Python's filter(): Extract Values From Iterables

Extracting Even Numbers

As a first example, say you need to process a list of integer numbers and build a new list containing the even numbers. Your first approach to this problem might be to use a for loop like this:

>>> numbers = [1, 3, 10, 45, 6, 50]

>>> def extract_even(numbers):
...     even_numbers = []
...     for number in numbers:
...         if number % 2 == 0:  # Filtering condition
...             even_numbers.append(number)
...     return even_numbers
...

>>> extract_even(numbers)
[10, 6, 50]

Here, extract_even() takes an iterable of integer numbers and returns a list containing only those that are even. The conditional statement plays the role of a filter that tests every number to find out if it’s even or not.

When you run into code like this, you can extract the filtering logic into a small predicate function and use it with filter(). This way, you can perform the same computation without using an explicit loop:

>>> numbers = [1, 3, 10, 45, 6, 50]

>>> def is_even(number):
...     return number % 2 == 0  # Filtering condition
...

>>> list(filter(is_even, numbers))
[10, 6, 50]

Here, is_even() takes an integer and returns True if it’s even and False otherwise. The call to filter() does the hard work and filters out the odd numbers. As a result, you get a list of the even numbers. This code is shorter and more efficient than its equivalent for loop.

Finding Prime Numbers

Another interesting example might be to extract all the prime numbers in a given interval. To do that, you can start by coding a predicate function that takes an integer as an argument and returns True if the number is prime and False otherwise. Here’s how you can do that:

>>> import math

>>> def is_prime(n):
...     if n <= 1:
...         return False
...     for i in range(2, int(math.sqrt(n)) + 1):
...         if n % i == 0:
...             return False
...     return True
...

>>> is_prime(5)
True
>>> is_prime(12)
False

The filtering logic is now in is_prime(). The function iterates through the integers between 2 and the square root of n. Inside the loop, the conditional statement checks if the current number is divisible by any other in the interval. If so, then the function returns False because the number isn’t prime. Otherwise, it returns True to signal that the input number is prime.

With is_prime() in place and tested, you can use filter() to extract prime numbers from an interval like this:

>>> list(filter(is_prime, range(1, 51)))
[2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

This call to filter() extracts all the prime numbers in the range between 1 and 50. The algorithm used in is_prime() comes from Wikipedia’s article about primality tests. You can check out that article if you need more efficient approaches.

Removing Outliers in a Sample

When you’re trying to describe and summarize a sample of data, you probably start by finding its mean, or average. The mean is a quite popular central tendency measurement and is often the first approach to analyzing a dataset. It gives you a quick idea of the center, or location, of the data.

In some cases, the mean isn’t a good enough central tendency measure for a given sample. Outliers are one of the elements that affect how accurate the mean is. Outliers are data points that differ significantly from other observations in a sample or population. Other than that, there is no unique mathematical definition for them in statistics.

However, in normally distributed samples, outliers are often defined as data points that lie more than two standard deviations from the sample mean.

Now suppose you have a normally distributed sample with some outliers that are affecting the mean accuracy. You’ve studied the outliers, and you know they’re incorrect data points. Here’s how you can use a couple of functions from the statistics module along with filter() to clean up your data:

>>> import statistics as st
>>> sample = [10, 8, 10, 8, 2, 7, 9, 3, 34, 9, 5, 9, 25]

>>> # The mean before removing outliers
>>> mean = st.mean(sample)
>>> mean
10.692307692307692

>>> stdev = st.stdev(sample)
>>> low = mean - 2 * stdev
>>> high = mean + 2 * stdev

>>> clean_sample = list(filter(lambda x: low <= x <= high, sample))
>>> clean_sample
[10, 8, 10, 8, 2, 7, 9, 3, 9, 5, 9, 25]

>>> # The mean after removing outliers
>>> st.mean(clean_sample)
8.75

In the highlighted line, the lambda function returns True if a given data point lies between the mean and two standard deviations. Otherwise, it returns False. When you filter the sample with this function, 34 is excluded. After this cleanup, the mean of the sample has a significantly different value.

Validating Python Identifiers

You can also use filter() with iterables containing nonnumeric data. For example, say you need to process a list of strings and extract those that are valid Python identifiers. After doing some research, you find out that Python’s str provides a method called .isidentifier() that can help you out with that validation.

Here’s how you can use filter() along with str.isidentifier() to quickly validate identifiers:

>>> words = ["variable", "file#", "header", "_non_public", "123Class"]

>>> list(filter(str.isidentifier, words))
['variable', 'header', '_non_public']

In this case, filter() runs .isidentifier() on every string in words. If the string is a valid Python identifier, then it’s included in the final result. Otherwise, the word is filtered out. Note that you need to use str to access .isidentifier() in the call to filter().

Finally, an interesting exercise might be to take the example further and check if the identifier is also a keyword. Go ahead and give it a try! Hint: you can use .kwlist from the keyword module.

Finding Palindrome Words

An exercise that often arises when you’re getting familiar with Python strings is to find palindrome words in a list of strings. A palindrome word reads the same backward as forward. Typical examples are “madam” and “racecar.”

To solve this problem, you’ll start by coding a predicate function that takes a string and checks if it reads the same in both directions, backward and forward. Here’s a possible implementation:

>>> def is_palindrome(word):
...     reversed_word = "".join(reversed(word))
...     return word.lower() == reversed_word.lower()
...

>>> is_palindrome("Racecar")
True
>>> is_palindrome("Python")
False

In is_palindrome(), you first reverse the original word and store it in reversed_word. Then you return the result of comparing both words for equality. In this case, you use .lower() to prevent case-related differences. If you call the function with a palindrome word, then you get True. Otherwise, you get False.

You already have a working predicate function to identify palindrome words. Here’s how you can use filter() to do the hard work:

>>> words = ("filter", "Ana", "hello", "world", "madam", "racecar")

>>> list(filter(is_palindrome, words))
['Ana', 'madam', 'racecar']

Cool! Your combination of filter() and is_palindrome() works properly. It’s also concise, readable, and efficient. Good job!

The Square of Even Numbers: filter() and map()

Sometimes you need to take an iterable, process each of its items with a transformation function, and produce a new iterable with the resulting items. In that case, you can use map(). The function has the following signature:

map(function, iterable[, iterable1, ..., iterableN])

The arguments work like this:

1. function holds the transformation function. This function should take as many arguments as iterables you pass into map().
2. iterable holds a Python iterable. Note that you can provide several iterables to map(), but that’s optional.

map() applies function to each item in iterable to transform it into a different value with additional features. Then map() yields each transformed item on demand.

To illustrate how you can use filter() along with map(), say you need to compute the square value of all the even numbers in a given list. In that case, you can use filter() to extract the even numbers and then map() to calculate the square values:

>>> numbers = [1, 3, 10, 45, 6, 50]

>>> def is_even(number):
...     return number % 2 == 0
...

>>> even_numbers = list(filter(is_even, numbers))
>>> even_numbers
[10, 6, 50]

>>> list(map(lambda n: n ** 2, even_numbers))
[100, 36, 2500]

>>> list(map(lambda n: n ** 2, filter(is_even, numbers)))
[100, 36, 2500]

First, you get the even numbers using filter() and is_even() just like you’ve done so far. Then you call map() with a lambda function that takes a number and returns its square value. The call to map() applies the lambda function to each number in even_numbers, so you get a list of square even numbers. The final example shows how to combine filter() and map() in a single expression.

The Sum of Even Numbers: filter() and reduce()

Another functional programming tool in Python is reduce(). Unlike filter() and map(), which are still built-in functions, reduce() was moved to the functools module. This function is useful when you need to apply a function to an iterable and reduce it to a single cumulative value. This kind of operation is commonly known as a reduction or folding.

The signature of reduce() is like this:

reduce(function, iterable, initial)

Here’s what the arguments mean:

1. function holds any Python callable that accepts two arguments and returns a single value.
2. iterable holds any Python iterable.
3. initial holds a value that serves as a starting point for the first partial computation or reduction. It’s an optional argument.

A call to reduce() starts by applying function to the first two items in iterable. This way, it computes the first cumulative result, called an accumulator. Then reduce() uses the accumulator and the third item in iterable to compute the next cumulative result. The process continues until the function returns with a single value.

If you supply a value to initial, then reduce() runs the first partial computation using initial and the first item of iterable.

Here’s an example that combines filter() and reduce() to cumulatively calculate the total sum of all the even numbers in a list:

>>> from functools import reduce
>>> numbers = [1, 3, 10, 45, 6, 50]

>>> def is_even(number):
...     return number % 2 == 0
...

>>> even_numbers = list(filter(is_even, numbers))
>>> reduce(lambda a, b: a + b, even_numbers)
66

>>> reduce(lambda a, b: a + b, filter(is_even, numbers))
66

Here, the first call to reduce() computes the sum of all the even numbers that filter() provides. To do that, reduce() uses a lambda function that adds two numbers at a time.

The final example shows how to chain filter() and reduce() to produce the same result you got before.

Extracting Odd Numbers

You already coded a predicate function called is_even() to check if a number is even or not. With that function and the help of filterfalse(), you can build an iterator that yields odd numbers without having to code an is_odd() function:

>>> from itertools import filterfalse
>>> numbers = [1, 3, 10, 45, 6, 50]

>>> def is_even(number):
...     return number % 2 == 0
...

>>> list(filterfalse(is_even, numbers))
[1, 3, 45]

In this example, filterfalse() returns an iterator that yields the odd numbers from the input iterator. Note that the call to filterfalse() is straightforward and readable.

Filtering Out NaN Values

Sometimes when you’re working with floating-point arithmetic, you can face the issue of having NaN (not a number) values. For example, say you’re calculating the mean of a sample of data that contains NaN values. If you use Python’s statistics module for this computation, then you get the following result:

>>> import statistics as st

>>> sample = [10.1, 8.3, 10.4, 8.8, float("nan"), 7.2, float("nan")]
>>> st.mean(sample)
nan

In this example, the call to mean() returns nan, which isn’t the most informative value you can get. NaN values can have different origins. They can be due to invalid inputs, corrupted data, and so on. You should find the right strategy to deal with them in your applications. One alternative might be to remove them from your data.

The math module provides a convenient function called isnan() that can help you out with this problem. The function takes a number x as an argument and returns True if x is a NaN and False otherwise. You can use this function to provide the filtering criteria in a filterfalse() call:

>>> import math
>>> import statistics as st
>>> from itertools import filterfalse

>>> sample = [10.1, 8.3, 10.4, 8.8, float("nan"), 7.2, float("nan")]

>>> st.mean(filterfalse(math.isnan, sample))
8.96

Using math.isnan() along with filterfalse() allows you to exclude all the NaN values from the mean computation. Note that after the filtering, the call to mean() returns a value that provides a better description of your sample data.

Replacing filter() With a List Comprehension

You can use the following pattern to quickly replace a call to filter() with an equivalent list comprehension:

# Generating a list with filter()
list(filter(function, iterable))

# Generating a list with a list comprehension
[item for item in iterable if function(item)]

In both cases, the final purpose is to create a list object. The list comprehension approach is more explicit than its equivalent filter() construct. A quick read through the comprehension reveals the iteration and also the filtering functionality in the if clause.

Using list comprehensions instead of filter() is probably the path most Python developers take nowadays. However, list comprehensions have some drawbacks compared to filter(). The most notable one is the lack of lazy evaluation. Also, when developers start reading code that uses filter(), they immediately know that the code is performing filtering operations. However, that’s not so evident in code that uses list comprehensions with the same goal.

A detail to notice when turning a filter() construct into a list comprehension is that if you pass None to the first argument of filter(), then the equivalent list comprehension looks like this:

# Generating a list with filter() and None
list(filter(None, iterable))

# Equivalent list comprehension
[item for item in iterable if item]

In this case, the if clause in the list comprehension tests item for its truth value. This test follows the standard Python rules about truth values you already saw.

Here’s an example of replacing filter() with a list comprehension to build a list of even numbers:

>>> numbers = [1, 3, 10, 45, 6, 50]

>>> # Filtering function
>>> def is_even(x):
...     return x % 2 == 0
...

>>> # Use filter()
>>> list(filter(is_even, numbers))
[10, 6, 50]

>>> # Use a list comprehension
>>> [number for number in numbers if is_even(number)]
[10, 6, 50]

In this example, you can see that the list comprehension variant is more explicit. It reads almost like plain English. The list comprehension solution also avoids having to call list() to build the final list.

Replacing filter() With a Generator Expression

The natural replacement for filter() is a generator expression. That’s because filter() returns an iterator that yields items on demand just like a generator expression does. Python iterators are known to be memory efficient. That’s why filter() now returns an iterator instead of a list.

Here’s how you can use generator expressions to write the example in the above section:

>>> numbers = [1, 3, 10, 45, 6, 50]

>>> # Filtering function
>>> def is_even(x):
...     return x % 2 == 0
...

>>> # Use filter()
>>> even_numbers = filter(is_even, numbers)
>>> even_numbers
<filter object at 0x7f58691de4c0>
>>> list(even_numbers)
[10, 6, 50]

>>> # Use a generator expression
>>> even_numbers = (number for number in numbers if is_even(number))
>>> even_numbers
<generator object <genexpr> at 0x7f586ade04a0>
>>> list(even_numbers)
[10, 6, 50]

A generator expression is as efficient as a call to filter() in terms of memory consumption. Both tools return iterators that yield items on demand. Using either one might be a question of taste, convenience, or style. So, you’re in charge!