Why does setting a key/value pair of list of dictionaries fill all dictionaries?

Question

test = [{}]*5 # Creates [{}, {}, {}, {}, {}]
print(test[1]) # outputs "{}"

test[1]['asdf'] = 5

print(test)

This gives test as [{'asdf': 5}, {'asdf': 5}, {'asdf': 5}, {'asdf': 5}, {'asdf': 5}]. Somehow, all the dictionaries in the list have been set to the same value

But this doesn't happen if we initialize the list a different way:

test2 = [{} for i in range(5)] # Also [{}, {}, {}, {}, {}]
test2[1]['asdf'] = 1

print(test2)

test2 now equals [{}, {'asdf': 1}, {}, {}, {}], the expected result.

See also: https://stackoverflow.com/questions/240178/list-of-lists-changes-reflected-across-sublists-unexpectedly — Carcigenicate, Jul 26 '21 at 23:10
Because `[{}]*5` creates a list with five references to the same list* — juanpa.arrivillaga, Jul 26 '21 at 23:15

James Wright · Answer 1 · 2021-07-27T23:15:54.353

0

The difference between the two is that for [{}]*5, Python creates a 5 element list containing the dictionary given in the square brackets. Ie. it's 5 different references to the same dictionary.

While the other method creates a new dictionary for every element in the list.

More detailed explanation can be found in the duplicate question link

edited Jul 27 '21 at 23:15

answered Jul 26 '21 at 23:08

James Wright

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Why does setting a key/value pair of list of dictionaries fill all dictionaries?

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