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Creating a dictionary from a list in Python

Here's the original list:

['name', 'value', 'name', 'value', 'name', 'value']

Etc. I need to extract name / value pairs into a dictionary:

{'name': 'value', 'name': 'value', 'name': 'value'}

Can anyone describe the easiest way to do this?

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4 answers


If L

is your original list, you can use zip(*[iter(L)]*2)

to group items into pairs. The dict constructor can accept the iterable of such pairs directly

>>> L = ['name1', 'value1', 'name2', 'value2', 'name3', 'value3']
>>> dict(zip(*[iter(L)]*2))
{'name1': 'value1', 'name2': 'value2', 'name3': 'value3'}


I'm not sure what you mean by simpler (easier to understand?). It's hard to guess if you think this is easier as I don't know what level you are at. Here is the way without using iter

or zip

. If you don't already know what enumerate

, you should watch it.

>>> d = {}
>>> for i, item in enumerate(L):
...     if i % 2 == 0:
...         key = item
...     else:
...         d[key] = item
... 
>>> d
{'name1': 'value1', 'name2': 'value2', 'name3': 'value3'}
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Don't take anyone away. I think it might be a little easier to understand:

dict (zip (L[::2] , L[1::2] ))


Though less efficient for a large list than gnibbler's answer.

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You need a function to group a sequence into fixed length chunks. Unfortunately this is missing from the Python core. You can use partition

in the library funcy . (Note: in other languages ​​this is called chunksOf or grouped ).

The itertools documentation offers this feature:

from itertools import zip_longest

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

And the zip documentation suggests another way to accomplish the same thing (quoted in gnibbler's answer ):

def grouper(iterable, n):
    return zip(*[iter(iterable)]*n)

Once it is available, the rest of the assignment will be trivial.

>>> dict(grouper(range(8), 2))
{0: 1, 2: 3, 4: 5, 6: 7}
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You can try this

>>> a = ['a', 'b', 'c', 'd', 'e', 'f']
>>> dict(zip([x for x in a[::2]], [x for x in a[1::2]]))
{'a': 'b', 'c': 'd', 'e': 'f'}

a[::2]

will get the entire second element from the beginning of the list from the element 0th

.

a[1::2]

will get the entire second element from the beginning of the list from the element 1st

.

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