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If I have a large list that I want to create a dictionary out of, what would be the most efficient way of doing this supposing I just want to assign the value as so:

{'item1':'0','item2':'1','item3':'2','itemn':'n-1'}

I've seen a lot on here about just assigning same value to all the keys, but nothing about how to assign the values as I need.

Thanks.

EDIT: The reason I want to do this is because I've been handed someone's code that is atrocious...(not that I'm a skilled programmer by any means), and we have a list of objects that are represented by ID numbers: 5432, 8976, etc etc.

There's a few hundred of them. Well rather than, as the original code does, treat the list as an array, then find its range(len(my_list)) to get an indicial value for each object, I was thinking it might be better to just create a dictionary with keys/values, declare that once and refer to it later rather than recalling the array or -in this case- recreating the array every time. Is that a reasonable idea? I don't know, but I wanted to try it.

Matt
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    please post what your input and expected output are for a real list – Joran Beasley Jul 18 '13 at 22:45
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    Although this is a fine question, you might want to consider why you want to do this transformation. The reason is that you've added precisely zero information to what you already have; this makes me wonder what larger purpose you are hoping to accomplish. – msw Jul 18 '13 at 23:08
  • @msw Gave more info, see edit. – Matt Jul 19 '13 at 01:15
  • Thanks for the additional information. A dictionary **maps** something to something else, and it still isn't clear what you are trying to map; in particular the phrase "list of objects" is unclear as you use it next to the list `[5432, 8976, ...]`. In Python jargon, you can have a list of indices, or a list of objects and your question leaves what you actually have ambiguous. (Technically, a list of indices is a list of objects, but sort of a degenerate one). – msw Jul 19 '13 at 01:43
  • Or, put another way, given index `5342` what would you do with it? – msw Jul 19 '13 at 01:45

6 Answers6

2

Try this command:

d = dict(zip(your_list, range(len(your_list)))
Ryan Haining
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Frodon
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dict(x[i:i+2] for i in range(0, len(x), 2))
Ran Eldan
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  • This just assigns the next item in the list as the value. I assume you got that from [here](http://stackoverflow.com/questions/2597166/make-dictionary-from-list-with-python) – Matt Jul 18 '13 at 22:52
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{item: str(i) for i, item in enumerate(my_list)}
ijmarshall
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The answer seems to depend on which version of Python you are using.

These are my results with IPython 0.13 on Python 2.7.3:

In [11]: mylist = ["item{}".format(i) for i in range(10**7)]

In [12]: %timeit d = dict(zip(mylist, range(len(mylist))))
1 loops, best of 3: 6.97 s per loop

In [13]: %timeit d = {item: i for i, item in enumerate(mylist)}
1 loops, best of 3: 3.68 s per loop

In [14]: # Edit: xrange is faster than range in Python 2    

In [15]: %timeit d = dict(zip(mylist, xrange(len(mylist))))
1 loops, best of 3: 5.58 s per loop

And this is what I get with IPython 0.13 on Python 3.3.2:

In [5]: mylist = ["item{}".format(i) for i in range(10**7)]

In [6]: %timeit d = dict(zip(mylist, range(len(mylist))))
1 loops, best of 3: 2.62 s per loop

In [7]: %timeit d = {item: i for i, item in enumerate(mylist)}
1 loops, best of 3: 2.92 s per loop

This is probably because zip doesn't return a list in Python 3 but only an iterator…

sjakobi
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dict(("item%d"%i,val) for i,val in enumerate(my_list,1))

presumably

Joran Beasley
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If you don't care about the assigned values for now, the fastest way to build a dict from a list is dict.fromkeys(mylist):

In [13]: mylist = ["stuff", "more stuff"]

In [14]: dict.fromkeys(mylist)
Out[14]: {'more stuff': None, 'stuff': None}
sjakobi
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