


Simplifying and Accelerating List Element Equality Checks
Background Information
Python provides convenient mechanisms to check whether all elements in a list satisfy a specific condition. Existing approaches utilize the built-in function all() to perform this task efficiently. Additionally, for conditions involving membership in another container, optimized solutions are available.
Using all() for Equality Verification
The simplest and fastest method to check if all elements of a list match a condition is to employ the all() function. This function evaluates if the condition holds true for every element in the sequence. For example, to ascertain whether each last element in a sublist is 0:
import all my_list = [[1, 2, 3], [4, 5, 0], [7, 8, 0]] result = all(item[2] == 0 for item in my_list) print(result) # True
Leveraging Generator Expression for Efficiency
To further enhance efficiency, generator expressions can be combined with all(). This combination generates the elements in the list lazily, providing a streamlined evaluation process.
result = all(flag == 0 for (_, _, flag) in my_list)
Utilizing any() for Inequality Verification
Conversely, to check if at least one element of a list matches a condition, any() can be employed. This function determines whether any element in the sequence satisfies the condition.
result = any(flag == 0 for (_, _, flag) in my_list)
Alternative Approaches for Element Filtering
In scenarios where an element needs to be filtered based on a condition, list comprehensions offer an effective solution:
filtered_list = [x for x in my_list if x[2] == 0]
This comprehension extracts all sublists where the last element is 0. Similarly, one can use filter():
filtered_list = filter(lambda x: x[2] == 0, my_list)
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