Python: min(None, x)

Question

I would like to perform the following:

a=max(a,3)
b=min(b,3)

However sometimes a and b may be None.
I was happy to discover that in the case of max it works out nicely, giving my required result 3, however if b is None, b remains None...

Anyone can think of an elegant little trick to make min return the number in case one of the arguments in None?

It doesn't do the right thing. It happens to give the result you expect in one of two cases because the nonsensical comparision between `NoneType` and `int` returns a fixed value regardless of the integer value. In Python 3, you get a `TypeError` when you do things like that (comparing types that have no meaningful ordering). — , Jun 06 '11 at 16:11
Seems like an inconsistency in Python, more than anything else. — Rafe Kettler, Jun 06 '11 at 16:12
http://stackoverflow.com/questions/2214194/is-everything-greater-than-none — Ben Jackson, Jun 06 '11 at 16:13
if you are ok with replacing None to a default int value (like 0), you can do something like this: `max(a,3,key=lambda x: x or 0)` — Shay Tsadok, Jan 23 '20 at 08:31

score 53 · Accepted Answer · answered Jun 06 '11 at 16:16

53

Why don't you just create a generator without None values? It's simplier and cleaner.

>>> l=[None ,3]
>>> min(i for i in l if i is not None)
3

answered Jun 06 '11 at 16:16

utdemir

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No need for the list comprehension. Still +1 (in advance - seriously, please remove the brackets), it's a clean and simple solution. – Jun 06 '11 at 16:18
Thanks, without the braces, does Python interprets it like a generator expression? – utdemir Jun 06 '11 at 16:22
3

Yes. The parens around generator expressions are optional if the genexpr is the sole argument to a function call. – Jun 06 '11 at 16:27
16

One potential problem with a solution like this is that it works fine for the listed example but if you have a list with `[None, None]`, the `min()` function will fail because you're not giving it a valid argument. – Kevin London May 12 '16 at 17:29
6

@KevinLondon, specify the default value for the min as 0 or None depending upon the behavior you want. ```min((x for x in [None,None] if x is not None), default=None)``` – rahul.deshmukhpatil Dec 28 '20 at 08:56

score 9 · Answer 2 · answered Nov 08 '18 at 16:32

9

My solution for Python 3 (3.4 and greater):

min((x for x in lst if x is not None), default=None)
max((x for x in lst if x is not None), default=None)

answered Nov 08 '18 at 16:32

R1tschY

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score 7 · Answer 3 · edited Nov 13 '18 at 05:09

A solution for the Python 3

Code:

# variable lst is your sequence

min(filter(lambda x: x is not None, lst)) if any(lst) else None

Examples:

In [3]: lst = [None, 1, None]

In [4]: min(filter(lambda x: x is not None, lst)) if any(lst) else None
Out[4]: 1

In [5]: lst = [-4, None, 11]

In [6]: min(filter(lambda x: x is not None, lst)) if any(lst) else None
Out[6]: -4

In [7]: lst = [0, 7, -79]

In [8]: min(filter(lambda x: x is not None, lst)) if any(lst) else None
Out[8]: -79

In [9]: lst = [None, None, None]

In [10]: min(filter(lambda x: x is not None, lst)) if any(lst) else None

In [11]: print(min(filter(lambda x: x is not None, lst)) if any(lst) else None)
None

Notes:

Worked in sequence presents as numbers as well as None. If all values is None min() raise exception

ValueError: min() arg is an empty sequence

This code resolve this problem at all

Pros:

Worked if None presents in sequence
Worked on Python 3
max() will be work also

Cons

Need more than one non-zero variable in the list. i.e. [0,None] fails.
Need a variable (example lst) or need duplicate the sequence

This also works for Python 2.7. Since it also resolves the min(None) problem of the other solutions, it gets my +1 — UlrichWuenstel, Mar 23 '18 at 15:23
A boundary case that this does not handle is if it is possible for `lst` to contain only 0 and None values. You would want it to return 0, but it will return None. — BazookaDave, Dec 03 '20 at 21:46

PaulMcG · Answer 4 · 2022-06-30T16:26:56.757

4

Here is a decorator that you can use to filter out None values that might be passed to a function:

def no_nones(fn):
    def _inner(*args):
        return fn(a for a in args if a is not None)
    return _inner

print no_nones(min)(None, 3)
print no_nones(max)(None, 3)

prints:

3
3

edited Jun 30 '22 at 16:26

answered Jun 06 '11 at 17:25

PaulMcG

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Cleaned up to use def instead of lambda, also changed to a PEP8-compliant name. – PaulMcG Jun 30 '22 at 16:27

Steve Mayne · Answer 5 · 2012-04-12T09:01:24.123

3

def max_none(a, b):
    if a is None:
        a = float('-inf')
    if b is None:
        b = float('-inf')
    return max(a, b)

def min_none(a, b):
    if a is None:
        a = float('inf')
    if b is None:
        b = float('inf')
    return min(a, b)

max_none(None, 3)
max_none(3, None)
min_none(None, 3)
min_none(3, None)

edited Apr 12 '12 at 09:01

answered Jun 06 '11 at 16:11

Steve Mayne

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Note that this only works if the default is 0. Passing 0 instead of `None` still triggers passing the default but doesn't change the value since the default is 0 (and `0 == 0.0 == 0j`). – Jun 06 '11 at 16:16
1

The OP said max was not a problem. Also what happens if b is negative ? – Pavan Yalamanchili Jun 06 '11 at 16:16

Blender · Answer 6 · 2011-06-06T16:19:17.967

2

You can use an inline if and an infinity as the default, as that will work for any value:

a = max(a if a is not None else float('-inf'), 3)
b = min(b if b is not None else float('inf'), 3)

edited Jun 06 '11 at 16:19

answered Jun 06 '11 at 16:14

Blender

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Okay, then I'll make it a bit more explicit – Blender Jun 06 '11 at 16:15
Then this will pass `True` on occasion ;) – Jun 06 '11 at 16:17

Steve Mitchell · Answer 7 · 2015-04-28T00:56:59.520

1

a=max(a,3) if a is not None else 3
b=min(b,3) if b is not None else 3

edited Apr 28 '15 at 00:56

answered Apr 27 '15 at 20:27

Steve Mitchell

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I always try to avoid negations. So I prefer `b = 3 if b is None else min(b, 3)` – Martin Thoma Jan 03 '21 at 18:45

Kevin London · Answer 8 · 2016-05-12T17:42:20.563

@utdemir's answer works great for the provided example but would raise an error in some scenarios.

One issue that comes up is if you have a list with only None values. If you provide an empty sequence to min(), it will raise an error:

>>> mylist = [None, None]
>>> min(value for value in mylist if value)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: min() arg is an empty sequence

As such, this snippet would prevent the error:

def find_minimum(minimums):
    potential_mins = (value for value in minimums if value is not None)
    if potential_mins:
        return min(potential_mins)

score 0 · Answer 9 · answered Aug 08 '19 at 17:26

0

I believe the cleanest way is to use filter built-in function

a = max(filter(None, [a, 3]))
b = min(filter(None, [b, 3]))

answered Aug 08 '19 at 17:26

Sam

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3

This filters out zeros. – rrauenza Aug 30 '19 at 23:44

score 0 · Answer 10 · answered Feb 02 '22 at 12:40

0

If you can use np.nan instead of None:

import numpy as np
float(np.nanmax([a, np.nan]))
float(np.nanmin([a, np.nan]))

answered Feb 02 '22 at 12:40

alEx

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Aldhra · Answer 11 · 2022-08-09T14:24:22.050

Based on @PADYMKO's answer you can also use numpy array .any() to filter zeros from None if your array contains both :

>>> lst = np.asarray(lst)
>>> min(filter(lambda x: x is not None, lst)) if (lst != None).any() else None

Results :

>>> lst = [None, None]
>>> lst= np.array(lst)
>>> min(filter(lambda x: x is not None, lst)) if (lst != None).any() else None

>>> lst = [0, None]
>>> lst= np.array(lst)
>>> min(filter(lambda x: x is not None, lst)) if (lst != None).any() else None
0

>>> lst = [0, -5, 2, None]
>>> lst= np.array(lst)
>>> min(filter(lambda x: x is not None, lst)) if (lst != None).any() else None
-5

Python: min(None, x)

11 Answers11