Python for loop with condition

Question

I am quite new to Python (have a C++ background). I have a function in my Python code that returns an array of some objects (on which I don't have any control) like below:

def _get_object_array():
    object_array[]
    # Magically get array contents from somewhere
    # The objects have some fields like id, name etc.
    return object_array

Then in another function I call _get_object_array() to retrieve only the id's of the objects returned.

What I have is following:

id_array = []
id_array = [x.id for x in _get_object_array()] #<--

But I don't want the ids of all the objects in the returned array. Instead I only want to store the ids for array elements which meet some criteria (e.g. id is an even number?)

Is there a way I can do that in the same line marked with <-- like below (which btw is just a pseudo code)?

id_array = [if x.id % 2 == 0: x.id for x in _get_object_array()]

Thanks in advance.

/R

Morgan Harris · Accepted Answer · 2013-09-05 03:43:40Z

5

Yep: same words, you just got the syntax backwards.

id_array = [x.id for x in _get_object_array() if x.id % 2 == 0]

List comprehensions are amazing, and can do a lot more work than most people realise; have a read of the relevant PEP.

answered Sep 5, 2013 at 3:43

Morgan Harris

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2 Comments

Perkins Over a year ago

Also remember if you are going to iterate over the array a single time later, you can use () instead of [] to create a generator that calculates the values on the fly instead of all at once, it saves memory that way.

Morgan Harris Over a year ago

Yeah, that's also great for using any and all.

Tadeck · Accepted Answer · 2013-09-05 03:44:56Z

2

Yes, and you were very close:

id_array = [x.id for x in _get_object_array() if x.id % 2 == 0]

answered Sep 5, 2013 at 3:44

Tadeck

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