# Numpy array, how to select indices satisfying multiple conditions?

### Question

Suppose I have a numpy array `x = [5, 2, 3, 1, 4, 5]`, `y = ['f', 'o', 'o', 'b', 'a', 'r']`. I want to select the elements in `y` corresponding to elements in `x` that are greater than 1 and less than 5.

I tried

``````x = array([5, 2, 3, 1, 4, 5])
y = array(['f','o','o','b','a','r'])
output = y[x > 1 & x < 5] # desired output is ['o','o','a']
``````

but this doesn't work. How would I do this?

1
121
2/28/2012 10:14:50 PM

``````>>> y[(1 < x) & (x < 5)]
array(['o', 'o', 'a'],
dtype='|S1')
``````
186
6/13/2010 12:50:32 AM

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.