ndarray每行中有N个最大值

wal*_*ol1 4 arrays algorithm numpy python-2.7

我有一个ndarray,每行是一个单独的直方图.对于每一行,我希望找到前N个值.

我知道全局前N个值的解决方案(在numpy数组中找到最大N个元素的快速方法),但我没有看到如何获得每行的前N个.

我可以遍历每一行并应用1D解决方案,但我不能用numpy广播做到这一点吗?

xnx*_*xnx 8

您可以使用np.partition与链接的问题相同的方式:排序已经沿着最后一个轴:

In [2]: array([[ 5,  4,  3,  2,  1],
               [10,  9,  8,  7,  6]])
In [3]: b = np.partition(a, 3)    # top 3 values from each row
In [4]: b[:,-3:]
Out[4]: 
array([[ 3,  4,  5],
       [ 8,  9, 10]])
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Div*_*kar 6

您可以像这样np.argsort沿着行使用axis = 1-

import numpy as np

# Find sorted indices for each row
sorted_row_idx = np.argsort(A, axis=1)[:,A.shape[1]-N::]

# Setup column indexing array
col_idx = np.arange(A.shape[0])[:,None]

# Use the column-row indices to get specific elements from input array. 
# Please note that since the column indexing array isn't of the same shape 
# as the sorted row indices, it will be broadcasted
out = A[col_idx,sorted_row_idx]
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样品运行 -

In [417]: A
Out[417]: 
array([[0, 3, 3, 2, 5],
       [4, 2, 6, 3, 1],
       [2, 1, 1, 8, 8],
       [6, 6, 3, 2, 6]])

In [418]: N
Out[418]: 3

In [419]: sorted_row_idx = np.argsort(A, axis=1)[:,A.shape[1]-N::]

In [420]: sorted_row_idx
Out[420]: 
array([[1, 2, 4],
       [3, 0, 2],
       [0, 3, 4],
       [0, 1, 4]], dtype=int64)

In [421]: col_idx = np.arange(A.shape[0])[:,None]

In [422]: col_idx
Out[422]: 
array([[0],
       [1],
       [2],
       [3]])

In [423]: out = A[col_idx,sorted_row_idx]

In [424]: out
Out[424]: 
array([[3, 3, 5],
       [3, 4, 6],
       [2, 8, 8],
       [6, 6, 6]])
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如果您希望元素按降序排列,可以使用此附加步骤 -

In [425]: out[:,::-1]
Out[425]: 
array([[5, 3, 3],
       [6, 4, 3],
       [8, 8, 2],
       [6, 6, 6]])
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