On 05/04/15 15:12, Narci Edson Venturini wrote:
The next code has an unexpected result:
a=3*[3*[0]]
Note that this makes three references to
the list of 3 references to 0.
In other words you reference the same list 3 times.
So when you change the first copy you change the
other 2 also.
Put another way:
>>> mylist = [0,0,0]
>>> a = 3 * [mylist]
'a' now contains 3 references to mylist, making
4 references altogether to the same list object.
If I change the list via any of them it will
change in all the places it is referenced:
>>> mylist[0] = 1 # replace a zero with a one
>>> mylist
[1, 0, 0]
>>> a
[[1, 0, 0], [1, 0, 0], [1, 0, 0]]
When the followind code is ran, them the correct result is obtained:
a=[[0 for i in range(3)] for j in range(3)]
This created 3 new lists. In fact you could simplify it to:
a = [[0,0,0] for i in range(3)]
HTH
--
Alan G
Author of the Learn to Program web site
http://www.alan-g.me.uk/
http://www.amazon.com/author/alan_gauld
Follow my photo-blog on Flickr at:
http://www.flickr.com/photos/alangauldphotos
_______________________________________________
Tutor maillist - Tutor@python.org
To unsubscribe or change subscription options:
https://mail.python.org/mailman/listinfo/tutor
