假设我有两个形状为(1.000.000,??,50(,(???=见下文(的三维数组(a和b(。
如何将它们合并,因此结果将是(1.000.000,{a的形状+b的第二维度},50(?
以下是示例,如下所示:(np.arrays也是可能的(
编辑:添加可用代码,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]
]
]
你知道一种有效的方法吗?
编辑:尝试连接(不起作用(:
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
维度中不能有长度可变的数组。CCD_ 1和CCD_。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)])
EDIT:如果a
和b
是数组列表,则基于提供的注释:
np.array([np.vstack((x,y)) for x,y in zip(a,b)])
示例的输出如下:
[[[ 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.]]]
要执行串联,请运行:
result = np.concatenate([a, b], axis=1)
为了测试此代码,我将a和b创建为:
a = np.stack([ np.full((2, 3), i) for i in range(1, 6)], axis=1)
b = np.stack([ np.full((2, 3), i + 10) for i in range(1, 4)], axis=1)
所以它们包含:
array([[[1, 1, 1], array([[[11, 11, 11],
[2, 2, 2], [12, 12, 12],
[3, 3, 3], [13, 13, 13]],
[4, 4, 4],
[5, 5, 5]], [[11, 11, 11],
[12, 12, 12],
[[1, 1, 1], [13, 13, 13]]])
[2, 2, 2],
[3, 3, 3],
[4, 4, 4],
[5, 5, 5]]])
其形状为:(2,5,3(和
我的串联结果是:
array([[[ 1, 1, 1],
[ 2, 2, 2],
[ 3, 3, 3],
[ 4, 4, 4],
[ 5, 5, 5],
[11, 11, 11],
[12, 12, 12],
[13, 13, 13]],
[[ 1, 1, 1],
[ 2, 2, 2],
[ 3, 3, 3],
[ 4, 4, 4],
[ 5, 5, 5],
[11, 11, 11],
[12, 12, 12],
[13, 13, 13]]])
并且形状是(2,8,3(,正如它应该的那样。
在19:56Z的评论后编辑
我试过了你评论中的代码。在执行CCD_ 5之后,结果是:
[array([[1, 1]], dtype=int64),
array([[1, 1],
[1, 1]], dtype=int64),
array([[1, 1],
[1, 1],
[1, 1]], dtype=int64),
array([[1, 1],
[1, 1],
[1, 1],
[1, 1]], dtype=int64),
array([[1, 1],
[1, 1],
[1, 1],
[1, 1],
[1, 1]], dtype=int64),
...
因此a是大小不等的数组的列表。
你构建数据的方式有问题。首先检查您的两个数组是否为3-D,并且它们的形状是否不同仅在轴1中。只有这样你才能在它们上运行我的代码。目前,a和b都是普通的蟒蛇列表,而不是Numpy数组!