doc - 1 year ago 39

Python Question

This works as intended:

`if c1 in r':6' or c2 in r':6':`

subst_weight = 1

This doesn't:

`if (c1 or c2) in r':6':`

subst_weight = 1

Why and what is the differnce? The goal is to find out whether

`c1`

`c2`

Same here:

Works:

`if c1 == '6' or c2 == '6':`

Doesn't work:

`if (c1 or c2) == '6':`

Thanks

Answer Source

The statement `one or two`

will return `one`

if defined, and `two`

if `one`

is not truthy, and I don't see a reason it wouldn't work with your logic because if either is truthy, and is contained in string, it will evaluate to `True`

```
>>> a = "6"
>>> b = ":"
>>> (a or b)
'6'
>>> (b or a)
':'
>>> a = None
>>> (a or b)
':'
>>> b
':'
```

For more than 2 variables it will return the first that's truthy

```
>>> a = None
>>> b = None
>>> c = 6
>>> a or b or c
6
```

But note your logic, you want to check if any of the variables is in that string, you can't use this, because the `c1`

ca be equal to `"adasd"`

and that's what's gonna be returned to that `if`

statement and it's not in `":6"`