Question

我有一个简单的numpy数组。

array([[10,   0,  10,  0],
       [ 1,   1,   0,  0]
       [ 9,   9,   9,  0]
       [ 0,  10,   1,  0]])

我想分别取这个数组的每列的中位数。

但是，在计算中位数时，我想忽略各个0个值。

为了进一步复杂化，我想将仅包含0条目的列保留为0的中位数。以这种方式，这些列将作为占位符的一部分，保持矩阵的尺寸相同。

numpy文档中没有任何可以满足我想要的参数（也许我被R获得的许多开关所破坏了！）

numpy.median(a, axis=None, out=None, overwrite_input=False)[source]

有人可以说明一个有效的方法来做到这一点，这符合numpy的精神？我可以把它搞砸，但在那种情况下，我觉得我首先打败了使用numpy的目的。

提前致谢。

Answer 1

Masked array总是很方便，但是懒散地说：

In [14]:

%timeit np.ma.median(y, axis=0).filled(0)
1000 loops, best of 3: 1.73 ms per loop
In [15]:

%%timeit
ans=np.apply_along_axis(lambda v: np.median(v[v!=0]), 0, x)
ans[np.isnan(ans)]=0.
1000 loops, best of 3: 402 µs per loop

In [16]:

ans=np.apply_along_axis(lambda v: np.median(v[v!=0]), 0, x)
ans[np.isnan(ans)]=0.; ans
Out[16]:
array([ 9.,  9.,  9.,  0.])

np.nonzero甚至更快：

In [25]:

%%timeit
ans=np.apply_along_axis(lambda v: np.median(v[np.nonzero(v)]), 0, x)
ans[np.isnan(ans)]=0.
1000 loops, best of 3: 384 µs per loop

Answer 2

使用蒙面数组和np.ma.median(axis=0).filled(0)来获取列的中位数。

In [1]: x = np.array([[10, 0, 10, 0], [1, 1, 0, 0], [9, 9, 9, 0], [0, 10, 1, 0]])
In [2]: y = np.ma.masked_where(x == 0, x)
In [3]: x
Out[3]: 
array([[10,  0, 10, 0],
       [ 1,  1,  0, 0],
       [ 9,  9,  9, 0],
       [ 0, 10,  1, 0]])
In [4]: y
Out[4]: 
masked_array(data =
 [[10 -- 10 --]
 [1 1 -- --]
 [9 9 9 --]
 [-- 10 1 --]],
             mask =
 [[False  True False True]
 [False False  True True]
 [False False False True]
 [ True False False True]],
       fill_value = 999999)
In [6]: np.median(x, axis=0)
Out[6]: array([ 5.,  5.,  5., 0.])
In [7]: np.ma.median(y, axis=0).filled(0)
Out[7]: 
array(data = [ 9.  9.  9., 0.])

Answer 3

您可以使用masked arrays。

a = np.array([[10, 0, 10, 0], [1, 1, 0, 0],[9,9,9,0],[0,10,1,0]])
m = np.ma.masked_equal(a, 0)

In [44]: np.median(a)
Out[44]: 1.0

In [45]: np.ma.median(m)
Out[45]: 9.0

In [46]: m
Out[46]:
masked_array(data =
 [[10 -- 10 --]
 [1 1 -- --]
 [9 9 9 --]
 [-- 10 1 --]],
             mask =
 [[False  True False  True]
 [False False  True  True]
 [False False False  True]
 [ True False False  True]],
       fill_value = 0)

Answer 4

这可能会有所帮助。获得非零数组后，您可以直接从[非零（a）]

获得中值

numpy.nonzero

numpy.nonzero的（a）[源]

Return the indices of the elements that are non-zero.

Returns a tuple of arrays, one for each dimension of a, containing the indices of the non-zero elements in that dimension. The corresponding non-zero values can be obtained with:

a[nonzero(a)]

To group the indices by element, rather than dimension, use:

transpose(nonzero(a))

The result of this is always a 2-D array, with a row for each non-zero element.
Parameters :    

a : array_like

    Input array.

Returns :   

tuple_of_arrays : tuple

    Indices of elements that are non-zero.

See also

flatnonzero
    Return indices that are non-zero in the flattened version of the input array.
ndarray.nonzero
    Equivalent ndarray method.
count_nonzero
    Counts the number of non-zero elements in the input array.

Examples

>>> x = np.eye(3)
>>> x
array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])
>>> np.nonzero(x)
(array([0, 1, 2]), array([0, 1, 2]))

>>> x[np.nonzero(x)]
array([ 1.,  1.,  1.])
>>> np.transpose(np.nonzero(x))
array([[0, 0],
       [1, 1],
       [2, 2]])

A common use for nonzero is to find the indices of an array, where a condition is True. Given an array a, the condition a > 3 is a boolean array and since False is interpreted as 0, np.nonzero(a > 3) yields the indices of the a where the condition is true.

>>> a = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> a > 3
array([[False, False, False],
       [ True,  True,  True],
       [ True,  True,  True]], dtype=bool)
>>> np.nonzero(a > 3)
(array([1, 1, 1, 2, 2, 2]), array([0, 1, 2, 0, 1, 2]))

The nonzero method of the boolean array can also be called.

>>> (a > 3).nonzero()
(array([1, 1, 1, 2, 2, 2]), array([0, 1, 2, 0, 1, 2]))

当我在数组的列上取中值时，如何忽略零？

4 个答案: