Create list from values of a dictionary using a list containing the order of the keys

Question

I have a dictionary which contains a certain number of key/value pairs such as:

d = { "v1": 1, "v2": 2, "v3":3, "v4": 4 }

and a list containing the keys in a certain order:

l = [ "v2", "v3", "v1", "v4" ]

how can I obtain a list of values using the order of l in an elegant way ?

[ 2, 3, 1, 4]

The only thing I could think of is:

[ d[k] for k in l ]

Same question if I wanted to obtain a tuple.

Anand S Kumar · Answer 1 · 2015-06-26T17:31:23.313

3

For Python 2.x , try using -

>>> d = { "v1": 1, "v2": 2, "v3":3, "v4": 4 }
>>> l = [ "v2", "v3", "v1", "v4" ]
>>> map(d.get,l)
[2, 3, 1, 4]

If you are using Python 3.x , map function returns a map object which you can iterate over, for converting to list use list function , Example -

>>> list(map(d.get,l))
[2, 3, 1, 4]

edited Jun 26 '15 at 17:31

answered Jun 26 '15 at 17:27

Anand S Kumar

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Remove the list() to make this even better in Python 2. – garnertb Jun 26 '15 at 17:28
map returns a `map object` over which we can iterate, tried in Python 3.x though – Anand S Kumar Jun 26 '15 at 17:29
`map()` returns a `list` in Python 2 and a `map` in Python 3. – TigerhawkT3 Jun 26 '15 at 17:30
Edited the answer to include that as well. – Anand S Kumar Jun 26 '15 at 17:31
Since the question is for python 2.7 , update the 2.x version first and then the 3.x version as a side note – Anand S Kumar Jun 26 '15 at 17:33

Create list from values of a dictionary using a list containing the order of the keys

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