How to know the position of items in a Python ordered dictionary
Question:
Can we know the position of items in Python’s ordered dictionary?
For example:
If I have dictionary:
// Ordered_dict is OrderedDictionary
Ordered_dict = {"fruit": "banana", "drinks": "water", "animal": "cat"}
Now how do I know in which position cat belongs to?
Is it possible to get an answer like:
position (Ordered_dict["animal"]) = 2 ? or in some other way?
Answers:
You may get a list of keys with the keys property:
In [20]: d=OrderedDict((("fruit", "banana"), ("drinks", 'water'), ("animal", "cat")))
In [21]: d.keys().index('animal')
Out[21]: 2
Better performance could be achieved with the use of iterkeys() though.
For those using Python 3:
>>> list(d.keys()).index('animal')
2
Think first that you need to read documentation. If you open a Python tutorial and then try to find information about OrderedDict you will see the following:
class collections.OrderedDict([items]) – Return an instance of a dict
subclass, supporting the usual dict methods. An OrderedDict is a dict
that remembers the order that keys were first inserted. If a new entry
overwrites an existing entry, the original insertion position is left
unchanged. Deleting an entry and reinserting it will move it to the
end.
New in version 2.7.
So in case you are using an ordered dictionary and you are not going to delete keys – then ‘animal’ will be always in the position you add – e.g. index 2.
Also to get an index of a ‘cat’ you can simply use:
from collections import OrderedDict
d = OrderedDict((("fruit", "banana"), ("drinks", "water"), ("animal", "cat")))
d.keys()
>>> ['fruit', 'drinks', 'animal']
d.values()
>>> ['banana', 'water', 'cat']
# So
d.values().index('cat')
>>> 2
For Python3: tuple(d).index('animal')
This is almost the same as Marein’s answer above, but uses an immutable tuple instead of a mutable list. So it should run a little bit faster (~12% faster in my quick sanity check).
Can we know the position of items in Python’s ordered dictionary?
For example:
If I have dictionary:
// Ordered_dict is OrderedDictionary
Ordered_dict = {"fruit": "banana", "drinks": "water", "animal": "cat"}
Now how do I know in which position cat belongs to?
Is it possible to get an answer like:
position (Ordered_dict["animal"]) = 2 ? or in some other way?
You may get a list of keys with the keys property:
In [20]: d=OrderedDict((("fruit", "banana"), ("drinks", 'water'), ("animal", "cat")))
In [21]: d.keys().index('animal')
Out[21]: 2
Better performance could be achieved with the use of iterkeys() though.
For those using Python 3:
>>> list(d.keys()).index('animal')
2
Think first that you need to read documentation. If you open a Python tutorial and then try to find information about OrderedDict you will see the following:
class collections.OrderedDict([items]) – Return an instance of a dict
subclass, supporting the usual dict methods. An OrderedDict is a dict
that remembers the order that keys were first inserted. If a new entry
overwrites an existing entry, the original insertion position is left
unchanged. Deleting an entry and reinserting it will move it to the
end.New in version 2.7.
So in case you are using an ordered dictionary and you are not going to delete keys – then ‘animal’ will be always in the position you add – e.g. index 2.
Also to get an index of a ‘cat’ you can simply use:
from collections import OrderedDict
d = OrderedDict((("fruit", "banana"), ("drinks", "water"), ("animal", "cat")))
d.keys()
>>> ['fruit', 'drinks', 'animal']
d.values()
>>> ['banana', 'water', 'cat']
# So
d.values().index('cat')
>>> 2
For Python3: tuple(d).index('animal')
This is almost the same as Marein’s answer above, but uses an immutable tuple instead of a mutable list. So it should run a little bit faster (~12% faster in my quick sanity check).