It looks like the lists returned by keys() and values() methods of a dictionary are always a 1-to-1 mapping (assuming the dictionary is not altered between calling the 2 methods).
For example:
>>> d = {'one':1, 'two': 2, 'three': 3}
>>> k, v = d.keys(), d.values()
>>> for i in range(len(k)):
print d[k[i]] == v[i]
True
True
True
If you do not alter the dictionary between calling keys() and calling values(), is it wrong to assume the above for-loop will always print True? I could not find any documentation confirming this.
-
3In CPython 3.7 (and up, presumably) you may rely on the iteration order of a dictionary matching insertion order. mail.python.org/pipermail/python-dev/2017-December/151283.htmlBallpointBen– BallpointBen2018年04月19日 18:47:32 +00:00Commented Apr 19, 2018 at 18:47
-
2@BallpointBen it's in CPython 3.6 and up and all other Python implementations starting with Python 3.7user3064538– user30645382020年05月19日 14:43:31 +00:00Commented May 19, 2020 at 14:43
9 Answers 9
Found this:
If
items(),keys(),values(),iteritems(),iterkeys(), anditervalues()are called with no intervening modifications to the dictionary, the lists will directly correspond.
On 2.x documentation and 3.x documentation.
5 Comments
values() after keys(), it is guaranteed that the order of values() is the same as the order of keys()?Yes, what you observed is indeed a guaranteed property -- keys(), values() and items() return lists in congruent order if the dict is not altered. iterkeys() &c also iterate in the same order as the corresponding lists.
1 Comment
dataframe offers an alternative where items can be updated or deleted, and the order and the index locations of the items in a dict-like structure remain fixed.Yes it is guaranteed in python 2.x:
If keys, values and items views are iterated over with no intervening modifications to the dictionary, the order of items will directly correspond.
Comments
Yes. Starting with CPython 3.6, dictionaries return items in the order you inserted them.
Ignore the part that says this is an implementation detail. This behaviour is guaranteed in CPython 3.6 and is required for all other Python implementations starting with Python 3.7.
Comments
Good references to the docs. Here's how you can guarantee the order regardless of the documentation / implementation:
k, v = zip(*d.iteritems())
1 Comment
According to http://docs.python.org/dev/py3k/library/stdtypes.html#dictionary-view-objects , the keys(), values() and items() methods of a dict will return corresponding iterators whose orders correspond. However, I am unable to find a reference to the official documentation for python 2.x for the same thing.
So as far as I can tell, the answer is yes, but only in python 3.0+
Comments
For what it's worth, some heavy used production code I have written is based on this assumption and I never had a problem with it. I know that doesn't make it true though :-)
If you don't want to take the risk I would use iteritems() if you can.
for key, value in myDictionary.iteritems():
print key, value
Comments
I would agree with others that in python 3.6+ it should stay the same if unaffected by user.
an example from my code few days ago:
ips = { '001' : '199.250.178.14', '002' : '199.18.2.89', '003' : '109.251.63.21' }
def run(self):
for x, y in self.ips.items():
try:
subprocess.check_call(
['ping', '-n', '1', y],
stdout=DEVNULL, # suppress output
stderr=DEVNULL
)
except subprocess.CalledProcessError:
nextServer = ('HUB ' + x + ' is OFFLINE ' + " IP: " + y)
and the output is always of the same order, exactly as i inputted it i.e. only offline servers will show up in order: HUB 003 is OFFLINE IP: 109.251.63.21 becasue first two are reachable
Hope that clears it out
Comments
I wasn't satisfied with these answers since I wanted to ensure the exported values had the same ordering even when using different dicts.
Here you specify the key order upfront, the returned values will always have the same order even if the dict changes, or you use a different dict.
keys = dict1.keys()
ordered_keys1 = [dict1[cur_key] for cur_key in keys]
ordered_keys2 = [dict2[cur_key] for cur_key in keys]