Return True if item is empty string

Question

Please consider this snippet:

>>> i = ["", 1, 2]
>>> all([x for x in i])
False

What would be Pythonic way to make this snippet return True even if item in iterable is empty string?

Standard restrictions should apply - if item is None or False etc, it should return False as expected.

Do you mean you want the snippet to return True unless at least one item is falsey, but count an empty string as truthy? I.e. follow Python's rules on what is counted True and False except for an empty string? — Rory Daulton
– Rory Daulton, Commented Aug 28, 2016 at 18:44

Kasravnd · Accepted Answer · 2016-08-28 18:42:42Z

2

>>> lst = ["", 1, 2]
>>> 
>>> all(True if i=='' else i for i in lst)
True

If you want True if there is at least one item that evaluates to True use any().

>>> any(lst)
True

Note that in general any() and all() accept iterable argument and you don't need to loop over them.

answered Aug 28, 2016 at 18:42

Kasravnd

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4 Comments

vlad Over a year ago

As both your and @two-bit-alchemist's answers are correct, I'll accept his answer because he has much less score points compared to you. I hope you don't mind. Thanks

Kasravnd Over a year ago

@vlad First off, The answers are different secondly you always should consider to accepting the correct answer, I don't think this imaginary points gotta do something for us :-D

vlad Over a year ago

Ok, thanks for your reply. Can you point to single scenario where your snippet will give different result then @two-bit-alchemist's? Or what makes your snippet different? As a side note: my points are not imaginary, but driven from common human ethics, something lot of people unfortunately lack.

Kasravnd Over a year ago

@vlad It's correct in this case but actually it denies all the strings. For example if you want to get False for a string like "''" or another ones you can't use that answer. In that case you can use all(True if i not it {set of special strings} else i for i in lst)

Darth Kotik · Accepted Answer · 2016-08-28 18:55:30Z

2

This option looks nice to me.

all(x or x=="" for x in i)

answered Aug 28, 2016 at 18:55

Darth Kotik

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Comments

Two-Bit Alchemist · Accepted Answer · 2016-08-28 18:43:56Z

1

all([x for x in i if not isinstance(x, str)])

The only falsy string is the empty string, so if you don't want to test for it, then filter out the strings.

answered Aug 28, 2016 at 18:43

Two-Bit Alchemist

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7 Comments

bereal Over a year ago

A nitpick: [] are redundant.

reticentroot Over a year ago

all(x for x in filter(None, i)) -slightly cleaner if you only have to worry about empty strings

Two-Bit Alchemist Over a year ago

@reticentroot No, the default filter (None) removes all falsy, which would get rid of False and None as well.

reticentroot Over a year ago

Oh your right, totally forget about that. So if the OP needs to filter those items out that's still cleaner than an isinstance, though an isinstance is more flexible. :-) thanks for the reminder.

Two-Bit Alchemist Over a year ago

@bereal I can't find my handy link comparing the behavior, but it basically depends. Generators consume less memory but must be remade each time they are used. List comprehensions generate all the values but are more efficient for reuse. It's not redundant; it's different. Which is better depends on context.

|

BPL · Accepted Answer · 2016-08-28 18:58:33Z

-1

You can use all, for instance:

tests = [
    [None, False, "", 1, 2],
    ["", 1, 2],
    [1, 2],
    []
]

for i in tests:
    print i
    print all(filter(lambda x: x != '', i))
    print all(True if x == '' else x for x in i)
    print '-' * 80

edited Aug 28, 2016 at 18:58

answered Aug 28, 2016 at 18:45

BPL

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1 Comment

Two-Bit Alchemist Over a year ago

Your first solution fails the requirement of catching other falsy values like None or False because it filters them out. Your third solution fails because it bails after it finds one truthy value, which is not what is wanted.

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