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Laszlo Nagy wrote:

> Given this iterator:
>
> class SomeIterableObject(object):
> ....
> ....
>
> def __iter__(self):
> ukeys = self.updates.keys()
> for key in ukeys:
> if self.updates.has_key(key):
> yield self.updates[key]
> for rec in self.inserts:
> yield rec
> ....
> ....
>
> How can I get this exception:
>
> RuntimeError: dictionary changed size during iteration
>
>
> It is true that self.updates is being changed during the iteration. But
> I have created the "ukeys" variable solely to prevent this kind of
> error. Here is a proof of correctness:
>
>>>> d = {1:1,2:2}
>>>> k = d.keys()
>>>> del d[1]
>>>> k
> [1, 2]
>>>> k is d.keys()
> False
>
> So what is wrong with this iterator? Why am I getting this error message?

The keys() method which used to return a list in 2.x was changed in 3.x to
return a view object and to become more or less the equivalent of the old
dict.iterkeys():

>>> d = dict(a=1)
>>> keys = d.keys()
>>> keys
dict_keys(['a'])
>>> for k in keys:
.... d["b"] = 42
....
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
RuntimeError: dictionary changed size during iteration
>>> keys
dict_keys(['a', 'b'])

You now have to create the list explicitly to avoid the error:

>>> d = dict(a=1)
>>> keys = list(d.keys())
>>> for k in keys:
.... d["b"] = 42
....
>>> d
{'a': 1, 'b': 42}
>>> keys
['a']

In article <mailman.644.1303306435.9059.python->,
Peter Otten <__peter__@web.de> wrote:

> You now have to create the list explicitly to avoid the error:
>
> >>> d = dict(a=1)
> >>> keys = list(d.keys())
> >>> for k in keys:
> ... d["b"] = 42
> ...

That works, but if d is large, it won't be very efficient because it has
to generate a large list.

If d is large, and the number of keys to be mutated is relatively small,
a better solution may be to do it in two passes. The first loop
traverses the iterator and builds a list of things to be changed. The
second loop changes them.

changes = [ ]
for key in d.iterkeys():
if is_bad(key):
changes.append(key)
for key in changes:
d[key] = "I'm not dead yet"

Both solutions are O(n), but the second may run significantly faster and
use less memory.

>> ...
> That works, but if d is large, it won't be very efficient because it has
> to generate a large list.
It is not large. But I'm using Python 2.6 , not Python 3.

I did not get this error again in the last two days. I'll post a new
reply if I encounter it again. (It happened just a few times out of many
thousand program invocations)

Roy Smith <> writes:
> changes = [ ]
> for key in d.iterkeys():
> if is_bad(key):
> changes.append(key)

changes = list(k for k in d if is_bad(k))

is a little bit more direct.

In article <>,
Paul Rubin <> wrote:

> Roy Smith <> writes:
> > changes = [ ]
> > for key in d.iterkeys():
> > if is_bad(key):
> > changes.append(key)
>
> changes = list(k for k in d if is_bad(k))
>
> is a little bit more direct.

This is true. I still file list comprehensions under "new fangled
toys". While I use them, and appreciate their value, I admit they're
not always the first thing that comes to my mind.

OBTW,

> changes = [k for k in d if is_bad(k)]

is even more direct

On 08/05/2011 00:12, Roy Smith wrote:
> In article<>,
> Paul Rubin<> wrote:
>
>> Roy Smith<> writes:
>>> changes = [ ]
>>> for key in d.iterkeys():
>>> if is_bad(key):
>>> changes.append(key)
>>
>> changes = list(k for k in d if is_bad(k))
>>
>> is a little bit more direct.
>
> This is true. I still file list comprehensions under "new fangled
> toys". While I use them, and appreciate their value, I admit they're
> not always the first thing that comes to my mind.
>
> OBTW,
>
>> changes = [k for k in d if is_bad(k)]
>
> is even more direct

How about:

changes = filter(is_bad, d)

Or would that be too compact?

-- HansM

Hans Mulder <> writes:
> How about:
> changes = filter(is_bad, d)
> Or would that be too compact?

I thought of writing something like that but filter in python 3 creates
an iterator that would have the same issue of walking the dictionary
while the dictionary is mutating.

changes = list(filter(is_bad, d))

should work.