How do I select elements of an array given condition?
Solution 1:
Your expression works if you add parentheses:
>>> y[(1 < x) & (x < 5)]
array(['o', 'o', 'a'],
dtype='|S1')
Solution 2:
IMO OP does not actually want np.bitwise_and()
(aka &
) but actually wants np.logical_and()
because they are comparing logical values such as True
and False
- see this SO post on logical vs. bitwise to see the difference.
>>> x = array([5, 2, 3, 1, 4, 5])
>>> y = array(['f','o','o','b','a','r'])
>>> output = y[np.logical_and(x > 1, x < 5)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
dtype='|S1')
And equivalent way to do this is with np.all()
by setting the axis
argument appropriately.
>>> output = y[np.all([x > 1, x < 5], axis=0)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
dtype='|S1')
by the numbers:
>>> %timeit (a < b) & (b < c)
The slowest run took 32.97 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.15 µs per loop
>>> %timeit np.logical_and(a < b, b < c)
The slowest run took 32.59 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.17 µs per loop
>>> %timeit np.all([a < b, b < c], 0)
The slowest run took 67.47 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.06 µs per loop
so using np.all()
is slower, but &
and logical_and
are about the same.
Solution 3:
Add one detail to @J.F. Sebastian's and @Mark Mikofski's answers:
If one wants to get the corresponding indices (rather than the actual values of array), the following code will do:
For satisfying multiple (all) conditions:
select_indices = np.where( np.logical_and( x > 1, x < 5) )[0] # 1 < x <5
For satisfying multiple (or) conditions:
select_indices = np.where( np.logical_or( x < 1, x > 5 ) )[0] # x <1 or x >5
Solution 4:
I like to use np.vectorize
for such tasks. Consider the following:
>>> # Arrays
>>> x = np.array([5, 2, 3, 1, 4, 5])
>>> y = np.array(['f','o','o','b','a','r'])
>>> # Function containing the constraints
>>> func = np.vectorize(lambda t: t>1 and t<5)
>>> # Call function on x
>>> y[func(x)]
>>> array(['o', 'o', 'a'], dtype='<U1')
The advantage is you can add many more types of constraints in the vectorized function.
Hope it helps.