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) 3  1174 
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.
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
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 This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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