How to check if all elements of a list match a condition?

Efficient element checks with generators

You can improve efficiency using generator expressions in all(), eliminating the need to create temporary lists:

# Efficient zero-dodging check
all_non_zero = all(element != 0 for element in lst)
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This approach evaluates items in a lazy manner, ensuring minimal memory usage, unlike list comprehensions which evaluate everything instantaneously.

Use built-in function magic with all()

When your checks or transformations become more complex, judiciously combine all() with built-in functions like map() or filter():

# Using `map()` with a lambda for positive reckoning
all_positive = all(map(lambda x: x > 0, lst))
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This approach keeps your expression clean and readable, something your future self will thank you for!

Dealing with nested list mystery

When handling nested lists and you need to verify conditions against elements in these sublists, employ processed-elements "flag":

sublists = [[1, 0], [2, 0], [3, 1]]
all_sublists_have_zero_flag = all(sublist[-1] == 0 for sublist in sublists) 
# mischief managed!
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Advancing with itertools

When built-in iteration patterns aren't fitting the bill, it might be time to call in heavy artillery like the itertools module:

import itertools

# Condition-satiating `takewhile` breaks at the first sign of trouble
all(items) == all(itertools.takewhile(lambda x: x > 10, lst))
# Good dog, itertools!
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Make boolean checks great again

Use keyword heroes not in and in for swift and effective boolean checks:

# Quick "all True" check. Bet you didn't see that coming!
no_false_values = (False not in lst)
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Diversify condition checks: Enter any()

any() is a close kin to all(), timely checking if at least one element fulfils the condition:

# Inverse checks have never felt so good
any_negative = any(x < 0 for x in lst)
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Troubleshooting common pitfalls

Manipulating the list while iterating can be a dangerous game. Be smart and use flags or progress the iterator with itertools to the rescue:

# Using `dropwhile` to carry on 'til false
all(itertools.dropwhile(lambda x: x == 0, lst))
# Here's an iterator, iterate like a king.
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Level up with ifilter() (Python 2) and Python 3's champion filter(). This duo aids in condition checking without annoying list mutations:

# Continue as long as 0 leads the way
all(filter(lambda x: x == 0, lst))
# 'filter' – for when the going gets tough and the tough gets going.
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Code that is lean and mean

Juggling generators and itertools can lead to code that's both feature-rich and speedy Gonzales. Aim for laid-back implementations that clearly express intent without sacrificing speed.