`elif` in list comprehension conditionals

Question

Python's conditional expressions were designed exactly for this sort of use-case:

>>> l = [1, 2, 3, 4, 5]
>>> ['yes' if v == 1 else 'no' if v == 2 else 'idle' for v in l]
['yes', 'no', 'idle', 'idle', 'idle']

>>> d = {1: 'yes', 2: 'no'}
>>> [d.get(x, 'idle') for x in l]
['yes', 'no', 'idle', 'idle', 'idle']

You can, sort of.

Note that when you use sytax like:

['yes' if v == 1 else 'no' for v in l]

You are using the ternary form of the if/else operator (if you're familiar with languages like C, this is like the ?: construct: (v == 1 ? 'yes' : 'no')).

The ternary form of the if/else operator doesn't have an 'elif' built in, but you can simulate it in the 'else' condition:

['yes' if v == 1 else 'no' if v == 2 else 'idle' for v in l]

This is like saying:

for v in l:
    if v == 1 :
        print 'yes'
    else:
        if v == 2:
            print 'no'
        else:
            print 'idle'

So there's no direct 'elif' construct like you asked about, but it can be simulated with nested if/else statements.

You can use list comprehension is you are going to create another list from original.

>>> l = [1, 2, 3, 4, 5]
>>> result_map = {1: 'yes', 2: 'no'}
>>> [result_map[x] if x in result_map else 'idle' for x in l]
['yes', 'no', 'idle', 'idle', 'idle']

Another easy way is to use conditional list comprehension like this:

l=[1,2,3,4,5]
print [[["no","yes"][v==1],"idle"][v!=1 and v!=2] for v in l]

gives you the correct anwer:

['yes', 'no', 'idle', 'idle', 'idle']

Maybe you want this:

l = [1, 2, 3, 4, 5] 

print ([['idle','no','yes'][2*(n==1)+(n==2)] for n in l])