I need your help to solve a problem.
I want to convert a dictionary d = {key1:value1, key2:value2} into list= [keys1, keys1, ... (value1 times), keys2, ... (value2 times)] without using a nested loop.
Example:
d1 = {4: 1, 3: 2, 12: 2}
The code should produce the output:
l = [4, 3, 3, 12, 12]
This is what I have:
for key, value in nums1.items():
temp = (str(key))*value
nums2.append(int(temp))
print(nums2)
Which gives: [4, 33, 1212], but should give [4, 3, 3, 12, 12].
The complexity should be O(n).
The easiest solution is to use collections.Counter. It features an elements() method that yields all elements with the correct count:
>>> from collections import Counter
>>> list(Counter(d1).elements())
[4, 3, 3, 12, 12]
If you want to implement this yourself, I think the most readable version is this for loop:
from itertools import repeat
result = []
for k, count in d1.items():
result += repeat(k, count)
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