问题
I have a simple numpy array.
array([[10, 0, 10, 0],
[ 1, 1, 0, 0]
[ 9, 9, 9, 0]
[ 0, 10, 1, 0]])
I would like to take the median of each column, individually, of this array.
However, there are a few 0 values in various places which I would like to ignore in the calculation of the medians.
To further complicate, I would like to keep the columns with only 0 entries as having the median of 0. In this manner, those columns would serve as a bit of a place holder, keeping the dimensions of the matrix the same.
The numpy documentation doesn't have any argument that would work for what I want (maybe I am spoiled by the many switches we get with R!)
numpy.median(a, axis=None, out=None, overwrite_input=False)[source]
Can someone please shed some light on an effective way to do this, which is in line with the spirit of numpy? I could hack it out but in that case I feel like I've defeated the purpose of using numpy in the first place.
Thanks in advance.
回答1:
Use masked arrays and np.ma.median(axis=0).filled(0) to get the medians of the columns.
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.])
回答2:
Masked array is always handy, but slooooooow:
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 is even faster:
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
回答3:
You can use 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)
回答4:
This may help. Once you get the nonzero array, you can obtain the median directly from a[nonzero(a)]
numpy.nonzero
numpy.nonzero(a)[source]
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]))
来源：https://stackoverflow.com/questions/22049140/how-can-i-ignore-zeros-when-i-take-the-median-on-columns-of-an-array