How to "else" in a list comprehension when the condition is non-boolean?

Question

Given a list e.g.

a = ["No", 1, "No"]

I want to convert this to a new list (or actually just re-assign the list) such that each "No" gets converted to a 0 e.g.

[0,1,0]

If I try the following list comprehension:

[0 for i in a if i =="No"]

This obviously results in a list that omits every element that is not "No"

[0,0]

If I follow the lead given here and try:

[0 if "No" else i for i in a]

This gives me:

[0, 0, 0]

And I guess this is because "No" equates to True.

What am I misunderstanding about the usage of if/else in a list comprehension?

Also related: [Inconsistent comprehension syntax?](https://stackoverflow.com/q/20025280) — Martijn Pieters, Jun 27 '15 at 22:11

Bhargav Rao · Accepted Answer · 2015-06-27T22:19:19.387

4

Make it a boolean expression by using == operator

>>> a = ["No", 1, "No"]
>>> [0 if i == "No" else i for i in a]
[0, 1, 0]

Or another way is

>>> [[0, i][i != "No"] for i in a]
[0, 1, 0]

This works because i != "No" returns a boolean value which can be then interpreted as 1 for True and 0 for false. This then is used to access that particular element from the list [0, i]

edited Jun 27 '15 at 22:19

answered Jun 27 '15 at 22:09

Bhargav Rao

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2

or `[int(i != "No") for i in a]` or `[[1,0 ][i == "No"] for i in a]` ;) – Padraic Cunningham Jun 27 '15 at 22:14
1

Hehe. Nice trick @Padraic. But the OP wants to print that item in the original list, so can't use it here. :( – Bhargav Rao Jun 27 '15 at 22:15
2

I guess `[[0, i][i != "No"] for i in a]` should do the trick – Padraic Cunningham Jun 27 '15 at 22:17
1

@Padraic Yup. Thanks for that – Bhargav Rao Jun 27 '15 at 22:19

How to "else" in a list comprehension when the condition is non-boolean?

1 Answers1