I have a dictionary that I declared in a particular order and want to keep it in that order all the time. The keys/values can't really be kept in order based on their value, I just want it in the order that I declared it.
So if I have the dictionary:
d = {'ac': 33, 'gw': 20, 'ap': 102, 'za': 321, 'bs': 10}
It isn't in that order if I view it or iterate through it. Is there any way to make sure Python will keep the explicit order that I declared the keys/values in?
From Python 3.6 onwards, the standard dict type maintains insertion order by default.
Defining
d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}
will result in a dictionary with the keys in the order listed in the source code.
This was achieved by using a simple array with integers for the sparse hash table, where those integers index into another array that stores the key-value pairs (plus the calculated hash). That latter array just happens to store the items in insertion order, and the whole combination actually uses less memory than the implementation used in Python 3.5 and before. See the original idea post by Raymond Hettinger for details.
In 3.6 this was still considered an implementation detail; see the What's New in Python 3.6 documentation:
The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5).
Python 3.7 elevates this implementation detail to a language specification, so it is now mandatory that dict preserves order in all Python implementations compatible with that version or newer. See the pronouncement by the BDFL. As of Python 3.8, dictionaries also support iteration in reverse.
You may still want to use the collections.OrderedDict() class in certain cases, as it offers some additional functionality on top of the standard dict type. Such as as being reversible (this extends to the view objects), and supporting reordering (via the move_to_end() method).
from collections import OrderedDict
OrderedDict((word, True) for word in words)
contains
OrderedDict([('He', True), ('will', True), ('be', True), ('the', True), ('winner', True)])
If the values are True (or any other immutable object), you can also use:
OrderedDict.fromkeys(words, True)
Rather than explaining the theoretical part, I'll give a simple example.
>>> from collections import OrderedDict
>>> my_dictionary=OrderedDict()
>>> my_dictionary['foo']=3
>>> my_dictionary['aol']=1
>>> my_dictionary
OrderedDict([('foo', 3), ('aol', 1)])
>>> dict(my_dictionary)
{'foo': 3, 'aol': 1}
Note that this answer applies to python versions prior to python3.7. CPython 3.6 maintains insertion order under most circumstances as an implementation detail. Starting from Python3.7 onward, it has been declared that implementations MUST maintain insertion order to be compliant.
python dictionaries are unordered. If you want an ordered dictionary, try collections.OrderedDict.
Note that OrderedDict was introduced into the standard library in python 2.7. If you have an older version of python, you can find recipes for ordered dictionaries on ActiveState.
Dictionaries will use an order that makes searching efficient, and you cant change that,
You could just use a list of objects (a 2 element tuple in a simple case, or even a class), and append items to the end. You can then use linear search to find items in it.
Alternatively you could create or use a different data structure created with the intention of maintaining order.
I came across this post while trying to figure out how to get OrderedDict to work. PyDev for Eclipse couldn't find OrderedDict at all, so I ended up deciding to make a tuple of my dictionary's key values as I would like them to be ordered. When I needed to output my list, I just iterated through the tuple's values and plugged the iterated 'key' from the tuple into the dictionary to retrieve my values in the order I needed them.
example:
test_dict = dict( val1 = "hi", val2 = "bye", val3 = "huh?", val4 = "what....")
test_tuple = ( 'val1', 'val2', 'val3', 'val4')
for key in test_tuple: print(test_dict[key])
It's a tad cumbersome, but I'm pressed for time and it's the workaround I came up with.
note: the list of lists approach that somebody else suggested does not really make sense to me, because lists are ordered and indexed (and are also a different structure than dictionaries).
You can't really do what you want with a dictionary. You already have the dictionary d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}created. I found there was no way to keep in order once it is already created. What I did was make a json file instead with the object:
{"ac":33,"gw":20,"ap":102,"za":321,"bs":10}
I used:
r = json.load(open('file.json'), object_pairs_hook=OrderedDict)
then used:
print json.dumps(r)
to verify.
from collections import OrderedDict
list1 = ['k1', 'k2']
list2 = ['v1', 'v2']
new_ordered_dict = OrderedDict(zip(list1, list2))
print new_ordered_dict
# OrderedDict([('k1', 'v1'), ('k2', 'v2')])
Another alternative is to use Pandas dataframe as it guarantees the order and the index locations of the items in a dict-like structure.
I had a similar problem when developing a Django project. I couldn't use OrderedDict, because I was running an old version of python, so the solution was to use Django's SortedDict class:
https://code.djangoproject.com/wiki/SortedDict
e.g.,
from django.utils.datastructures import SortedDict
d2 = SortedDict()
d2['b'] = 1
d2['a'] = 2
d2['c'] = 3
Note: This answer is originally from 2011. If you have access to Python version 2.7 or higher, then you should have access to the now standard collections.OrderedDict, of which many examples have been provided by others in this thread.
Generally, you can design a class that behaves like a dictionary, mainly be implementing the methods __contains__, __getitem__, __delitem__, __setitem__ and some more. That class can have any behaviour you like, for example prividing a sorted iterator over the keys ...
if you would like to have a dictionary in a specific order, you can also create a list of lists, where the first item will be the key, and the second item will be the value
and will look like this
example
>>> list =[[1,2],[2,3]]
>>> for i in list:
... print i[0]
... print i[1]
1
2
2
3
You can do the same thing which i did for dictionary.
Create a list and empty dictionary:
dictionary_items = {}
fields = [['Name', 'Himanshu Kanojiya'], ['email id', 'hima#gmail.com']]
l = fields[0][0]
m = fields[0][1]
n = fields[1][0]
q = fields[1][1]
dictionary_items[l] = m
dictionary_items[n] = q
print dictionary_items
Related
In Python 3.10, I am aware that a dictionary preserves insertion order. However when performing conditional list comprehensions, can this order still be guaranteed?
For example, given:
my_dict = {}
my_dict['a'] = 1
my_dict['b'] = 2
my_dict['c'] = 3
my_dict['d'] = 4
Can one guarantee that either (option A):
print([k for k in my_dict.keys() if k not in ['c']])
or (option B):
print([k for k in (my_dict.keys() - {'c'})])
will always return:
['a', 'b', 'd']
Iterating over dict or dict.keys() should give the same results for any version of Python, since the language guarantees the current order will always be stable, even if it doesn't necessarily match the insertion order. In Python 3, the keys() method provides a dynamic view of the dictionary's entries, so it will directly reflect the current state of the dict. The views themselves may be "set-like", but that does not imply they are unordered (or independently ordered).
The problem with the examples in the question is that they don't compare like with like. The keys() method returns a view (or a list in earlier versions), whereas keys() - {'a'} evaluates to a set (i.e. an object with no guaranteed order). So it is safe to assume option A will always give the same results, but not option B.
I think the short answer is yes, the "preserves insertion order" clause gives you a proper order of keys whenever you go through them (be it via for k in my_dict or my_dict.keys()), and together with the one that #Larry pointed out gives you what you ask for.
However the downwotes on this question are probably due to the fact that if you need an answer to this question for a coding problem, you should either learn more about list comprehensions or just rethink your solution and sort the keys based on insertion order or whatever way of guaranteing you'd imagine
I have a dictionary that I declared in a particular order and want to keep it in that order all the time. The keys/values can't really be kept in order based on their value, I just want it in the order that I declared it.
So if I have the dictionary:
d = {'ac': 33, 'gw': 20, 'ap': 102, 'za': 321, 'bs': 10}
It isn't in that order if I view it or iterate through it. Is there any way to make sure Python will keep the explicit order that I declared the keys/values in?
From Python 3.6 onwards, the standard dict type maintains insertion order by default.
Defining
d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}
will result in a dictionary with the keys in the order listed in the source code.
This was achieved by using a simple array with integers for the sparse hash table, where those integers index into another array that stores the key-value pairs (plus the calculated hash). That latter array just happens to store the items in insertion order, and the whole combination actually uses less memory than the implementation used in Python 3.5 and before. See the original idea post by Raymond Hettinger for details.
In 3.6 this was still considered an implementation detail; see the What's New in Python 3.6 documentation:
The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5).
Python 3.7 elevates this implementation detail to a language specification, so it is now mandatory that dict preserves order in all Python implementations compatible with that version or newer. See the pronouncement by the BDFL. As of Python 3.8, dictionaries also support iteration in reverse.
You may still want to use the collections.OrderedDict() class in certain cases, as it offers some additional functionality on top of the standard dict type. Such as as being reversible (this extends to the view objects), and supporting reordering (via the move_to_end() method).
from collections import OrderedDict
OrderedDict((word, True) for word in words)
contains
OrderedDict([('He', True), ('will', True), ('be', True), ('the', True), ('winner', True)])
If the values are True (or any other immutable object), you can also use:
OrderedDict.fromkeys(words, True)
Rather than explaining the theoretical part, I'll give a simple example.
>>> from collections import OrderedDict
>>> my_dictionary=OrderedDict()
>>> my_dictionary['foo']=3
>>> my_dictionary['aol']=1
>>> my_dictionary
OrderedDict([('foo', 3), ('aol', 1)])
>>> dict(my_dictionary)
{'foo': 3, 'aol': 1}
Note that this answer applies to python versions prior to python3.7. CPython 3.6 maintains insertion order under most circumstances as an implementation detail. Starting from Python3.7 onward, it has been declared that implementations MUST maintain insertion order to be compliant.
python dictionaries are unordered. If you want an ordered dictionary, try collections.OrderedDict.
Note that OrderedDict was introduced into the standard library in python 2.7. If you have an older version of python, you can find recipes for ordered dictionaries on ActiveState.
Dictionaries will use an order that makes searching efficient, and you cant change that,
You could just use a list of objects (a 2 element tuple in a simple case, or even a class), and append items to the end. You can then use linear search to find items in it.
Alternatively you could create or use a different data structure created with the intention of maintaining order.
I came across this post while trying to figure out how to get OrderedDict to work. PyDev for Eclipse couldn't find OrderedDict at all, so I ended up deciding to make a tuple of my dictionary's key values as I would like them to be ordered. When I needed to output my list, I just iterated through the tuple's values and plugged the iterated 'key' from the tuple into the dictionary to retrieve my values in the order I needed them.
example:
test_dict = dict( val1 = "hi", val2 = "bye", val3 = "huh?", val4 = "what....")
test_tuple = ( 'val1', 'val2', 'val3', 'val4')
for key in test_tuple: print(test_dict[key])
It's a tad cumbersome, but I'm pressed for time and it's the workaround I came up with.
note: the list of lists approach that somebody else suggested does not really make sense to me, because lists are ordered and indexed (and are also a different structure than dictionaries).
You can't really do what you want with a dictionary. You already have the dictionary d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}created. I found there was no way to keep in order once it is already created. What I did was make a json file instead with the object:
{"ac":33,"gw":20,"ap":102,"za":321,"bs":10}
I used:
r = json.load(open('file.json'), object_pairs_hook=OrderedDict)
then used:
print json.dumps(r)
to verify.
from collections import OrderedDict
list1 = ['k1', 'k2']
list2 = ['v1', 'v2']
new_ordered_dict = OrderedDict(zip(list1, list2))
print new_ordered_dict
# OrderedDict([('k1', 'v1'), ('k2', 'v2')])
Another alternative is to use Pandas dataframe as it guarantees the order and the index locations of the items in a dict-like structure.
I had a similar problem when developing a Django project. I couldn't use OrderedDict, because I was running an old version of python, so the solution was to use Django's SortedDict class:
https://code.djangoproject.com/wiki/SortedDict
e.g.,
from django.utils.datastructures import SortedDict
d2 = SortedDict()
d2['b'] = 1
d2['a'] = 2
d2['c'] = 3
Note: This answer is originally from 2011. If you have access to Python version 2.7 or higher, then you should have access to the now standard collections.OrderedDict, of which many examples have been provided by others in this thread.
Generally, you can design a class that behaves like a dictionary, mainly be implementing the methods __contains__, __getitem__, __delitem__, __setitem__ and some more. That class can have any behaviour you like, for example prividing a sorted iterator over the keys ...
if you would like to have a dictionary in a specific order, you can also create a list of lists, where the first item will be the key, and the second item will be the value
and will look like this
example
>>> list =[[1,2],[2,3]]
>>> for i in list:
... print i[0]
... print i[1]
1
2
2
3
You can do the same thing which i did for dictionary.
Create a list and empty dictionary:
dictionary_items = {}
fields = [['Name', 'Himanshu Kanojiya'], ['email id', 'hima#gmail.com']]
l = fields[0][0]
m = fields[0][1]
n = fields[1][0]
q = fields[1][1]
dictionary_items[l] = m
dictionary_items[n] = q
print dictionary_items
I have a dictionary that I declared in a particular order and want to keep it in that order all the time. The keys/values can't really be kept in order based on their value, I just want it in the order that I declared it.
So if I have the dictionary:
d = {'ac': 33, 'gw': 20, 'ap': 102, 'za': 321, 'bs': 10}
It isn't in that order if I view it or iterate through it. Is there any way to make sure Python will keep the explicit order that I declared the keys/values in?
From Python 3.6 onwards, the standard dict type maintains insertion order by default.
Defining
d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}
will result in a dictionary with the keys in the order listed in the source code.
This was achieved by using a simple array with integers for the sparse hash table, where those integers index into another array that stores the key-value pairs (plus the calculated hash). That latter array just happens to store the items in insertion order, and the whole combination actually uses less memory than the implementation used in Python 3.5 and before. See the original idea post by Raymond Hettinger for details.
In 3.6 this was still considered an implementation detail; see the What's New in Python 3.6 documentation:
The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5).
Python 3.7 elevates this implementation detail to a language specification, so it is now mandatory that dict preserves order in all Python implementations compatible with that version or newer. See the pronouncement by the BDFL. As of Python 3.8, dictionaries also support iteration in reverse.
You may still want to use the collections.OrderedDict() class in certain cases, as it offers some additional functionality on top of the standard dict type. Such as as being reversible (this extends to the view objects), and supporting reordering (via the move_to_end() method).
from collections import OrderedDict
OrderedDict((word, True) for word in words)
contains
OrderedDict([('He', True), ('will', True), ('be', True), ('the', True), ('winner', True)])
If the values are True (or any other immutable object), you can also use:
OrderedDict.fromkeys(words, True)
Rather than explaining the theoretical part, I'll give a simple example.
>>> from collections import OrderedDict
>>> my_dictionary=OrderedDict()
>>> my_dictionary['foo']=3
>>> my_dictionary['aol']=1
>>> my_dictionary
OrderedDict([('foo', 3), ('aol', 1)])
>>> dict(my_dictionary)
{'foo': 3, 'aol': 1}
Note that this answer applies to python versions prior to python3.7. CPython 3.6 maintains insertion order under most circumstances as an implementation detail. Starting from Python3.7 onward, it has been declared that implementations MUST maintain insertion order to be compliant.
python dictionaries are unordered. If you want an ordered dictionary, try collections.OrderedDict.
Note that OrderedDict was introduced into the standard library in python 2.7. If you have an older version of python, you can find recipes for ordered dictionaries on ActiveState.
Dictionaries will use an order that makes searching efficient, and you cant change that,
You could just use a list of objects (a 2 element tuple in a simple case, or even a class), and append items to the end. You can then use linear search to find items in it.
Alternatively you could create or use a different data structure created with the intention of maintaining order.
I came across this post while trying to figure out how to get OrderedDict to work. PyDev for Eclipse couldn't find OrderedDict at all, so I ended up deciding to make a tuple of my dictionary's key values as I would like them to be ordered. When I needed to output my list, I just iterated through the tuple's values and plugged the iterated 'key' from the tuple into the dictionary to retrieve my values in the order I needed them.
example:
test_dict = dict( val1 = "hi", val2 = "bye", val3 = "huh?", val4 = "what....")
test_tuple = ( 'val1', 'val2', 'val3', 'val4')
for key in test_tuple: print(test_dict[key])
It's a tad cumbersome, but I'm pressed for time and it's the workaround I came up with.
note: the list of lists approach that somebody else suggested does not really make sense to me, because lists are ordered and indexed (and are also a different structure than dictionaries).
You can't really do what you want with a dictionary. You already have the dictionary d = {'ac':33, 'gw':20, 'ap':102, 'za':321, 'bs':10}created. I found there was no way to keep in order once it is already created. What I did was make a json file instead with the object:
{"ac":33,"gw":20,"ap":102,"za":321,"bs":10}
I used:
r = json.load(open('file.json'), object_pairs_hook=OrderedDict)
then used:
print json.dumps(r)
to verify.
from collections import OrderedDict
list1 = ['k1', 'k2']
list2 = ['v1', 'v2']
new_ordered_dict = OrderedDict(zip(list1, list2))
print new_ordered_dict
# OrderedDict([('k1', 'v1'), ('k2', 'v2')])
Another alternative is to use Pandas dataframe as it guarantees the order and the index locations of the items in a dict-like structure.
I had a similar problem when developing a Django project. I couldn't use OrderedDict, because I was running an old version of python, so the solution was to use Django's SortedDict class:
https://code.djangoproject.com/wiki/SortedDict
e.g.,
from django.utils.datastructures import SortedDict
d2 = SortedDict()
d2['b'] = 1
d2['a'] = 2
d2['c'] = 3
Note: This answer is originally from 2011. If you have access to Python version 2.7 or higher, then you should have access to the now standard collections.OrderedDict, of which many examples have been provided by others in this thread.
Generally, you can design a class that behaves like a dictionary, mainly be implementing the methods __contains__, __getitem__, __delitem__, __setitem__ and some more. That class can have any behaviour you like, for example prividing a sorted iterator over the keys ...
if you would like to have a dictionary in a specific order, you can also create a list of lists, where the first item will be the key, and the second item will be the value
and will look like this
example
>>> list =[[1,2],[2,3]]
>>> for i in list:
... print i[0]
... print i[1]
1
2
2
3
You can do the same thing which i did for dictionary.
Create a list and empty dictionary:
dictionary_items = {}
fields = [['Name', 'Himanshu Kanojiya'], ['email id', 'hima#gmail.com']]
l = fields[0][0]
m = fields[0][1]
n = fields[1][0]
q = fields[1][1]
dictionary_items[l] = m
dictionary_items[n] = q
print dictionary_items
r_dict={'answer1': "value1",'answer11': "value11",'answer2': "value2",'answer3': "value3",'answer4': "value4",}
for i in r_dict:
if("answer" in i.lower()):
print i
Result is answer11,answer2,snswer4,answer3
I am using Python 2.4.3. I there any way to get the order in which it is populated?
Or is there a way to do this by regular expression since I am using the older Python version?
Dictionaries are unordered - that is, they do have some order, but it's influenced in nonobvious ways by the order of insertion and the hash of the keys. However, there is another implementation that remembers the order of insertion, collections.OrderedDict.
Edit: For Python 2.4, there are several third party implementations. I haven't used any, but since the one from voidspace looks promising.
A dictionary is by construction unordered. If you want an ordered one, use a collections.OrderedDict:
import collections
r_dict = collections.OrderedDict( [ ( 'answer1', "value1"), ('answer11', "value11"), ('answer2', "value2"), ('answer3', "value3"), ('answer4', "value4") ] )
for i in r_dict:
if("answer" in i.lower()):
print i
Not just by using the dictionary by itself. Dictionaries in Python (and a good portion of equivalent non-specialized data structures that involve mapping) are not sorted.
You could potentially subclass dict and override the __setitem__ and __delitem__ methods to add/remove each key to an internal list where you maintain your own sorting. You'd probably then have to override other methods, such as __iter__ to get the sorting you want out of your for loop.
...or just use the odict module as #delnan suggested
Short answer: no. Python dictionaries are fundamentally unordered.
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.
Found this:
If items(), keys(), values(),
iteritems(), iterkeys(), and
itervalues() are called with no
intervening modifications to the
dictionary, the lists will directly
correspond.
On 2.x documentation and 3.x documentation.
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.
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.
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.
Good references to the docs. Here's how you can guarantee the order regardless of the documentation / implementation:
k, v = zip(*d.iteritems())
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+
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
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
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]