Jan Danielsson <[EMAIL PROTECTED]> wrote in news:[EMAIL PROTECTED]:
> Hello all, > > Behold: > > ---------- > a = [ 'Foo', 'Bar' ] > b = [ 'Boo', 'Far' ] > q = [ a, b ] > > print q[0][0] > print q[1][1] > > a[0] = 'Snoo' > b[1] = 'Gnuu' > > print q[0][0] > print q[1][1] > ---------- > > This will output: > Foo > Far > Snoo > Gnuu > > I assume it does so because q stores _references_ to a and b. > How > would do I do if I want to copy the list? I.e. I want the output > from the code above to be: > If your original assigment to q were: q = [ a[:], b[:] ] ... you would have copies of a and b, so that: >>> a = [ 'Foo', 'Bar' ] >>> b = [ 'Boo', 'Far' ] >>> q = [ a[:], b[:] ] >>> print q[0][0] Foo >>> print q[1][1] Far >>> >>> a[0] = 'Snoo' >>> b[1] = 'Gnuu' >>> >>> print q[0][0] Foo >>> print q[1][1] Far > Foo > Far > Foo > Far > > ..even if a[0] = 'Snoo' and b[1] = 'Gnuu' remain where they are. > > Or, better yet, how do I store a and b in q, and then tell > Python > that I want a and b to point to new lists, without touching the > contents in q? That's easier: >>> a = [ 'Foo', 'Bar' ] >>> b = [ 'Boo', 'Far' ] >>> q = [a,b] >>> a = ['Splee','Hoongk'] >>> b = ['Blik','Poit'] >>> print q[0][0] Foo >>> print q[1][1] Far You've stuck the 'a' and 'b' labels on new objects this way. The original objects would vanish except that there is still a reference to them through the 'q' list. > > C equivalent of what I want to do: > ----------- > a = calloc(n, size); > prepare(a) > > q[0] = a; > > a = calloc(n, size); // new list; 'q' is unaffected if I change > 'a' ----------- -- rzed -- http://mail.python.org/mailman/listinfo/python-list
