Why does this boolean expression evaluate to False?

Question

Can someone explain to me why the Python interpreter evaluates this expression to be False?

1 in [1] == True

I would expect that 1 in [1] would evaluate to True, and obviously True == True would be True. However this isn't what happens - the expression is False. Why does this happen?

Try doing `(1 in [1]) == True` – Kunal Mukherjee May 16 '19 at 17:33 — Kunal Mukherjee, May 16 '19 at 17:33
(1 in [1]) == True evaluates to True as expected – rickyrichboy May 16 '19 at 17:35 — rickyrichboy, May 16 '19 at 17:35

score 11 · Accepted Answer · answered May 16 '19 at 17:36

11

== and in are both comparison operators. And when you have multiple comparison operators like this, Python considers it a chained comparison. For example, 1 < x < 10 is equivalent to 1 < x and x < 10.

In your case, 1 in [1] == True is equivalent to (1 in [1]) and ([1] == True), which evaluates to False.

answered May 16 '19 at 17:36

interjay

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Ohhh now that makes sense. Thanks! – rickyrichboy May 16 '19 at 17:37

score 1 · Answer 2 · answered May 16 '19 at 17:37

1

If you have an expression like this python split it to more statements. In fact:

1 in [1] == True equals to: (1 in [1]) and ([1] == True)

Right side is false since [1] != True and whole sentence is false

answered May 16 '19 at 17:37

Relandom

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Why does this boolean expression evaluate to False?

2 Answers2