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Removing from a List in Place

I'm going to assume that it's supposed to work like this, but could
someone tell me the reasoning behind it? I.E. why is 3 skipped?
>>alist=[1,2,3]
for item in alist:
.... print item
.... if item==2:
.... alist.remove(it em)
....
1
2
>>>
Bonus Question:
Can we make this behave more intuitiviely in Python 3000?

-Greg
--
Gregory Piñero
Chief Innovation Officer
Blended Technologies
(www.blendedtechnologies.com)
Sep 5 '06 #1
3 1161
I'm going to assume that it's supposed to work like this, but could
someone tell me the reasoning behind it? I.E. why is 3 skipped?
Because:
>>alist[2]
3

You are removing the third item, not the second.

Sep 5 '06 #2
bayerj wrote:
I'm going to assume that it's supposed to work like this, but could
someone tell me the reasoning behind it? I.E. why is 3 skipped?

Because:
>alist[2]
3

You are removing the third item, not the second.
This is incorrect.
You may need to remind yourself that the arg of remove is a value to be
searched for and then removed, not an index. del alist[2] would remove
the third item.

You may have been confused by the OP's obfuscatory example alist = [1,
2, 3].
Consider this equivalent:
| >>alist = ['foo', 'bar', 'zot']
| >>for item in alist:
| ... print item
| ... if item == 'bar':
| ... alist.remove(it em)
| ...
| foo
| bar
| >>alist
| ['foo', 'zot']
| >>>

HTH,
John

Sep 5 '06 #3
On 5 Sep 2006 16:05:36 -0700, bayerj <ba****@in.tum. dewrote:
I'm going to assume that it's supposed to work like this, but could
someone tell me the reasoning behind it? I.E. why is 3 skipped?

Because:
>alist[2]
3

You are removing the third item, not the second.
Actually, he's removing 2 from the list, but then the length of the
list shrinks by 1 and iteration stops.

The example would have been better if alist = ['a','b','c'] and 'b' was removed.

L.remove(value) -- remove first occurrence of value

you were possibly thinking of alist.pop(2), which removes the item
alist[2] from alist

HTH :)
--

Tim Williams
Sep 5 '06 #4

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