Multiple IF conditions in a python list comprehension

Question

I was wondering, is it possible to put multiple if conditions in a list comprehension? I didn\'t find anything like this in the docs.

I want to be able

Answer 1

How about

ar = [('four' if i % 4 == 0 else ('six' if i % 6 == 0 else i)) for i in range(1, n)]

For example, if n = 30 this is

[1, 2, 3, 'four', 5, 'six', 7, 'four', 9, 10, 11, 'four', 13, 14, 15, 'four', 17, 'six', 19, 'four', 21, 22, 23, 'four', 25, 26, 27, 'four', 29]

ETA: Here's how you could apply a list of conditions:

CONDITIONS = [(lambda i: i % 4 == 0, "four"), (lambda i: i % 6 == 0, "six"),
              (lambda i: i % 7 == 0, "seven")]

def apply_conditions(i):
    for condition, replacement in CONDITIONS:
        if condition(i):
            return replacement
    return i

ar = map(apply_conditions, range(0, n))

Answer 2

ar = ["four" if i%4==0 else "six" if i%6==0  else i for i in range(1,30)]

Answer 3

You can put you logic in a separate function, and then have the elegance of the list comprehension along with the readability of the function:

def cond(i):
    if i % 4 == 0: return 'four'
    elif i % 6 == 0: return 'six'

    return i

l=[cond(i) for i in range(1,n)]