The Order of Elements in Dictionary

The order of elements in Dictionary

The order of elements in a dictionary is non-deterministic. The notion of order simply is not defined for hashtables. So don't rely on enumerating in the same order as elements were added to the dictionary. That's not guaranteed.

Quote from the doc:

For purposes of enumeration, each item in the dictionary is treated as a KeyValuePair<TKey, TValue> structure representing a value and its key. The order in which the items are returned is undefined.

Order of keys in dictionaries in old versions of Python

You could use OrderedDict (requires Python 2.7) or higher.

Also, note that OrderedDict({'a': 1, 'b':2, 'c':3}) won't work since the dict you create with {...} has already forgotten the order of the elements. Instead, you want to use OrderedDict([('a', 1), ('b', 2), ('c', 3)]).

As mentioned in the documentation, for versions lower than Python 2.7, you can use this recipe.

Setting order in Dictionary

Dictionaries are unordered and there is not much you can do to change that, short of writing your own subclass of dict (not recommended). Using OrderedDict will work, as shown in hiro protagonists's comment, as long as you keep in mind that the order in an OrderedDict is the order of insertion and is not related to the values of the keys. (The same is true of the ordering of Python 3 dicts.) If a and b have the same elements but in a different order then the resulting OrderedDicts will not be equal, unless you sort a and b beforehand.

If you want to compare two dicts where you can't easily determine the order of insertion, you need a data structure that supports unordered comparisons. I didn't have your data so I had to make some up. I started with this input

>>> a
[[1, 3, 5, 7, 9], [4, 9, 14, 19, 24, 29, 34, 39], [8, 17, 26, 35, 44, 53, 62, 71], [9, 19, 29, 39, 49, 59, 69, 79, 89]]
>>> b
[[1, 3, 5, 7, 9], [4, 9, 14, 19, 24, 29, 34, 39], [8, 17, 26, 35, 44, 53, 62, 71]]

As you can see, a has one extra element beginning 9.

Modify your dictionary construction code to make the values tuples not lists, because the dict values need to be hashable for the next step to work:

>>> d = { k[0]: tuple(k[1:]) for k in a }
>>> d2= { k[0]: tuple(k[1:]) for k in b }

Then you can convert the dictionaries to sets to do an unordered comparison:

>>> s = set(d.items())
>>> s2 = set(d2.items())
>>> s == s2
False

And you can use the set operators to discover what the difference is:

>>> s2 <= s
True
>>> s - s2
set([(9, (19, 29, 39, 49, 59, 69, 79, 89))])

In what order does a dictionary in python store data?

The short answer is: in an implementation-defined order. You can't rely and shouldn't expect any particular order, and it can change after changing the dictionary in a supposedly-irrelevant manner.

Although not directly, it's somehow explained in Dictionary view objects:

Keys and values are iterated over in an arbitrary order which is non-random, varies across Python implementations, and depends on the dictionary’s history of insertions and deletions. If keys, values and items views are iterated over with no intervening modifications to the dictionary, the order of items will directly correspond.

How do I sort a dictionary by value?

Python 3.7+ or CPython 3.6

Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.

>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

or

>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

Older Python

It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.

For instance,

import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(1))

sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.

And for those wishing to sort on keys instead of values:

import operator
x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=operator.itemgetter(0))

In Python3 since unpacking is not allowed we can use

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])

If you want the output as a dict, you can use collections.OrderedDict:

import collections

sorted_dict = collections.OrderedDict(sorted_x)

How do I sort a dictionary by key?

Note: for Python 3.7+, see this answer

Standard Python dictionaries are unordered (until Python 3.7). Even if you sorted the (key,value) pairs, you wouldn't be able to store them in a dict in a way that would preserve the ordering.

The easiest way is to use OrderedDict, which remembers the order in which the elements have been inserted:

In [1]: import collections

In [2]: d = {2:3, 1:89, 4:5, 3:0}

In [3]: od = collections.OrderedDict(sorted(d.items()))

In [4]: od
Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])

Never mind the way od is printed out; it'll work as expected:

In [11]: od[1]
Out[11]: 89

In [12]: od[3]
Out[12]: 0

In [13]: for k, v in od.iteritems(): print k, v
   ....: 
1 89
2 3
3 0
4 5

Python 3

For Python 3 users, one needs to use the .items() instead of .iteritems():

In [13]: for k, v in od.items(): print(k, v)
   ....: 
1 89
2 3
3 0
4 5

Order of keys in dictionary

Dictionaries are not ordered. So there is no way to do it.

If you have python2.7+, you can use collections.OrderedDict - in this case you could retrieve the item list using .items() and then reverse it and create a new OrderedDict from the reversed list:

>>> od = OrderedDict((('a', 'first'), ('b', 'second')))
>>> od
OrderedDict([('a', 'first'), ('b', 'second')])
>>> items = od.items()  # list(od.items()) in Python3
>>> items.reverse()
>>> OrderedDict(items)
OrderedDict([('b', 'second'), ('a', 'first')])

If you are using an older python version you can get a backport from http://code.activestate.com/recipes/576693/

Why items order in a dictionary changed in Python?

May I know why this is happening?

It is because of the way dicts are organized internally.

In short, this works via a hash-table which puts the keys into buckets according to their hash() value.

If I use dict.keys() to extract the keys from a dictionary and iterate it in an order that I suppose it to be, will that cause dismatch problem?

Depending on how you do it.

k = list(d.keys())
k.sort()
for i in k: print i, d[i]

should exactly work how you want it to work.

How can I get the keys in a dictionary in the order that they appear in the dictionary?

Try from this source: Key Order in Python Dictionaries

OrderedDict([('a', 1), ('b', 2), ('c', 3)])

Sadly, OrderedDict({'a': 1, 'b':2, 'c':3}) won't work because The {} has already forgotten order of the elements.

Your code would be:

Dict = OrderedDict([('Name', 'Zara'), ('Age', 7), ('Class', 'First)])

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