vinc...@vincentdavis.net wrote:
I have a short peace of code that is not doing what I expect. when I
assign a value to a list in a list alist[2][4]=z this seems replace
all the 4 elements in all the sub lists. I assume it is supposed to
but this is not what I expect. How would I assign a value to the 4th
element in the 2nd sublist. here is the code I have. All the printed
values are what I would expect except that all sublist values are
replaced.
Thanks for your help
Vincent
on the first iteration I get ;
new_list [[None, 0, 1, None], [None, 0, 1, None], [None, 0, 1, None],
[None, 0, 1, None], [None, 0, 1, None], [None, 0, 1, None]]
and expected this;
new_list [[None, 0, 1, None], [None, None, None, None],
[None, None, None, None], [None, None, None, None], [None, None, None,
None], [None, None, None, None]]
Code;
list1=[[1,2],[0,3,2,1],[0,1,3],[2,0,1],[3],[2,3]]
new_list=[[None]*4]*6
print 'new_list',new_list
for sublist in range(6): # 6 becuase it is the # of rows lists1
print 'sublist', sublist
for x in list1[sublist]:
print list1[sublist]
print 'new_list[sublist][x]', new_list[sublist][x]
new_list[sublist][x]=list1[sublist].index(x)
print 'sublist', sublist, 'x', x
print new_list[sublist][x]
print 'new_list', new_list
------------------------------------------------------------------------
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The problem is likely this right here:
[[None]*4]*6
This first creates an inner list that has 4 Nones, then the outer list
contains 6 references to that same list, so (new_list[0] is new_list[1])
and (new_list[1] is new_list[2]). I make this mistake a lot myself.
l=[[None]*4]*6
print id(l[0]) # -1210893364
print id(l[1]) # -1210893364
l = [list([None]*4) for x in range(6)]
print id(l[0]) # -1210893612
print id(l[1]) # -1210893580
Works better
Brian Vanderburg II
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