list to map in python

Question

I have a list and I want to convert this list into map

mylist = ["a",1,"b",2,"c",3]

mylist is equivalent to

mylist = [Key,Value,Key,Value,Key,Value]

So Input:

mylist = ["a",1,"b",2,"c",3]

Output:

mymap = {"a":1,"b":2,"c":3}

P.S: I already have written following function that do same work, but I want to use iterator tools of python:

def fun():
    mylist = ["a",1,"b",2,"c",3]
    mymap={}
    count = 0
    for value in mylist:
        if not count%2:
            mymap[value] = mylist[count+1]
        count = count+1
    return mymap

score 11 · Accepted Answer · edited May 23 '17 at 12:28

11

Using iter and dict-comprehension:

>>> mylist = ["a",1,"b",2,"c",3]
>>> it = iter(mylist)
>>> {k: next(it) for k in it}
{'a': 1, 'c': 3, 'b': 2}

Using zip and iter:

>>> dict(zip(*[iter(mylist)]*2)) #use `itertools.izip` if the list is huge.
{'a': 1, 'c': 3, 'b': 2}

Related: How does zip(*[iter(s)]*n) work in Python

edited May 23 '17 at 12:28

Community

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answered Sep 03 '13 at 17:33

Ashwini Chaudhary

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1

That second solution is beautiful. – Slater Victoroff Sep 03 '13 at 17:36
1

Ashwini, Can you please explain your second solution ?? – Anurag Sep 03 '13 at 17:43
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@Anurag [How does `zip(*[iter(s)]*n)` work in Python](http://stackoverflow.com/questions/2233204/how-does-zipitersn-work-in-python) – Ashwini Chaudhary Sep 03 '13 at 17:44
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`it = iter(mylist); dict(zip(it, it))` if you want the second solution as a two-liner that mirrors the first solution. – Steven Rumbalski Sep 03 '13 at 17:58
Also a slicing solution: `dict(zip(mylist[::2], mylist[1::2]))` – Steven Rumbalski Sep 03 '13 at 18:00

score 4 · Answer 2 · edited Sep 03 '13 at 19:42

4

>>> mylist = ["a",1,"b",2,"c",3]
>>> zippedlist = zip(mylist [0::2],mylist [1::2]) #slicing magic
>>> zippedlist
[('a', 1), ('b', 2), ('c', 3)]
>>> dictedlist = dict(zippedlist)
>>> dictedlist 
{'a': 1, 'c': 3, 'b': 2}

This works because of slicing [start:stop:skip]

list starting at 0, skipping 2

zipped with

list starting at 1, skipping 2

edited Sep 03 '13 at 19:42

Ashwini Chaudhary

244,495
58
464
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answered Sep 03 '13 at 18:04

TehTris

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score 1 · Answer 3 · edited Sep 03 '13 at 19:42

To get all the keys from the list you can do:

keys = mylist[::2]

where the mylist[::2] creates a new list by iterating over every second element in mylist starting at index 0.

To get all the values from the list you can do:

vals = mylist[1::2]

where the mylist[1::2] creates a new list by iterating over every second element in mylist starting at index 1.

and then you can just use dict and zip:

dict(zip(keys,vals))  # or all in one line
dict(zip(mylist[::2], mylist[1::2]))

zip takes two lists like ["a","b","c"] and [1,2,3] and "zips" them together to give a list like

[("a",1), ("b", 2), ("c", 3)]

and dict takes an iterable of pairs and turns them into a dictionary.

score 0 · Answer 4 · answered Sep 03 '13 at 17:41

Using Compress and Cycle:-

from itertools import compress
from itertools import cycle

def fun(mylist):
    keys = compress(mylist,cycle([1,0]))
    values = compress(mylist,cycle([0,1]))
    mymap = dict(zip(keys,values))
    return mymap

mylist = ["a",1,"b",2,"c",3]

fun(mylist)
{'a': 1, 'c': 3, 'b': 2}

list to map in python

4 Answers4