Python merge two lists with different lengths - python-2.7

Question

Hi please help me develop a logic which does following.
list_1 = [1,2,3]
list_2 = [a,b,c,d,e,f,g,h,i]
Required output (List of dictionaries):
output = [{1:a,2:b,3:c}, {1:d,2:e,3:f}, {1:g,2:h,3:i}]
My script:
return_list = []
k = 0
temp_dict = {}
for i, value in enumerate(list_2):
if k <= len(list_1)-1:
temp_dict[list_1[k]] = value
if k == len(list_1)-1:
k = 0
print temp_dict
return_list.append(temp_dict)
print return_list
print '\n'
else:
k = k + 1
print return_list
My output:
{1: 'a', 2: 'b', 3: 'c'}
[{1: 'a', 2: 'b', 3: 'c'}]
{1: 'd', 2: 'e', 3: 'f'}
[{1: 'd', 2: 'e', 3: 'f'}, {1: 'd', 2: 'e', 3: 'f'}]
{1: 'g', 2: 'h', 3: 'i'}
[{1: 'g', 2: 'h', 3: 'i'}, {1: 'g', 2: 'h', 3: 'i'}, {1: 'g', 2: 'h', 3: 'i'}]
[{1: 'g', 2: 'h', 3: 'i'}, {1: 'g', 2: 'h', 3: 'i'}, {1: 'g', 2: 'h', 3: 'i'}]
As you can see, temp_dict is getting printed correctly, but return_list is the last temp_dict 3 times.
please help to solve.

Answer 1

return_list = []
k = 0
temp_dict = {}
for i, value in enumerate(list_2):
if k <= len(list_1)-1:
temp_dict[list_1[k]] = value
if k == len(list_1)-1:
k = 0
print temp_dict
return_list.append(temp_dict)
temp_dict = {}
else:
k = k + 1

Answer 2

For simple, you could use zip
list_1 = [1,2,3]
list_2 = ['a','b','c','d','e','f','g','h','i']
chunks = [list_2[idx:idx+3] for idx in range(0, len(list_2), 3)]
output = []
for each in chunks:
output.append(dict(zip(list_1, each)))
print(output)

Answer 3

from itertools import cycle
list_1 = [1, 2, 3]
list_2 = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
def chunks(l, n):
for i in range(0, len(l), n):
yield l[i:i + n]
zipped = tuple(zip(cycle(list_1), list_2))
out = list(map(dict, chunks(zipped, len(list_1))))
print(out)
Gives you:
[{1: 'a', 2: 'b', 3: 'c'}, {1: 'd', 2: 'e', 3: 'f'}, {1: 'g', 2: 'h', 3: 'i'}]