if else for None in Python

Question

Any way to write this in one line in Python, or even better, as an expression instead of a statement?

parts = ['0', '1', 'None', '5', '4']
[int(p) for p in parts]

This of course gives an error,

ValueError: invalid literal for int() with base 10: 'None'

So:

[p=='None' and None or int(p) for p in parts]

Doesn't work because you can't use None with the and/or construct reliably. (Even if p=='None' is true, None is equivalent to false so it goes to the or part.)

[int(p) if p=='None' else None for p in parts]

Also doesn't work. I guess it evaluates int(p) before the conditions? (Seems odd semantics.)

a = []
for p in parts:
    if p=='None': k = None; else: k = int(p)
    a.append(k)

Nope, invalid syntax.

a = []
for p in parts:
    if p=='None':
        k = None;
    else:
        k = int(p)
    a.append(k)

Ah! Finally. But isn't there a shorter way to write such a simple loop?

score 14 · Accepted Answer · answered May 19 '11 at 22:20

14

You almost got it right:

[None if p == 'None' else int(p) for p in parts]

answered May 19 '11 at 22:20

Sven Marnach

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+1 This is the Python form of the [conditional ternary operator](http://en.wikipedia.org/wiki/Conditional_%28programming%29#As_a_ternary_operator): `when-true if expr else when-false` which maps to the C form of `expr ? when-true : when-false` (the operator is lazy in both languages) – May 19 '11 at 22:22
Or the more powerful (and dangerous) `map(eval, parts)` – 6502 May 19 '11 at 22:24
3

@6502: Or the less dangerous `map(ast.literal_eval, parts)`. – Sven Marnach May 19 '11 at 22:25
Reference: [5.11 Conditional Expressions](http://docs.python.org/reference/expressions.html#conditional-expressions) – May 19 '11 at 22:38
@Sven Marnach: Is there a place to submit a "comment of the day" award? that's brilliant? – SingleNegationElimination May 20 '11 at 01:05
A second thought: this seems to count on the fact that None is trivial to evaluate, but I still think it's weird that the part before 'if' gets evaluated first. If I wasn't using None, but calling a function or something, then it would be evaluated regardless of the condition, no? – Steve May 20 '11 at 03:19
@Steve: The condition gets evaluated first, and after it only the part the needs to be evaluated. – Sven Marnach May 20 '11 at 10:36

score 7 · Answer 2 · answered May 19 '11 at 22:22

7

Almost had it:

[int(p) if p != 'None' else None for p in parts]

answered May 19 '11 at 22:22

bluepnume

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A more legible improvement on Sven Marnach's accepted version. Better to put the common case first. – smci Jul 23 '11 at 00:57

score 4 · Answer 3 · answered May 19 '11 at 22:24

4

[ [int, eval] [p=='None'] (p) for p in parts]

Should be save as eval is only called on string 'None'.

answered May 19 '11 at 22:24

Hyperboreus

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I had no idea you could index arrays with booleans. I'd say the syntax is pretty unpythonic, but it's still pretty cool. – Lucas Wiman May 19 '11 at 23:56
Nice idea, thanks. I'll probably stick to if-else for clarity but it's a good answer too. – Steve May 20 '11 at 03:16
1

@intractelicious: See http://stackoverflow.com/questions/3174392/is-it-pythonic-to-use-bools-as-ints/3175293#3175293 for a discussion if this is "Pythonic". – Sven Marnach May 20 '11 at 10:45
Actually I personally use these constructs excessively. Let sendSMTP, sendVoice and SensGSM be functions. So I call {'mail': sendSMTP, 'sip': sendVoice, 'sms': SendGSM} [protocol] (message). To send a message depending on its protocol (being in this case one of the dictionary keys.) – Hyperboreus May 20 '11 at 13:29
Same, I've always used this construct as a switch-statement replacement. – Matt Luongo Sep 28 '11 at 20:35

score 0 · Answer 4 · answered Jun 27 '13 at 07:47

A simpler solution will be:

parts = ['0', '1', 'None', '5', '4']
[int(p) for p in parts if p.isdigit()]

However, float is not accepted here. If float string(?? I dont know what to call that) to integer is a necessity, you can consider the following:

parts = ['0', '1', 'None', '5', '4']
[int(p.split(".")[0]) for p in parts if p.replace(".", "", 1).isdigit()]

score 0 · Answer 5 · answered Sep 06 '11 at 16:22

Do you really only have 1 place in your code where you need this transform? If not, define a function and then make it multiple lines, commented, and easy to follow. That way you hide the complexity of your code, and keep the main code simple to read.

if else for None in Python

5 Answers5