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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

Abdulrahman Bres
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0rt
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    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 is tired 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 is tired Dec 14 '17 at 00:58
  • @NickA, does it omit them? – 0rt Dec 14 '17 at 00:59
  • @0rt Does it omit what? – Nick is tired Dec 14 '17 at 01:00
  • @NickA, braces around expression ('x' and 'y') – 0rt Dec 14 '17 at 01:01
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    @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 is tired Dec 14 '17 at 01:01

4 Answers4

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']):
Kilian Obermeier
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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.

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

Olga
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