Numpy重复2d数组

Adi*_*369 5 python arrays numpy vectorization

比如两个阵列

arr = array([10, 24, 24, 24,  1, 21,  1, 21,  0,  0], dtype=int32)
rep = array([3, 2, 2, 0, 0, 0, 0, 0, 0, 0], dtype=int32)
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np.repeat(arr,rep)返回

array([10, 10, 10, 24, 24, 24, 24], dtype=int32)
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有没有办法为一组2D数组复制此功能?

这是给定的

arr = array([[10, 24, 24, 24,  1, 21,  1, 21,  0,  0],
            [10, 24, 24,  1, 21,  1, 21, 32,  0,  0]], dtype=int32)
rep = array([[3, 2, 2, 0, 0, 0, 0, 0, 0, 0],
            [2, 2, 2, 0, 0, 0, 0, 0, 0, 0]], dtype=int32)
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是否可以创建一个矢量化的函数?

PS:每行中的重复次数不必相同.我填充每个结果行以确保它们具有相同的大小.

def repeat2d(arr, rep):
    # Find the max length of repetitions in all the rows. 
    max_len = rep.sum(axis=-1).max()  
    # Create a common array to hold all results. Since each repeated array will have 
    # different sizes, some of them are padded with zero.
    ret_val = np.empty((arr.shape[0], maxlen))  
    for i in range(arr.shape[0]):
        # Repeated array will not have same num of cols as ret_val.
        temp = np.repeat(arr[i], rep[i])
        ret_val[i,:temp.size] = temp
    return ret_val 
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我知道np.vectorize,我知道它不会比普通版本带来任何性能优势.

hpa*_*ulj 4

那么每行都有不同的重复数组?但每行的总重复次数是相同的吗?

只需repeat对展平的数组执行此操作,然后将其重新调整回正确的行数即可。

In [529]: np.repeat(arr,rep.flat)
Out[529]: array([10, 10, 10, 24, 24, 24, 24, 10, 10, 24, 24, 24, 24,  1])
In [530]: np.repeat(arr,rep.flat).reshape(2,-1)
Out[530]: 
array([[10, 10, 10, 24, 24, 24, 24],
       [10, 10, 24, 24, 24, 24,  1]])
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如果每行的重复次数不同,我们就会遇到填充可变长度行的问题。这在其他问题中也出现过。我不记得所有细节,但我认为解决方案是这样的:

进行更改rep以使数字不同:

In [547]: rep
Out[547]: 
array([[3, 2, 2, 0, 0, 0, 0, 0, 0, 0],
       [2, 2, 2, 1, 0, 2, 0, 0, 0, 0]])
In [548]: lens=rep.sum(axis=1)
In [549]: lens
Out[549]: array([7, 9])
In [550]: m=np.max(lens)
In [551]: m
Out[551]: 9
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创建目标:

In [552]: res = np.zeros((arr.shape[0],m),arr.dtype)
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创建一个索引数组 - 需要解决细节:

In [553]: idx=np.r_[0:7,m:m+9]
In [554]: idx
Out[554]: array([ 0,  1,  2,  3,  4,  5,  6,  9, 10, 11, 12, 13, 14, 15, 16, 17])
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平面索引分配:

In [555]: res.flat[idx]=np.repeat(arr,rep.flat)
In [556]: res
Out[556]: 
array([[10, 10, 10, 24, 24, 24, 24,  0,  0],
       [10, 10, 24, 24, 24, 24,  1,  1,  1]])
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