How Can I Ignore Zeros When I Take The Median On Columns Of An Array?

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 co

Solution 1:

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

Solution 2:

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 --]
 [11-- --]
 [999--]
 [-- 10 1 --]],
             mask =
 [[FalseTrueFalseTrue]
 [FalseFalseTrueTrue]
 [FalseFalseFalseTrue]
 [ TrueFalseFalseTrue]],
       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.])

Solution 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]: 1.0In [45]: np.ma.median(m)
Out[45]: 9.0In [46]: m
Out[46]:
masked_array(data =
 [[10-- 10 --]
 [11-- --]
 [999--]
 [-- 10 1 --]],
             mask =
 [[FalseTrueFalseTrue]
 [FalseFalseTrueTrue]
 [FalseFalseFalseTrue]
 [ TrueFalseFalseTrue]],
       fill_value =0)

Solution 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, oneforeach 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)]

Togroup the indices by element, rather than dimension, use:

transpose(nonzero(a))

The resultof this is always a 2-D array, with a rowforeach 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 isto find the indices of an array, where a conditionis True. Given an array a, the condition a >3is a booleanarrayand since Falseis interpreted as0, np.nonzero(a >3) yields the indices of the a where the conditionis true.

>>> a = np.array([[1,2,3],[4,5,6],[7,8,9]])
>>> a >3array([[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 methodof the booleanarray can also be called.

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

Solution 5:

I perefer to use

# replace 0.0 with nan to exclude 0.0 from medianzero_to_nan = numpy.where(a == 0.0, numpy.nan, a)
n = numpy.nanmedian(zero_to_nan, ....)