# GOTCHA with list comprehension

Saran Ahluwalia ahlusar.ahluwalia at gmail.com
Fri Aug 7 15:08:05 CEST 2015

@ChrisA

You, my friend, certainly put the nail in the coffin!

Sent from my iPhone

> On Aug 5, 2015, at 5:52 AM, Chris Angelico <rosuav at gmail.com> wrote:
>
>> On Wed, Aug 5, 2015 at 7:01 PM, Marko Rauhamaa <marko at pacujo.net> wrote:
>> Chris Angelico <rosuav at gmail.com>:
>>
>>> You can chain 'for' and 'if' clauses as much as you like, and they
>>> behave exactly the way you'd expect.
>>
>> How do you know what I'd expect?
>>
>> I wouldn't know what to expect myself.
>
> A list comprehension can always be unwound into statement form. [1] For example:
>
> squares = [n*n for n in range(10)]
>
> can be unwound into:
>
> squares = []
> for n in range(10):
>    squares.append(n*n)
>
> You simply take the expression at the beginning and put that into the
> append() method call, and then the rest become statements. Here's a
> filtered version:
>
> odd_squares = [n*n for n in range(20) if n%2]
>
> Which becomes:
>
> odd_squares = []
> for n in range(20):
>    if n%2:
>        odd_squares.append(n*n)
>
> So what would you expect nested 'if' clauses to do? Well, they become
> nested 'if' statements:
>
> primes = [n for n in range(2,24) if n%2 if n%3]
>
> primes = []
> for n in range(2,24):
>    if n%2:
>        if n%3:
>            primes.append(n)
>
> What if we have multiple 'for' loops? Same thing!
>
> ways_to_get_seven = [(a,b) for a in range(1,7) for b in range(1,7) if a+b==7]
>
> ways_to_get_seven = []
> for a in range(1,7):
>    for b in range(1,7):
>        if a+b==7:
>            ways_to_get_seven.append((a,b))
>
> No matter what combination of 'if' and 'for' you use, it can be
> unwound like this, and it'll always behave the same way. Not sure what
> to expect? Follow the simple rules of unwinding, and then read the
> code that way.
>
> Of course, if you don't know how to predict what the statement form
> will do, then you won't understand what the comprehension will do. But
> that's not the comprehension's fault.
>
> peculiar = [a*x*x+b*x+c for a in range(1,10) for b in range(2,30) for
> c in range(-3,5) if b*b-4*a*c>=0 if (-b+math.sqrt(b*b-4*a*c))/2/a>0
> for x in (x*.01 for x in range(-500,500))]
>
> I suppose you could graph that. Or something. But that's just bad
> code, don't blame the list comp for that...
>
> ChrisA
>
> [1] To be technically correct, this unwinding should be done in a
> nested function, which you then call. There are some other minor
> technicalities too. But it's pretty much this.
> --
> https://mail.python.org/mailman/listinfo/python-list