CODEWITHSUNDEEP
pythondictionary
Talysin
Talysin

Reputation: 363

How to turn a python dictionary into a list with intentionally duplicate values

I have a dictionary in the following form:

{'string1': 3, 'string2': 4, 'string3':2}

I want to turn it into a list with instances of the key as a function of its corresponding value pair. Like so:

['string1','string1','string1','string2','string2','string2','string2','string3','string3']

I am having an annoyingly difficult time figuring this out, as I feel there should be an elegant solution. Here's what I have so far:

t2 = [dictionary[key]*key for key in dictionary.keys()]

but that yields the following mess:

['string2string2string2string2', 'string3string3', 'string1string1string1']

Upvotes: 2

Views: 65

Answers (3)

Andria
Andria

Reputation: 5075

The only thing you messed up was you forgot to put key in square brackets, here: 

dictionary = {'string1': 3, 'string2': 4, 'string3':2}
t2 = [dictionary[key]*[key] for key in dictionary.keys()]
print(t2)

result:

[['string2', 'string2', 'string2', 'string2'], ['string1', 'string1', 'string1'], ['string3', 'string3']]

* EDIT *

If you want a flattened version here's the modified code for that:

t2 = [''.join(j) for j in [[i]for arr in [dictionary[key]*[key] for key in dictionary.keys()] for i in arr]]

result:

['string3', 'string3', 'string2', 'string2', 'string2', 'string2', 'string1', 'string1', 'string1']

Upvotes: 1

AChampion
AChampion

Reputation: 30258

You can use collections.Counter() to do this:

>>> list(Counter(d).elements())
['string1', 'string1', 'string1', 'string3', 'string3', 'string2', 'string2', 'string2', 'string2']

Of course with a dictionary, order is not guaranteed.

Upvotes: 4

akuiper
akuiper

Reputation: 214927

You can try this, if the order of the keys is not important:

[k for k, v in d.items() for _ in range(v)]

#['string2',
# 'string2',
# 'string2',
# 'string2',
# 'string3',
# 'string3',
# 'string1',
# 'string1',
# 'string1']

Upvotes: 5

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