Best way to add a "position" value to each item in a list

Jerry Hill malaclypse2 at
Thu Jul 9 16:49:22 EDT 2009

On Thu, Jul 9, 2009 at 4:16 PM, Sean<sberry2a at> wrote:
> I have a huge list, 10,000,000+ items.  Each item is a dictionary with
> fields used to sort the list.  When I have completed sorting I want to
> grab a page of items, say 1,000 of them which I do easily by using
> list_data[x:x+1000]
> Now I want to add an additional key/value pair to each dictionary in
> the list, incrementing them by 1 each time.  So, if I grabbed page 2
> of the list I would get:
> [{'a':'a', 'b':'b', 'position':1001}, {'c':'c', 'd':'d', 'position':
> 1002}, ...]
> Any way to do that with list comprehension?  Any other good way to do
> it besides iterating over the list?

Normally you wouldn't mutate the items in a list with a list
comprehension.  Instead, you would use a for loop, like this:

for idx,item in enumerate(my_list_of_dicts):
    item['position'] = idx

Is there a particular reason you want to do this with a list
comprehension?  Using a list comp means you're going to create an
extra copy of your 10 million item list, just so you can add a key to
each member, then (probably) throw away the original list.  It doesn't
seem like the right tool for the job here.


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