Make dictionary from list with python

Question

I need to transform a list into dictionary as follows.
The odd elements has the key, and even number elements has the value.

x = (1,'a',2,'b',3,'c') -> {1: 'a', 2: 'b', 3: 'c'}

def set(self, val_): 
    i = 0 
    for val in val_: 
        if i == 0: 
            i = 1 
            key = val 
        else: 
            i = 0 
            self.dict[key] = val

A better way to get the same results?

ADDED

i = iter(k)
print dict(zip(i,i))

seems to be working

Asked By: prosseek

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Source

Answer 1

x = (1,'a',2,'b',3,'c') 
d = dict(x[n:n+2] for n in xrange(0, len(x), 2))
print d

Answered By: nosklo

Answer 2

dict(x[i:i+2] for i in range(0, len(x), 2))

Answered By: Alex Martelli

Answer 3

dict(zip(*[iter(val_)] * 2))

Answered By: Ignacio Vazquez-Abrams

Answer 4

Here are a couple of ways for Python3 using dict comprehensions

>>> x = (1,'a',2,'b',3,'c')
>>> {k:v for k,v in zip(*[iter(x)]*2)}
{1: 'a', 2: 'b', 3: 'c'}
>>> {x[i]:x[i+1] for i in range(0,len(x),2)}
{1: 'a', 2: 'b', 3: 'c'}

Answered By: John La Rooy

Answer 5

>>> x=(1,'a',2,'b',3,'c')
>>> dict(zip(x[::2],x[1::2]))
{1: 'a', 2: 'b', 3: 'c'}

Answered By: Mark Tolonen

Make dictionary from list with python

Question:

ADDED

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