Python idiom to return the only element or None

Question

I've got a list, which may or may not contain a unique element satisfying a given predicate. I am looking for an expression which evaluates to an item satisfying that predicate if it exists and is unique, and otherwise returns None. Something like

numbers = [4, 3, 9, 7, 1, 2, 8]
print(the(item for item in numbers if item > 10))     # None
print(the(item for item in numbers if item % 2 == 0)) # None
print(the(item for item in numbers if item % 7 == 0)) # 7

Is there a built-in idiom for this, or do I have to write my own the function?

Serge Ballesta · Accepted Answer · 2016-11-14 10:16:16Z

4

I do not know a single expression way for that, but that simple function should work:

def the(it, cond):
    l = [ i for i in it if cond(i) ]
    return l[0] if len(l) == 1 else None

Test:

>>> print(the(numbers,(lambda x: x > 10)))
None
>>> print(the(numbers,(lambda x: x % 7 == 0)))
7
>>> print(the(numbers,(lambda x: x % 2 == 0)))
None

answered Nov 14, 2016 at 10:16

Serge Ballesta

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1 Comment

Stefan Pochmann Over a year ago

Could be a single expression by using (l + [None])[::2][-1] or ((l + [None]) * 3)[2], but I don't really think they're good :-)

Stefan Pochmann · Accepted Answer · 2016-11-14 11:04:11Z

1

You could try asking for two elements with islice, would be a bit simpler:

def the(it):
    tmp = list(islice(it, 2))
    return tmp[0] if len(tmp) == 1 else None

Or a loop:

def the(it):
    value = None
    for i, value in enumerate(it):
        if i == 1:
            return None
    return value

Or yet another way to use next:

def the(it):
    first = next(it, None)
    o = object()
    if next(it, o) is o:
        return first

Or similar to yours:

def the(it):
    first = next(it, None)
    try:
        next(it)
    except:
        return first

edited Nov 14, 2016 at 11:04

answered Nov 14, 2016 at 10:27

Stefan Pochmann

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Comments

dumbass · Accepted Answer · 2016-11-14 10:00:03Z

0

For the record, I can just write the as

def the(it):
    it = iter(it)
    try:
        value = next(it)
    except StopIteration:
        return None
    try:
        next(it)
    except StopIteration:
        return value
    return None

answered Nov 14, 2016 at 10:00

dumbass

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Comments

Ébe Isaac · Accepted Answer · 2016-11-14 10:18:14Z

0

This should be a simple solution for this problem:

def the(items):
    return items[0] if (len(items) == 1) else None

numbers = [4, 3, 9, 7, 1, 2, 8]
print(the([item for item in numbers if item > 10]))     # None
print(the([item for item in numbers if item % 2 == 0])) # None
print(the([item for item in numbers if item % 7 == 0])) # 7

You could also apply the following format. It might not be any simpler in terms of code, but it certainly is faster when the item count is large.

print(the(list(filter(lambda x: x > 10, numbers))))     # None
print(the(list(filter(lambda x: x % 2 == 0, numbers)))) # None
print(the(list(filter(lambda x: x % 7 == 0, numbers)))) # 7

answered Nov 14, 2016 at 10:18

Ébe Isaac

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Comments

Moses Koledoye · Accepted Answer · 2016-11-14 10:51:45Z

0

You can use the following lazy approach. Get the next two items from the generator expression and return None if the second item from the generator is not None:

def func(lst, pred):
    gen = (i for i in lst if pred(i))
    v, w = next(gen, None), next(gen, None)
    return v if w is None else None


print(func([4, 3, 9, 7, 1, 2, 8], lambda x: x>10))
# None
print(func([4, 3, 9, 7, 1, 2, 8], lambda x: x % 2 == 0))
# None
print(func([4, 3, 9, 7, 1, 2, 8], lambda x: x % 7 == 0))
# 7
print(func([4, 3, 9, 0, 1, 2, 8], lambda x: x % 7 == 0))
# 0

edited Nov 14, 2016 at 10:51

answered Nov 14, 2016 at 10:00

Moses Koledoye

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6 Comments

TigerhawkT3 Over a year ago

Using a set will remove duplicates. I assume print(the(item for item in (1,1,2,2) if item % 2 == 0)) should be None.

Moses Koledoye Over a year ago

@TigerhawkT3 Didn't consider that. Updated

Stefan Pochmann Over a year ago

How about print(func([1, None], lambda x: True)) (prints 1 instead of None).

Moses Koledoye Over a year ago

@StefanPochmann They could use a sentinel instead of None and I strongly doubt they have a None in their list of numbers

Stefan Pochmann Over a year ago

That's only in their example, though. Their question title and text say "element" and their own answer as well as yours look general. Btw, you already have an object that you could use as sentinel: gen. I think that would be cool :-)

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