Python numpy filter two-dimensional array by condition
You can use a bool
index array that you can produce using np.in1d
.
You can index a np.ndarray
along any axis
you want using for example an array of bool
s indicating whether an element should be included. Since you want to index along axis=0
, meaning you want to choose from the outest index, you need to have 1D np.array
whose length is the number of rows. Each of its elements will indicate whether the row should be included.
A fast way to get this is to use np.in1d
on the second column of a
. You get all elements of that column by a[:, 1]
. Now you have a 1D np.array
whose elements should be checked against your filter. Thats what np.in1d
is for.
So the complete code would look like:
import numpy as np
a = np.asarray([[2,'a'],[3,'b'],[4,'c'],[5,'d']])
filter = np.asarray(['a','c'])
a[np.in1d(a[:, 1], filter)]
or in a longer form:
import numpy as np
a = np.asarray([[2,'a'],[3,'b'],[4,'c'],[5,'d']])
filter = np.asarray(['a','c'])
mask = np.in1d(a[:, 1], filter)
a[mask]
A somewhat elaborate pure numpy
vectorized solution:
>>> import numpy
>>> a = numpy.asarray([[2,'a'],[3,'b'],[4,'c'],[5,'d']])
>>> filter = numpy.array(['a','c'])
>>> a[(a[:,1,None] == filter[None,:]).any(axis=1)]
array([['2', 'a'],
['4', 'c']],
dtype='|S21')
None
in the index creates a singleton dimension, therefore we can compare the column of a
and the row of filter
, and then reduce the resulting boolean array
>>> a[:,1,None] == filter[None,:]
array([[ True, False],
[False, False],
[False, True],
[False, False]], dtype=bool)
over the second dimension with any
.