]
[8 1 3]
[1 2 3]
[1 2 NaN]# from b
],
[
[0 2 3]
[1 2 9]
[1 2 3]
[1 0 3]
[1 2 3]
]
]
你知道一种有效地做到这一点的方法吗?
编辑:尝试连接(没有工作):
DF_LEN, COL_LEN, cols = 20,5,['A', 'B']
a = np.asarray(pd.DataFrame(1, index=range(DF_LEN), columns=cols))
a = list((map(lambda i: a[:i], range(1,a.shape[0]+1))))
b = np.asarray(pd.DataFrame(np.nan, index=range(DF_LEN), columns=cols))
b = list((map(lambda i: b[:i], range(1,b.shape[0]+1))))
b = b[::-1]
a_first = a[0]; del a[0]
b_last = b[-1]; del b[-1]
result = np.concatenate([a, b], axis=1)
>>>AxisError: axis 1 is out of bounds for array of dimension 1
解决方案
您不能在维度中拥有可变长度的数组。a
并且b
很可能是列表列表而不是数组。您可以将列表理解与 zip 一起使用:
np.array([x+y for x,y in zip(a,b)])
编辑:或基于提供的评论,如果a
和b
是数组列表:
np.array([np.vstack((x,y)) for x,y in zip(a,b)])
您的示例的输出如下所示:
[[[ 7. 2. 3.]
[ 1. 2. 9.]
[ 1. 2. 3.]
[ 8. 0. 3.]
[ 1. 7. 3.]]
[[ 1. 2. 3.]
[ 3. 9. 0.]
[ 2. 2. 3.]
[ 8. 1. 3.]
[ 0. 2. 3.]]
[[ 0. 2. 7.]
[ 1. nan 3.]
[10. 0. 3.]
[ 0. nan 9.]
[10. nan 3.]]
[[10. 0. 3.]
[nan 9. 9.]
[10. nan 3.]
[ 0. 2. nan]
[ 1. nan 3.]]
[[ 8. 2. 0.]
[ 2. 2. 3.]
[ 8. 1. 3.]
[ 1. 2. 3.]
[ 1. 2. nan]]
[[ 0. 2. 3.]
[ 1. 2. 9.]
[ 1. 2. 3.]
[ 1. 0. 3.]
[ 1. 2. 3.]]]
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