Strange behavior

[Terry Reedy]

On 8/14/2012 11:59 AM, Alain Ketterlin wrote:
> light1quark at gmail.com writes:
>
>> However if you run the code you will notice only one of the strings
>> beginning with 'x' is removed from the startingList.
>
>>
>> def testFunc(startingList):
>> 	xOnlyList = [];
>> 	for str in startingList:
>> 		if (str[0] == 'x'):
>> 			print str;
>> 			xOnlyList.append(str)
>> 			startingList.remove(str) #this seems to be the problem
>> 	print xOnlyList;
>> 	print startingList
>> testFunc(['xasd', 'xjkl', 'sefwr', 'dfsews'])
>>
>> #Thanks for your help!
>
> Try with ['xasd', 'sefwr', 'xjkl', 'dfsews'] and you'll understand what
> happens. Also, have a look at:
>
> http://docs.python.org/reference/compound_stmts.html#the-for-statement
>
> You can't modify the list you're iterating on,

Except he obviously did ;-).
(Modifying set or dict raises SomeError.)

Indeed, people routine *replace* items while iterating.

def squarelist(lis):
     for i, n in enumerate(lis):
         lis[i] = n*n
     return lis

print(squarelist([0,1,2,3,4,5]))
# [0, 1, 4, 9, 16, 25]

Removals can be handled by iterating in reverse. This works even with 
duplicates because if the item removed is not the one tested, the one 
tested gets retested.

def removeodd(lis):
     for n in reversed(lis):
         if n % 2:
             lis.remove(n)
         print(n, lis)

ll = [0,1, 5, 5, 4, 5]
removeodd(ll)
 >>>
5 [0, 1, 5, 4, 5]
5 [0, 1, 4, 5]
5 [0, 1, 4]
4 [0, 1, 4]
1 [0, 4]
0 [0, 4]

> better use another list to collect the result.

If there are very many removals, a new list will be faster, even if one 
needs to copy the new list back into the original, as k removals from 
len n list is O(k*n) versus O(n) for new list and copy.

> P/S: str is a builtin, you'd better avoid assigning to it.

Agreed. People have actually posted code doing something like

...
list = [1,2,3]
...
z = list(x)
...
and wondered and asked why it does not work.

-- 
Terry Jan Reedy