Short answer: You can remove all empty lists from a list of lists by using the list comprehension statement
[x for x in list if x] to filter the list.
In the following, you’ll learn about the two methods using list comprehension and the filter() function to remove all empty lists from a list of lists.
But before that, feel free to play with the code yourself:
Method 1: List Comprehension
How can you remove all empty lists from a list of lists? Say, you’ve got a list of lists
[[1, 2, 3], [1, 2], , , , [1, 2, 3, 4], , ]
and you want all empty lists removed to obtain the list of lists
[[1, 2, 3], [1, 2], [1, 2, 3, 4]].
Solution: Use list comprehension
[x for x in list if x] to filter the list and remove all lists that are empty.
lst = [[1, 2, 3], [1, 2], , , , [1, 2, 3, 4], , ] print([x for x in lst if x]) # [[1, 2, 3], [1, 2], [1, 2, 3, 4]]
The condition if
x evaluates to
False only if the list
x is empty. In all other cases, it evaluates to
True and the element is included in the new list.
You can visualize the execution flow here by clicking the “Next” button:
Method 2: filter()
An alternative is to use the
filter() function to remove all empty lists from a list of lists:
lst = [[1, 2, 3], [1, 2], , , , [1, 2, 3, 4], , ] print(list(filter(lambda x: x, lst)))
The filter() function takes two arguments:
- the filter decision function to check for each element whether it should be included in the filtered iterable (it returns a Boolean value), and
- the iterable to be filtered.
As filter decision function, you use the identity function that just passes the list through. Why does this work? Because only an empty list will be evaluated to
False. All other lists will be evaluated to
True (and, thus, pass the filtering test).
Where to Go From Here?
Enough theory, let’s get some practice!
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