You misunderstand how boolean expressions work; they don’t work like an English sentence and guess that you are talking about the same comparison for all names here. You are looking for:

```
if x == 1 or y == 1 or z == 1:
```

`x`

and `y`

are otherwise evaluated on their own (`False`

if `0`

, `True`

otherwise).

You can shorten that using a containment test against a tuple:

```
if 1 in (x, y, z):
```

or better still:

```
if 1 in {x, y, z}:
```

using a `set`

to take advantage of the constant-cost membership test (i.e. `in`

takes a fixed amount of time whatever the left-hand operand is).

### Explanation

When you use `or`

, python sees each side of the operator as *separate* expressions. The expression `x or y == 1`

is treated as first a boolean test for `x`

, then if that is False, the expression `y == 1`

is tested.

This is due to operator precedence. The `or`

operator has a lower precedence than the `==`

test, so the latter is evaluated *first*.

However, even if this were *not* the case, and the expression `x or y or z == 1`

was actually interpreted as `(x or y or z) == 1`

instead, this would still not do what you expect it to do.

`x or y or z`

would evaluate to the first argument that is ‘truthy’, e.g. not `False`

, numeric 0 or empty (see boolean expressions for details on what Python considers false in a boolean context).

So for the values `x = 2; y = 1; z = 0`

, `x or y or z`

would resolve to `2`

, because that is the first true-like value in the arguments. Then `2 == 1`

would be `False`

, even though `y == 1`

would be `True`

.

The same would apply to the inverse; testing multiple values against a single variable; `x == 1 or 2 or 3`

would fail for the same reasons. Use `x == 1 or x == 2 or x == 3`

or `x in {1, 2, 3}`

.