CODEWITHSUNDEEP
pythonpython-3.xlist
Ahima
Ahima

Reputation: 83

How the short if with array works in python

Can someone explain me how this syntax works in Python? How is this expression being evaluated?

b = False
a = ['123', '456'][b == True]
print(a) => 123

b = True
a = ['123', '456'][b == True]
print(a) => 456

Upvotes: 1

Views: 158

Answers (2)

shellinator
shellinator

Reputation: 3

Adding to Selcuk's answer:

This is also an list indexing question. Let's make replace the values of False and True with its boolean equivalent 0 and 1. We will also solve for those booleans expression in the brackets.

>>> b = 0
>>> a = ['123', '456'][0]  # [b == True] is [0 == 1] is [False] 
>>> print(a)
'123'

>>> b = 1
>>> a = ['123', '456'][1]  # [b == True] is [1 == 1] is [True]
>>> print(a)
'456'

This looks far less intimidating. The first a wants the value of index 0 which is 123. The second a wants the value of index 1 which is 456.

You can try it out with a larger array to confirm:

>>> a = ['123','456','789'][2]
>>> a
'789'

More info on Python lists here.

Upvotes: 0

Selcuk
Selcuk

Reputation: 59315

Python booleans can implicitly be converted to integers where False is 0 and True is 1. You can see it more clearly in this example:

>>> ["foo", "bar"][True]
'bar'
>>> ["foo", "bar"][False]
'foo'

Since b == True returns a boolean its value can also be interpreted as either 0 or 1. ['123', '456'][b == True] simply returns the 0th or 1st element of the list depending on the value of b.

That being said, this is an obfuscated and unreadable way of writing conditionals. Prefer the proper ternary expression:

a = '123' if b else '456'

Upvotes: 3

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