在numpy中获得线性化指数

Question

我需要模拟MATLAB函数find,它返回数组非零元素的线性索引.例如:

>> a = zeros(4,4)
a =

     0     0     0     0
     0     0     0     0
     0     0     0     0
     0     0     0     0
>> a(1,1) = 1
>> a(4,4) = 1
>> find(a)
ans =

     1
    16

numpy具有类似的功能nonzero,但它返回索引数组的元组.例如:

In [1]: from numpy import *
In [2]: a = zeros((4,4))

In [3]: a[0,0] = 1

In [4]: a[3,3] = 1

In [5]: a
Out[5]: 
array([[ 1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.]])

In [6]: nonzero(a)
Out[6]: (array([0, 3]), array([0, 3]))

是否有一个函数给我线性索引而不自己计算它们？

Answer 1

numpy有你覆盖:

>>> np.flatnonzero(a)
array([ 0, 15])

在内部,它完全正是Sven Marnach所建议的.

>>> print inspect.getsource(np.flatnonzero)
def flatnonzero(a):
    """
    Return indices that are non-zero in the flattened version of a.

    This is equivalent to a.ravel().nonzero()[0].

    [more documentation]

    """
    return a.ravel().nonzero()[0]