How to append to Python list in a dict without having to initialize the list?

Question

In PHP, I can do this

$a['name'][] = 'el1';

In Python, I have to do this:

a = {}
a['name'] = []
a['name'].append('el1')

The problem with the above approach is that I'm losing the value of a['name'] in the next loop when iterating because I'm reinitializing a['name'] to an empty list .

Is there any way to append items to a list without having to initialize with an empty list first?

Answer 1

You can use the defaultdict module:

from collections import defaultdict
a = defaultdict(list)

With this setting, the module initializes to an empty list on first access, so you can just do:

 a['name'].append('el1')

and not worry about it.

Hope that helps!

Answer 2

So, the way that you've phrased it no, you can only append to something that exists. That said, you don't have to append to this at all, you could do this:

a = {'name': ['el1']}

No need to manually create your structure like that.

If you want to check for the existence of a key before trying to append to a list you can do this:

a.setdefault('name', []).append('el1')

The second argument to get also gives you a default, saying if name doesn't exist, append to an empty list instead.

Answer 3

I think you're asking for something like this:

a = {}

if 'name' not in a:
    a['name'] = []

a['name'].append('el1')

Also, you might want to look at Python's defaultdict datatype to see if that would be better for your purposes.