what is operater * really mean when use for list ？

Question

I'm trying to achieve something like C's structure, so I predefine a list :

__list = [0] * 1001

everything works well.

but when the list getting complicated (a sample)：

__list = [{"name": '', "score": 0}] * 3

for item in __list:
    name, score = input("input name and score split with blank:\n").split()    # raw_input for python2
    item['name'] = name
    item['score'] = int(score)
print(__list)

and I input this:

Lily 23
Paul 12
James 28

out:

[{"name": "James", "score": 28}, {"name": "James", "score": 28}, {"name": "James", "score": 28}]

why?

juanchopanza · Accepted Answer · 2015-06-18T06:59:18.857

2

__list = [{"name": '', "score": 0}] * 3

This will create a list with three references to a dictionary. Modifying any one of them modifies them all because they all refer to the same data.

If you want the list to hold references to different dictionaries, you may perform "deep copies" of a reference dict by using a list comprehension, for example,

__list = [{"name": '', "score": 0} for i in xrange(3)]

edited Jun 18 '15 at 06:59

answered Jun 18 '15 at 06:51

juanchopanza

223,364
34
402
480

2

At least in python3 you don't need the deepcopy with list comprehension. You can use `[{"name": "", "score": 0 } for i in range(3)]` – syntonym Jun 18 '15 at 06:58
@syntonym Of course, thanks, that's also true in python2. I updated the post. – juanchopanza Jun 18 '15 at 06:59

what is operater * really mean when use for list ？

1 Answers1