Python – Filter rows with required elements

When it is required to filter rows with required elements, a list comprehension and the all() operator can be used to check if all elements in each row are present in a reference list.

Example

The following example demonstrates how to filter rows where all elements are present in a check list ?

my_list = [[261, 49, 61], [27, 49, 3, 261], [261, 49, 85], [1, 1, 9]]

print("The list is :")
print(my_list)

check_list = [49, 61, 261, 85]

my_result = [row for row in my_list if all(element in check_list for element in row)]

print("The result is :")
print(my_result)

The list is :
[[261, 49, 61], [27, 49, 3, 261], [261, 49, 85], [1, 1, 9]]
The result is :
[[261, 49, 61], [261, 49, 85]]

How It Works

The solution uses a list comprehension combined with the all() function:

• A list of lists (rows) is defined and displayed on the console.

• A reference list (check_list) containing valid elements is defined.

• The list comprehension iterates over each row in the original list.

• The all() operator checks if every element in the current row exists in check_list.

• Only rows where all elements pass this condition are included in the result.

Alternative Approach Using Filter

You can also use the filter() function for the same result ?

my_list = [[261, 49, 61], [27, 49, 3, 261], [261, 49, 85], [1, 1, 9]]
check_list = [49, 61, 261, 85]

my_result = list(filter(lambda row: all(element in check_list for element in row), my_list))

print("The result is :")
print(my_result)

The result is :
[[261, 49, 61], [261, 49, 85]]

Conclusion

Use list comprehension with all() to filter rows where every element exists in a reference list. This approach is efficient and readable for checking multiple conditions across nested data structures.