Removing from a List in Place

[=?ISO-8859-1?Q?Gregory_Pi=F1ero?=]

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(item)
....
1
2
Bonus Question:
Can we make this behave more intuitiviely in Python 3000?

-Greg

 

[bayerj]

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.

 

[John Machin]

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(item)
| ...
| foo
| bar
| >>> alist
| ['foo', 'zot']
| >>>

HTH,
John

 

[Tim Williams]

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