# Filtering a list based on a list of booleans

## Question:

I have a list of values which I need to filter given the values in a list of booleans:

```
list_a = [1, 2, 4, 6]
filter = [True, False, True, False]
```

I generate a new filtered list with the following line:

```
filtered_list = [i for indx,i in enumerate(list_a) if filter[indx] == True]
```

which results in:

```
print filtered_list
[1,4]
```

The line works but looks (to me) a bit overkill and I was wondering if there was a simpler way to achieve the same.

## Advices

Summary of two good advices given in the answers below:

1- Don’t name a list `filter`

like I did because it is a built-in function.

2- Don’t compare things to `True`

like I did with `if filter[idx]==True..`

since it’s unnecessary. Just using `if filter[idx]`

is enough.

## Answers:

Like so:

```
filtered_list = [i for (i, v) in zip(list_a, filter) if v]
```

Using `zip`

is the *pythonic* way to iterate over multiple sequences in parallel, without needing any indexing. This assumes both sequences have the same length (zip stops after the shortest runs out). Using `itertools`

for such a simple case is a bit overkill …

One thing you do in your example you should really stop doing is comparing things to True, this is usually not necessary. Instead of `if filter[idx]==True: ...`

, you can simply write `if filter[idx]: ...`

.

You’re looking for `itertools.compress`

:

```
>>> from itertools import compress
>>> list_a = [1, 2, 4, 6]
>>> fil = [True, False, True, False]
>>> list(compress(list_a, fil))
[1, 4]
```

## Timing comparisons(py3.x):

```
>>> list_a = [1, 2, 4, 6]
>>> fil = [True, False, True, False]
>>> %timeit list(compress(list_a, fil))
100000 loops, best of 3: 2.58 us per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v] #winner
100000 loops, best of 3: 1.98 us per loop
>>> list_a = [1, 2, 4, 6]*100
>>> fil = [True, False, True, False]*100
>>> %timeit list(compress(list_a, fil)) #winner
10000 loops, best of 3: 24.3 us per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v]
10000 loops, best of 3: 82 us per loop
>>> list_a = [1, 2, 4, 6]*10000
>>> fil = [True, False, True, False]*10000
>>> %timeit list(compress(list_a, fil)) #winner
1000 loops, best of 3: 1.66 ms per loop
>>> %timeit [i for (i, v) in zip(list_a, fil) if v]
100 loops, best of 3: 7.65 ms per loop
```

_{Don’t use filter as a variable name, it is a built-in function.}

To do this using numpy, ie, if you have an array, `a`

, instead of `list_a`

:

```
a = np.array([1, 2, 4, 6])
my_filter = np.array([True, False, True, False], dtype=bool)
a[my_filter]
> array([1, 4])
```

With numpy:

```
In [128]: list_a = np.array([1, 2, 4, 6])
In [129]: filter = np.array([True, False, True, False])
In [130]: list_a[filter]
Out[130]: array([1, 4])
```

or see Alex Szatmary’s answer if list_a can be a numpy array but not filter

Numpy usually gives you a big speed boost as well

```
In [133]: list_a = [1, 2, 4, 6]*10000
In [134]: fil = [True, False, True, False]*10000
In [135]: list_a_np = np.array(list_a)
In [136]: fil_np = np.array(fil)
In [139]: %timeit list(itertools.compress(list_a, fil))
1000 loops, best of 3: 625 us per loop
In [140]: %timeit list_a_np[fil_np]
10000 loops, best of 3: 173 us per loop
```

```
filtered_list = [list_a[i] for i in range(len(list_a)) if filter[i]]
```

With python 3 you can use `list_a[filter]`

to get `True`

values. To get `False`

values use `list_a[~filter]`