How to know the position of items in a Python'

2019-04-17 21:07发布

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 to know in which position cat belongs to? Is it possible to get answer like:

position ( Ordered_dict["animal"]) = 2 ? or in some other way ?

3条回答
smile是对你的礼貌
2楼-- · 2019-04-17 21:26

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).

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三岁会撩人
3楼-- · 2019-04-17 21:33

Think first that you need is to read documentation. If you open 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
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我欲成王,谁敢阻挡
4楼-- · 2019-04-17 21:44

You may get 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

A better performance could be achieved with the use of iterkeys() though.

For those using Python 3

>>> list(x.keys()).index("c")
1
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