Python Iterators: Difference between revisions
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=Iterators and Dictionaries= | |||
Applying <code>iter()</code> to a dictionary returns a key iterator. | |||
<syntaxhighlight lang='py'> | |||
d = {'k1': 'v1', 'k2': 'v2'} | |||
i = iter(d) | |||
keys = set() | |||
keys.add(next(i)) | |||
keys.add(next(i)) | |||
assert len(keys) == 2 | |||
assert 'k1' in keys | |||
assert 'k2' in keys | |||
</syntaxhighlight> | |||
=Removing an Element while Iterating= | =Removing an Element while Iterating= |
Revision as of 02:28, 6 June 2023
External
Internal
TODO
TO PROCESS PyOOP "The Iterator Pattern" + "Iterators" + "The iterator protocol"
Overview
An iterator instance represents a stream of data.
The iterator instances are created from iterable objects with the built-in function iter()
:
l = ['a', 'b', 'c']
i = iter(l)
Once created, repeated invocations of the iterator's __next__()
method, or by passing it to the built-in function next()
, return successive items in the stream:
assert next(i) == 'a'
assert next(i) == 'b'
assert next(i) == 'c'
When no more data are available a StopIteration
exception is raised instead. At this point, the iterator object is exhausted and any further calls to its __next__()
method just raise StopIteration
again. There's no has_next()
method that tests the availability of a next item without consuming it. Unavailability of a next item is tested by checking the StopIteration
exception.
try:
while True:
arg = next(i)
print(arg)
except StopIteration:
pass
If you are OK to always consume the next item, the end of the iterator can be tested with:
...
next(i, None) is not None
Iterators and Dictionaries
Applying iter()
to a dictionary returns a key iterator.
d = {'k1': 'v1', 'k2': 'v2'}
i = iter(d)
keys = set()
keys.add(next(i))
keys.add(next(i))
assert len(keys) == 2
assert 'k1' in keys
assert 'k2' in keys
Removing an Element while Iterating
It does not seem that Python iterators can remove elements while iterating, similar to Java's Iterator.remove()
. To achieve the same behavior, use a list and remove an element while iterating the list.
itertools
To explore