Is there a way of returning the match sublist found in the any() function?-CodePudding

the code is:

list_ = [[1,2,3],[4,5,6],[7,8,9]]

if i try:

1 in list_

it will return false. But if i use the any() function it returns True!

any(1 in sublist for sublist in list_)

But i want it to return the sublist that the item '1' is found. I've tried things like:

if any(1 in sublist for sublist in list_):
    print(sublist)

it raise NameError: name 'sublist' is not defined

is there a way of doing it?? Thanks :)

CodePudding user response：

You can use the list comprehension syntax, which includes an expression for filtering in it.

In this case, you'd want:

[sublist for sublist in list_ if 1 in sublist]

The new list will be created dynamically, and just the elements in list_ that pass the guard expression `...if 1 in sublist``` will be included.

Just to be complete: there is no way to get all the elements from a call to any because it stops processing the iterator as soon as it finds the first match - that is the advantage of using it over a regular comprehension or generator expression: the syntax for those do not allow one to stop the processing of an iterator once a condition is met.

CodePudding user response：

any(...) returns a boolean value. It answers the question: "does the iterable have any element that matches the condition". So no, it cannot return the matched element itself.

If you want the matched element, it's better to write a simple loop. It's good to wrap this in a function:

def get_first_match(iter, cond):
    for item in iter:
        if cond(item):
            return item

Example usage:

print(get_first_match(list_, lambda item: 1 in item))

Note that when there's no matching item, the function will return None.

CodePudding user response：

Use an assignment expression to capture the last value evaluated by any.

if any(1 in (x := sublist) for sublist in list_):
    print(x)

x is repeatedly assigned the value of sublist as any iterates the generator expression, but since any stops as soon as 1 in sublist is true, the value of x after any returns will be the value that made 1 in sublist true.

The key is that sublist is local to the generator expression itself, but x is local to the scope any executes in.

This use-case is, in fact, one of the rationales provided for the scoping of the target of an assignment expression in PEP 572.