I'm messing around with dictionaries for the first time and something's coming up that's confusing me. Using two lists to create a new dictionary, the order of the list terms for the key part seems to be wrong. Here's my code:
list1 = ["a", "b", "c", "d"]
list2 = [5,3,7,3]
newDict = {list1[c]: number for c, number in enumerate(list2)}
print(newDict)
This gives me the following:
{'a': 5, 'd': 3, 'c': 7, 'b': 3}
Why is this happening? Surely the 'c' value getting terms from the list is going from 0 and upwards, so why isn't it creating the dictionary with the letters in the same order?
Thanks.
For purposes of efficiency, traditional python dictionaries are unordered. If you need order, then you need OrderedDict:
>>> from collections import OrderedDict
>>> newDict = OrderedDict((list1[c], number) for c, number in enumerate(list2))
>>> print(newDict)
OrderedDict([('a', 5), ('b', 3), ('c', 7), ('d', 3)])
In Python 3.7, ordinary python dictionaries, implemented using a new algorithm, will be ordered. Until then, if you need order, use OrderedDict.
Python dictionaries don't preserve their order, but there's another data type that does: OrderedDict, from the collections module.
Dictionaries are unordered. In fact, if you run your program on a different computer, you might get a different key ordering. This is an intentional feature of the built-in dictionary in python.
To understand why, take a look at this stackoverflow question.
As of Python 3.7, dictionaries are insertion ordered.
See this stackoverflow discussion Are dictionaries ordered in Python 3.6+?
Related
Code:
d = {'a': 0, 'b': 1, 'c': 2}
l = d.keys()
print l
This prints ['a', 'c', 'b']. I'm unsure of how the method keys() determines the order of the keywords within l. However, I'd like to be able to retrive the keywords in the "proper" order. The proper order of course would create the list ['a', 'b', 'c'].
Python 3.7+
In Python 3.7.0 the insertion-order preservation nature of dict objects has been declared to be an official part of the Python language spec. Therefore, you can depend on it.
Python 3.6 (CPython)
As of Python 3.6, for the CPython implementation of Python, dictionaries maintain insertion order by default. This is considered an implementation detail though; you should still use collections.OrderedDict if you want insertion ordering that's guaranteed across other implementations of Python.
Python >=2.7 and <3.6
Use the collections.OrderedDict class when you need a dict that
remembers the order of items inserted.
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.
>>> print sorted(d.keys())
['a', 'b', 'c']
Use the sorted function, which sorts the iterable passed in.
The .keys() method returns the keys in an arbitrary order.
From http://docs.python.org/tutorial/datastructures.html:
"The keys() method of a dictionary object returns a list of all the keys used in the dictionary, in arbitrary order (if you want it sorted, just apply the sorted() function to it)."
Just sort the list when you want to use it.
l = sorted(d.keys())
Although the order does not matter as the dictionary is hashmap. It depends on the order how it is pushed in:
s = 'abbc'
a = 'cbab'
def load_dict(s):
dict_tmp = {}
for ch in s:
if ch in dict_tmp.keys():
dict_tmp[ch]+=1
else:
dict_tmp[ch] = 1
return dict_tmp
dict_a = load_dict(a)
dict_s = load_dict(s)
print('for string %s, the keys are %s'%(s, dict_s.keys()))
print('for string %s, the keys are %s'%(a, dict_a.keys()))
output:
for string abbc, the keys are dict_keys(['a', 'b', 'c'])
for string cbab, the keys are dict_keys(['c', 'b', 'a'])
Code:
d = {'a': 0, 'b': 1, 'c': 2}
l = d.keys()
print l
This prints ['a', 'c', 'b']. I'm unsure of how the method keys() determines the order of the keywords within l. However, I'd like to be able to retrive the keywords in the "proper" order. The proper order of course would create the list ['a', 'b', 'c'].
Python 3.7+
In Python 3.7.0 the insertion-order preservation nature of dict objects has been declared to be an official part of the Python language spec. Therefore, you can depend on it.
Python 3.6 (CPython)
As of Python 3.6, for the CPython implementation of Python, dictionaries maintain insertion order by default. This is considered an implementation detail though; you should still use collections.OrderedDict if you want insertion ordering that's guaranteed across other implementations of Python.
Python >=2.7 and <3.6
Use the collections.OrderedDict class when you need a dict that
remembers the order of items inserted.
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.
>>> print sorted(d.keys())
['a', 'b', 'c']
Use the sorted function, which sorts the iterable passed in.
The .keys() method returns the keys in an arbitrary order.
From http://docs.python.org/tutorial/datastructures.html:
"The keys() method of a dictionary object returns a list of all the keys used in the dictionary, in arbitrary order (if you want it sorted, just apply the sorted() function to it)."
Just sort the list when you want to use it.
l = sorted(d.keys())
Although the order does not matter as the dictionary is hashmap. It depends on the order how it is pushed in:
s = 'abbc'
a = 'cbab'
def load_dict(s):
dict_tmp = {}
for ch in s:
if ch in dict_tmp.keys():
dict_tmp[ch]+=1
else:
dict_tmp[ch] = 1
return dict_tmp
dict_a = load_dict(a)
dict_s = load_dict(s)
print('for string %s, the keys are %s'%(s, dict_s.keys()))
print('for string %s, the keys are %s'%(a, dict_a.keys()))
output:
for string abbc, the keys are dict_keys(['a', 'b', 'c'])
for string cbab, the keys are dict_keys(['c', 'b', 'a'])
I have written a code which tries to sort a dictionary using the values rather than keys
""" This module sorts a dictionary based on the values of the keys"""
adict={1:1,2:2,5:1,10:2,44:3,67:2} #adict is an input dictionary
items=adict.items()## converts the dictionary into a list of tuples
##print items
list_value_key=[ [d[1],d[0]] for d in items] """Interchanges the position of the
key and the values"""
list_value_key.sort()
print list_value_key
key_list=[ list_value_key[i][1] for i in range(0,len(list_value_key))]
print key_list ## list of keys sorted on the basis of values
sorted_adict={}
*for key in key_list:
sorted_adict.update({key:adict[key]})
print key,adict[key]
print sorted_adict*
So when I print key_list i get the expected answer, but for the last part of the code where i try to update the dictionary, the order is not what it should be. Below are the results obtained. I am not sure why the "update" method is not working. Any help or pointers is appreciated
result:
sorted_adict={1: 1, 2: 2, 67: 2, 5: 1, 10: 2, 44: 3}
Python dictionaries, no matter how you insert into them, are unordered. This is the nature of hash tables, in general.
Instead, perhaps you should keep a list of keys in the order their values or sorted, something like: [ 5, 1, 44, ...]
This way, you can access your dictionary in sorted order at a later time.
Don't sort like that.
import operator
adict={1:1,2:2,5:1,10:2,44:3,67:2}
sorted_adict = sorted(adict.iteritems(), key=operator.itemgetter(1))
If you need a dictionary that retains its order, there's a class called OrderedDict in the collections module. You can use the recipes on that page to sort a dictionary and create a new OrderedDict that retains the sort order. The OrderedDict class is available in Python 2.7 or 3.1.
To sort your dictionnary, you could also also use :
adict={1:1,2:2,5:1,10:2,44:3,67:2}
k = adict.keys()
k.sort(cmp=lambda k1,k2: cmp(adict[k1],adict[k2]))
And by the way, it's useless to reuse a dictionnary after that because there are no order in dict (they are just mapping types - you can have keys of different types that are not "comparable").
One problem is that ordinary dictionaries can't be sorted because of the way they're implemented internally. Python 2.7 and 3.1 had a new class namedOrderedDictadded to theircollectionsmodule as #kindall mentioned in his answer. While they can't be sorted exactly either, they do retain or remember the order in which keys and associated values were added to them, regardless of how it was done (including via theupdate() method). This means that you can achieve what you want by adding everything from the input dictionary to anOrderedDictoutput dictionary in the desired order.
To do that, the code you had was on the right track in the sense of creating what you called thelist_value_keylist and sorting it. There's a slightly simpler and faster way to create the initial unsorted version of that list than what you were doing by using the built-inzip()function. Below is code illustrating how to do that:
from collections import OrderedDict
adict = {1:1, 2:2, 5:1, 10:2, 44:3, 67:2} # input dictionary
# zip together and sort pairs by first item (value)
value_keys_list = sorted(zip(adict.values(), adict.keys()))
sorted_adict = OrderedDict() # value sorted output dictionary
for pair in value_keys_list:
sorted_adict[pair[1]] = pair[0]
print sorted_adict
# OrderedDict([(1, 1), (5, 1), (2, 2), (10, 2), (67, 2), (44, 3)])
The above can be rewritten as a fairly elegant one-liner:
sorted_adict = OrderedDict((pair[1], pair[0])
for pair in sorted(zip(adict.values(), adict.keys())))
I am in simple doubt... I created the following dictionary:
>>> alpha={'a': 10, 'b': 5, 'c': 11}
But, when I want to see the dictionary keys and values I got:
>>> alpha
{'a': 10, 'c': 11, 'b': 5}
See that the "b" and "c" has swapped their position. How can I make the position be the same of the moment that the dictionary was created?
Dictionaries are unordered containers - if you want to preserve order, you can use collections.OrderedDict (Python 2.7 or later), or use another container type which is naturally order-preserving.
Generally if you have an access pattern that cares about ordered retrieval then a dictionary is solving a problem you don't have (fast access to random elements), while giving you a new one.
Dictonaries are not guaranting sorting of keys. You can find this information in python docs: http://docs.python.org/tutorial/datastructures.html#dictionaries
You can always sort dictionary keys or use other, more specialized collection.
I have this dict in python;
d={}
d['b']='beta'
d['g']='gamma'
d['a']='alpha'
when i print the dict;
for k,v in d.items():
print k
i get this;
a
b
g
it seems like python sorts the dict automatically! how can i get the original unsorted list?
Gath
Dicts don't work like that:
CPython implementation detail: Keys and values are listed in an arbitrary order which is non-random, varies across Python implementations, and depends on the dictionary’s history of insertions and deletions.
You could use a list with 2-tuples instead:
d = [('b', 'beta'), ('g', 'gamma'), ('a', 'alpha')]
A similar but better solution is outlined in Wayne's answer.
As has been mentioned, dicts don't order or unorder the items you put in. It's "magic" as to how it's ordered when you retrieve it. If you want to keep an order -sorted or not- you need to also bind a list or tuple.
This will give you the same dict result with a list that retains order:
greek = ['beta', 'gamma', 'alpha']
d = {}
for x in greek:
d[x[0]] = x
Simply change [] to () if you have no need to change the original list/order.
Don't use a dictionary. Or use the Python 2.7/3.1 OrderedDict type.
There is no order in dictionaries to speak of, there is no original unsorted list.
No, python does not sort dict, it would be too expensive. The order of items() is arbitrary. From python docs:
CPython implementation detail: Keys
and values are listed in an arbitrary
order which is non-random, varies
across Python implementations, and
depends on the dictionary’s history of
insertions and deletions.