Following the example here, I am able to find the column indices of a 2D numpy array and get back an array of column indices of all occurrences of the max value.
按照这里的示例,我能够找到2D numpy数组的列索引,并返回所有出现的max值的列索引数组。
But now I want to do the same thing but on a sparse csr_matrix.
但现在我想在稀疏的csr_matrix上做同样的事情。
x = np.array([[0,0,1,0,0,0,2],[0,0,0,4,0,0,0],[0,9,1,0,0,0,2],[0,0,1,0,0,9,2]])
max_col_inds = np.argwhere(x == np.max(x))[:,1]
# array([1, 5], dtype=int64)
Then I want to get the 1st and 5th elements of a 1D array using that result:
然后我想使用该结果获得1D数组的第1和第5个元素:
words[max_col_inds]
If x is a 2D numpy array and words is a 1D numpy array, this works.
如果x是2D numpy数组,而单词是1D numpy数组,则可行。
But now if I replace x with a scipy.sparse.csr.csr_matrix, I get this on the call to np.argwhere():
但是现在如果我用scipy.sparse.csr.csr_matrix替换x,我会在调用np.argwhere()时得到这个:
TypeError: tuple indices must be integers, not tuple
1 个解决方案
#1
1
In [804]: x = np.array([[0,0,1,0,0,0,2],[0,0,0,4,0,0,0],[0,9,1,0,0,0,2],[0,0,1,0,0,9,2]])
In [805]: np.max(x)
Out[805]: 9
In [806]: np.where(x == 9)
Out[806]: (array([2, 3], dtype=int32), array([1, 5], dtype=int32))
argwhere is just np.transpose(np.where(...)); that is, converts the tuple into a 2d array and transposes it:
argwhere只是np.transpose(np.where(...));也就是说,将元组转换为2d数组并对其进行转置:
In [807]: np.argwhere(x ==9)
Out[807]:
array([[2, 1],
[3, 5]], dtype=int32)
Doing the same thing with sparse
用稀疏做同样的事情
In [808]: xM = sparse.csr_matrix(x)
In [809]: xM == 9
Out[809]:
<4x7 sparse matrix of type '<class 'numpy.bool_'>'
with 2 stored elements in Compressed Sparse Row format>
np.where is the samething as np.nonzero:
np.where和np.nonzero一样:
In [810]: (xM==9).nonzero()
Out[810]: (array([2, 3], dtype=int32), array([1, 5], dtype=int32))
In [811]: np.transpose((xM==9).nonzero())
Out[811]:
array([[2, 1],
[3, 5]], dtype=int32)
Actually in the current numpy argwhere works with sparse. That's because np.nonzero delegates to the matrix method:
实际上在目前的numpy argwhere工作与稀疏。那是因为np.nonzero委托给矩阵方法:
In [813]: np.argwhere(xM==9)
Out[813]:
array([[2, 1],
[3, 5]], dtype=int32)
#1
1
In [804]: x = np.array([[0,0,1,0,0,0,2],[0,0,0,4,0,0,0],[0,9,1,0,0,0,2],[0,0,1,0,0,9,2]])
In [805]: np.max(x)
Out[805]: 9
In [806]: np.where(x == 9)
Out[806]: (array([2, 3], dtype=int32), array([1, 5], dtype=int32))
argwhere is just np.transpose(np.where(...)); that is, converts the tuple into a 2d array and transposes it:
argwhere只是np.transpose(np.where(...));也就是说,将元组转换为2d数组并对其进行转置:
In [807]: np.argwhere(x ==9)
Out[807]:
array([[2, 1],
[3, 5]], dtype=int32)
Doing the same thing with sparse
用稀疏做同样的事情
In [808]: xM = sparse.csr_matrix(x)
In [809]: xM == 9
Out[809]:
<4x7 sparse matrix of type '<class 'numpy.bool_'>'
with 2 stored elements in Compressed Sparse Row format>
np.where is the samething as np.nonzero:
np.where和np.nonzero一样:
In [810]: (xM==9).nonzero()
Out[810]: (array([2, 3], dtype=int32), array([1, 5], dtype=int32))
In [811]: np.transpose((xM==9).nonzero())
Out[811]:
array([[2, 1],
[3, 5]], dtype=int32)
Actually in the current numpy argwhere works with sparse. That's because np.nonzero delegates to the matrix method:
实际上在目前的numpy argwhere工作与稀疏。那是因为np.nonzero委托给矩阵方法:
In [813]: np.argwhere(xM==9)
Out[813]:
array([[2, 1],
[3, 5]], dtype=int32)