Why does python ignore the first part of a condition?

Question

Suppose I have the following code:

target = 'abc1234y'
if ('x' and 'y') in target:
    print('x and y are in target')

Why is if ('x' and 'y') in target true?

Why doesn't this code produce error?

Why do braces have no effect?

This does not seem logical, since if ('x' and 'y') = true then ('y' and 'x') should also be true, however, this is no the case.

At the same time expression if ('y' and 'x') in target is false

What do you mean *Why do braces have no effect?*. They **do**. The difference is between `('x' and 'y') in target` and `'x' and ('y' in target)`. They're very different! — Nick stands with Ukraine, Dec 14 '17 at 00:54
Can you please explain in words what you're trying to compare? Because it is not "x and y are in the target string" — OneCricketeer, Dec 14 '17 at 00:55
Braces have no effect means I expect ('x' and 'y') mean 'x' and 'y' not x and 'y in somewhere' — 0rt, Dec 14 '17 at 00:57
@0rt Unfortunately that's not how programming works, you'll need to say `'x' in target and 'y' in target` or something similar to the same effect — Nick stands with Ukraine, Dec 14 '17 at 00:58
@0rt No, read the first comment I posted and go ***thoroughly*** read up on operator precedence, and also probably [short-circuit evaluation](https://stackoverflow.com/questions/2580136/does-python-support-short-circuiting) — Nick stands with Ukraine, Dec 14 '17 at 01:01

Kilian Batzner · Accepted Answer · 2017-12-14T01:08:27.633

4

There are two things to consider here:

How `and` works with strings

'x' and 'y' is not True.

'x' and 'y' is 'y', which might seem confusing but it makes sense when looking at how and and or work in general:

a and b returns b if a is True, else returns a.
a or b returns a if a is True, else returns b.

Thus:

'x' and 'y' is 'y'
'x' or 'y' is 'x'

What the parentheses do

The brackets do have an effect in your if statement. in has a higher precedence than and meaning that if you write

if 'x' and 'y' in target:

it implicitly means the same as

if 'x' and ('y' in target):

which will only check if y is in the target string. So, using

if ('x' and 'y') in target:

will have an effect. You can see the whole table of operator precedences in the Python docs about Expressions.

Solution

To achieve what you are wanting to do, @Prem and @Olga have already given two good solutions:

if 'y' in target and 'x' in target:

or

if all(c in target for c in ['x','y']):

edited Dec 14 '17 at 01:08

answered Dec 14 '17 at 00:37

Kilian Batzner

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I agree the down votes are just random. The answers look correct. Not sure what is wrong with the answers. – Spandan Brahmbhatt Dec 14 '17 at 00:46
You have my upvote for the nice explanation on behaviour – IMCoins Dec 14 '17 at 00:48
so, in python and does not act like &? – 0rt Dec 14 '17 at 00:49
2

@0rt `and` and `&` are quite different things in python. `and` is logical, `&` is bitwise – Nick stands with Ukraine Dec 14 '17 at 00:50
Could you explain why do braces have no effect on the expression? – 0rt Dec 14 '17 at 00:54
@0rt They do have an effect! – Nick stands with Ukraine Dec 14 '17 at 00:56

Spandan Brahmbhatt · Answer 2 · 2017-12-14T00:37:48.430

2

You seem to be confusing how AND operator works.

'x' and 'y'
Out[1]: 'y'

and

'y' and 'x'
Out[1]: 'x'

So in your first case, you do check if y is present. In reverse, you check if x is present in string.

edited Dec 14 '17 at 00:37

answered Dec 14 '17 at 00:36

Spandan Brahmbhatt

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why does it ignore the braces then? – 0rt Dec 14 '17 at 00:37
@0rt it doesn't ignore braces, `('x' and 'y')` is `'y'` so all your statement is doing is `'y' in target` – Nick stands with Ukraine Dec 14 '17 at 00:38
1

The braces says, the condition should be evaluated first. – Spandan Brahmbhatt Dec 14 '17 at 00:38
maybe you want to test "if 'x' in target and 'y' in target". – Dec 14 '17 at 00:42
I'd like to understand the 'x' and 'y' behaviour. Is it a known and wanted behaviour or just some side effect to consider ? – IMCoins Dec 14 '17 at 00:45
@IMCoins it's well defined behaviour, it's called [short-circuit evaluation](https://stackoverflow.com/questions/2580136/does-python-support-short-circuiting) – Nick stands with Ukraine Dec 14 '17 at 01:04

score 0 · Answer 3 · answered Dec 14 '17 at 00:43

0

Because it is checking for just 'y' in target. You need something like this:

target = 'abc1234y'
if 'y' in target and 'x' in target:
    print('x and y are in target')

answered Dec 14 '17 at 00:43

Prem

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score 0 · Answer 4 · answered Dec 14 '17 at 00:50

0

You may want to use all instead if all(c in target for c in ['x','y']):...

answered Dec 14 '17 at 00:50

Olga

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1

this is not what I was asking for – 0rt Dec 14 '17 at 00:58

Why does python ignore the first part of a condition?

4 Answers4

How and works with strings

What the parentheses do

Solution

How `and` works with strings