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?

Your expression works if you add parentheses:

```
>>> y[(1 < x) & (x < 5)]
array(['o', 'o', 'a'],
dtype='|S1')
```

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.

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