Python 实用宝典

Question 1

I am having some trouble using the collections.OrderedDict class. I am using Python 2.7 on Raspbian, the Debian distro for Raspberry Pi. I am trying to print two dictionaries in order for comparison (side-by-side) for a text-adventure. The order is essential to compare accurately. No matter what I try the dictionaries print in their usual unordered way.

Here’s what I get when I do it on my RPi:

import collections

ship = {"NAME": "Albatross",
         "HP":50,
         "BLASTERS":13,
         "THRUSTERS":18,
         "PRICE":250}

ship = collections.OrderedDict(ship)

print ship
# OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)])

Obviously there is something not right because it is printing the function call and putting the keys and value groups into a nested list…

This is what I got by running something similar on my PC:

import collections

Joe = {"Age": 28, "Race": "Latino", "Job": "Nurse"}
Bob = {"Age": 25, "Race": "White", "Job": "Mechanic", "Random": "stuff"}

#Just for clarity:
Joe = collections.OrderedDict(Joe)
Bob = collections.OrderedDict(Bob)

print Joe
# OrderedDict([('Age', 28), ('Race', 'Latino'), ('Job', 'Nurse')])
print Bob
# OrderedDict([('Age', 25), ('Race', 'White'), ('Job', 'Mechanic'), ('Random', 'stuff')])

This time, it is in order, but it shouldn’t be printing the other things though right? (The putting it into list and showing function call.)

Where am I making my error? It shouldn’t be anything to do with the pi version of Python because it is just the Linux version.

Question 2

You are creating a dictionary first, then passing that dictionary to an OrderedDict. For Python versions < 3.6 ^(*), by the time you do that, the ordering is no longer going to be correct. dict is inherently not ordered.

Pass in a sequence of tuples instead:

ship = [("NAME", "Albatross"),
        ("HP", 50),
        ("BLASTERS", 13),
        ("THRUSTERS", 18),
        ("PRICE", 250)]
ship = collections.OrderedDict(ship)

What you see when you print the OrderedDict is it’s representation, and it is entirely correct. OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)]) just shows you, in a reproducable representation, what the contents are of the OrderedDict.

^(*): In the CPython 3.6 implementation, the dict type was updated to use a more memory efficient internal structure that has the happy side effect of preserving insertion order, and by extension the code shown in the question works without issues. As of Python 3.7, the Python language specification has been updated to require that all Python implementations must follow this behaviour. See this other answer of mine for details and also why you’d still may want to use an OrderedDict() for certain cases.

Question 3

If you can’t edit this part of code where your dict was defined you can still order it at any point in any way you want, like this:

from collections import OrderedDict

order_of_keys = ["key1", "key2", "key3", "key4", "key5"]
list_of_tuples = [(key, your_dict[key]) for key in order_of_keys]
your_dict = OrderedDict(list_of_tuples)

Question 4

Most of the time we go for OrderedDict when we required a custom order not a generic one like ASC etc.

Here is the proposed solution:

import collections
ship = {"NAME": "Albatross",
         "HP":50,
         "BLASTERS":13,
         "THRUSTERS":18,
         "PRICE":250}

ship = collections.OrderedDict(ship)

print ship


new_dict = collections.OrderedDict()
new_dict["NAME"]=ship["NAME"]
new_dict["HP"]=ship["HP"]
new_dict["BLASTERS"]=ship["BLASTERS"]
new_dict["THRUSTERS"]=ship["THRUSTERS"]
new_dict["PRICE"]=ship["PRICE"]


print new_dict

This will be output:

OrderedDict([('PRICE', 250), ('HP', 50), ('NAME', 'Albatross'), ('BLASTERS', 13), ('THRUSTERS', 18)])
OrderedDict([('NAME', 'Albatross'), ('HP', 50), ('BLASTERS', 13), ('THRUSTERS', 18), ('PRICE', 250)])

Note: The new sorted dictionaries maintain their sort order when entries are deleted. But when new keys are added, the keys are appended to the end and the sort is not maintained.(official doc)

Question 5

Use dict.items(); it can be as simple as following:

ship = collections.OrderedDict(ship.items())

Question 6

What’s the correct way to initialize an ordered dictionary (OD) so that it retains the order of initial data?

from collections import OrderedDict

# Obviously wrong because regular dict loses order
d = OrderedDict({'b':2, 'a':1}) 

# An OD is represented by a list of tuples, so would this work?
d = OrderedDict([('b',2), ('a', 1)])

# What about using a list comprehension, will 'd' preserve the order of 'l'
l = ['b', 'a', 'c', 'aa']
d = OrderedDict([(i,i) for i in l])

Question:

Will an OrderedDict preserve the order of a list of tuples, or tuple of tuples or tuple of lists or list of lists etc. passed at the time of initialization (2nd & 3rd example above)?
How does one go about verifying if OrderedDict actually maintains an order? Since a dict has an unpredictable order, what if my test vectors luckily have the same initial order as the unpredictable order of a dict? For example, if instead of d = OrderedDict({'b':2, 'a':1}) I write d = OrderedDict({'a':1, 'b':2}), I can wrongly conclude that the order is preserved. In this case, I found out that a dict is ordered alphabetically, but that may not be always true. What’s a reliable way to use a counterexample to verify whether a data structure preserves order or not, short of trying test vectors repeatedly until one breaks?

P.S. I’ll just leave this here for reference: “The OrderedDict constructor and update() method both accept keyword arguments, but their order is lost because Python’s function call semantics pass-in keyword arguments using a regular unordered dictionary”

P.P.S : Hopefully, in future, OrderedDict will preserve the order of kwargs also (example 1): http://bugs.python.org/issue16991

Question 7

The OrderedDict will preserve any order that it has access to. The only way to pass ordered data to it to initialize is to pass a list (or, more generally, an iterable) of key-value pairs, as in your last two examples. As the documentation you linked to says, the OrderedDict does not have access to any order when you pass in keyword arguments or a dict argument, since any order there is removed before the OrderedDict constructor sees it.

Note that using a list comprehension in your last example doesn’t change anything. There’s no difference between OrderedDict([(i,i) for i in l]) and OrderedDict([('b', 'b'), ('a', 'a'), ('c', 'c'), ('aa', 'aa')]). The list comprehension is evaluated and creates the list and it is passed in; OrderedDict knows nothing about how it was created.

Question 8

# An OD is represented by a list of tuples, so would this work?
d = OrderedDict([('b', 2), ('a', 1)])

Yes, that will work. By definition, a list is always ordered the way it is represented. This goes for list-comprehension too, the list generated is in the same way the data was provided (i.e. source from a list it will be deterministic, sourced from a set or dict not so much).

How does one go about verifying if OrderedDict actually maintains an order. Since a dict has an unpredictable order, what if my test vectors luckily has the same initial order as the unpredictable order of a dict?. For example, if instead of d = OrderedDict({'b':2, 'a':1}) I write d = OrderedDict({'a':1, 'b':2}), I can wrongly conclude that the order is preserved. In this case, I found out that a dict is order alphabetically, but that may not be always true. i.e. what’s a reliable way to use a counter example to verify if a data structure preserves order or not short of trying test vectors repeatedly until one breaks.

You keep your source list of 2-tuple around for reference, and use that as your test data for your test cases when you do unit tests. Iterate through them and ensure the order is maintained.

Question 9

I like the pprint module in Python. I use it a lot for testing and debugging. I frequently use the width option to make sure the output fits nicely within my terminal window.

It has worked fine until they added the new ordered dictionary type in Python 2.7 (another cool feature I really like). If I try to pretty-print an ordered dictionary, it doesn’t show nicely. Instead of having each key-value pair on its own line, the whole thing shows up on one long line, which wraps many times and is hard to read.

Does anyone here have a way to make it print nicely, like the old unordered dictionaries? I could probably figure something out, possibly using the PrettyPrinter.format method, if I spend enough time, but I am wondering if anyone here already knows of a solution.

UPDATE: I filed a bug report for this. You can see it at http://bugs.python.org/issue10592.

Question 10

As a temporary workaround you can try dumping in JSON format. You lose some type information, but it looks nice and keeps the order.

import json

pprint(data, indent=4)
# ^ugly

print(json.dumps(data, indent=4))
# ^nice

Question 11

The following will work if the order of your OrderedDict is an alpha sort, since pprint will sort a dict before print.

pprint(dict(o.items()))

Question 12

Here’s another answer that works by overriding and using the stock pprint() function internally. Unlike my earlier one it will handle OrderedDict‘s inside another container such as a list and should also be able to handle any optional keyword arguments given — however it does not have the same degree of control over the output that the other one afforded.

It operates by redirecting the stock function’s output into a temporary buffer and then word wraps that before sending it on to the output stream. While the final output produced isn’t exceptionalily pretty, it’s decent and may be “good enough” to use as a workaround.

Update 2.0

Simplified by using standard library textwrap module, and modified to work in both Python 2 & 3.

from collections import OrderedDict
try:
    from cStringIO import StringIO
except ImportError:  # Python 3
    from io import StringIO
from pprint import pprint as pp_pprint
import sys
import textwrap

def pprint(object, **kwrds):
    try:
        width = kwrds['width']
    except KeyError: # unlimited, use stock function
        pp_pprint(object, **kwrds)
        return
    buffer = StringIO()
    stream = kwrds.get('stream', sys.stdout)
    kwrds.update({'stream': buffer})
    pp_pprint(object, **kwrds)
    words = buffer.getvalue().split()
    buffer.close()

    # word wrap output onto multiple lines <= width characters
    try:
        print >> stream, textwrap.fill(' '.join(words), width=width)
    except TypeError:  # Python 3
        print(textwrap.fill(' '.join(words), width=width), file=stream)

d = dict((('john',1), ('paul',2), ('mary',3)))
od = OrderedDict((('john',1), ('paul',2), ('mary',3)))
lod = [OrderedDict((('john',1), ('paul',2), ('mary',3))),
       OrderedDict((('moe',1), ('curly',2), ('larry',3))),
       OrderedDict((('weapons',1), ('mass',2), ('destruction',3)))]

Sample output:

pprint(d, width=40)

» {'john': 1, 'mary': 3, 'paul': 2}

pprint(od, width=40)

» OrderedDict([('john', 1), ('paul', 2),
('mary', 3)])

pprint(lod, width=40)

» [OrderedDict([('john', 1), ('paul', 2),
   ('mary', 3)]), OrderedDict([('moe', 1),
   ('curly', 2), ('larry', 3)]),
   OrderedDict([('weapons', 1), ('mass',
   2), ('destruction', 3)])]

Question 13

To print an ordered dict, e.g.

from collections import OrderedDict

d=OrderedDict([
    ('a', OrderedDict([
        ('a1',1),
        ('a2','sss')
    ])),
    ('b', OrderedDict([
        ('b1', OrderedDict([
            ('bb1',1),
            ('bb2',4.5)])),
        ('b2',4.5)
    ])),
])

I do

def dict_or_OrdDict_to_formatted_str(OD, mode='dict', s="", indent=' '*4, level=0):
    def is_number(s):
        try:
            float(s)
            return True
        except ValueError:
            return False
    def fstr(s):
        return s if is_number(s) else '"%s"'%s
    if mode != 'dict':
        kv_tpl = '("%s", %s)'
        ST = 'OrderedDict([\n'; END = '])'
    else:
        kv_tpl = '"%s": %s'
        ST = '{\n'; END = '}'
    for i,k in enumerate(OD.keys()):
        if type(OD[k]) in [dict, OrderedDict]:
            level += 1
            s += (level-1)*indent+kv_tpl%(k,ST+dict_or_OrdDict_to_formatted_str(OD[k], mode=mode, indent=indent, level=level)+(level-1)*indent+END)
            level -= 1
        else:
            s += level*indent+kv_tpl%(k,fstr(OD[k]))
        if i!=len(OD)-1:
            s += ","
        s += "\n"
    return s

print dict_or_OrdDict_to_formatted_str(d)

Which yields

"a": {
    "a1": 1,
    "a2": "sss"
},
"b": {
    "b1": {
        "bb1": 1,
        "bb2": 4.5
    },
    "b2": 4.5
}

or

print dict_or_OrdDict_to_formatted_str(d, mode='OD')

which yields

("a", OrderedDict([
    ("a1", 1),
    ("a2", "sss")
])),
("b", OrderedDict([
    ("b1", OrderedDict([
        ("bb1", 1),
        ("bb2", 4.5)
    ])),
    ("b2", 4.5)
]))

Question 14

Here’s a way that hacks the implementation of pprint. pprint sorts the keys before printing, so to preserve order, we just have to make the keys sort in the way we want.

Note that this impacts the items() function. So you might want to preserve and restore the overridden functions after doing the pprint.

from collections import OrderedDict
import pprint

class ItemKey(object):
  def __init__(self, name, position):
    self.name = name
    self.position = position
  def __cmp__(self, b):
    assert isinstance(b, ItemKey)
    return cmp(self.position, b.position)
  def __repr__(self):
    return repr(self.name)

OrderedDict.items = lambda self: [
    (ItemKey(name, i), value)
    for i, (name, value) in enumerate(self.iteritems())]
OrderedDict.__repr__ = dict.__repr__

a = OrderedDict()
a[4] = '4'
a[1] = '1'
a[2] = '2'
print pprint.pformat(a) # {4: '4', 1: '1', 2: '2'}

Question 15

Here is my approach to pretty print an OrderedDict

from collections import OrderedDict
import json
d = OrderedDict()
d['duck'] = 'alive'
d['parrot'] = 'dead'
d['penguin'] = 'exploded'
d['Falcon'] = 'discharged'
print(d)
print(json.dumps(d,indent=4))

OutPut:

OrderedDict([('duck', 'alive'), ('parrot', 'dead'), ('penguin', 'exploded'), ('Falcon', 'discharged')])

{
    "duck": "alive",
    "parrot": "dead",
    "penguin": "exploded",
    "Falcon": "discharged"
}

If you want to pretty print dictionary with keys in sorted order

print(json.dumps(indent=4,sort_keys=True))
{
    "Falcon": "discharged",
    "duck": "alive",
    "parrot": "dead",
    "penguin": "exploded"
}

Question 16

This is pretty crude, but I just needed a way to visualize a data structure made up of any arbitrary Mappings and Iterables and this is what I came up with before giving up. It’s recursive, so it will fall through nested structures and lists just fine. I used the Mapping and Iterable abstract base classes from collections to handle just about anything.

I was aiming for almost yaml like output with concise python code, but didn’t quite make it.

def format_structure(d, level=0):
    x = ""
    if isinstance(d, Mapping):
        lenk = max(map(lambda x: len(str(x)), d.keys()))
        for k, v in d.items():
            key_text = "\n" + " "*level + " "*(lenk - len(str(k))) + str(k)
            x += key_text + ": " + format_structure(v, level=level+lenk)
    elif isinstance(d, Iterable) and not isinstance(d, basestring):
        for e in d:
            x += "\n" + " "*level + "- " + format_structure(e, level=level+4)
    else:
        x = str(d)
    return x

and some test data using OrderedDict and lists of OrderedDicts… (sheesh Python needs OrderedDict literals sooo badly…)

d = OrderedDict([("main",
                  OrderedDict([("window",
                                OrderedDict([("size", [500, 500]),
                                             ("position", [100, 900])])),
                               ("splash_enabled", True),
                               ("theme", "Dark")])),
                 ("updates",
                  OrderedDict([("automatic", True),
                               ("servers",
                                [OrderedDict([("url", "http://server1.com"),
                                              ("name", "Stable")]),
                                 OrderedDict([("url", "http://server2.com"),
                                              ("name", "Beta")]),
                                 OrderedDict([("url", "http://server3.com"),
                                              ("name", "Dev")])]),
                               ("prompt_restart", True)])),
                 ("logging",
                  OrderedDict([("enabled", True),
                               ("rotate", True)]))])

print format_structure(d)

yields the following output:

   main: 
               window: 
                         size: 
                             - 500
                             - 500
                     position: 
                             - 100
                             - 900
       splash_enabled: True
                theme: Dark
updates: 
            automatic: True
              servers: 
                     - 
                          url: http://server1.com
                         name: Stable
                     - 
                          url: http://server2.com
                         name: Beta
                     - 
                          url: http://server3.com
                         name: Dev
       prompt_restart: True
logging: 
       enabled: True
        rotate: True

I had some thoughts along the way of using str.format() for better alignment, but didn’t feel like digging into it. You’d need to dynamically specify the field widths depending on the type of alignment you want, which would get either tricky or cumbersome.

Anyway, this shows me my data in readable hierarchical fashion, so that works for me!

Question 17

def pprint_od(od):
    print "{"
    for key in od:
        print "%s:%s,\n" % (key, od[key]) # Fixed syntax
    print "}"

There you go ^^

for item in li:
    pprint_od(item)

or

(pprint_od(item) for item in li)

Question 18

I’ve tested this unholy monkey-patch based hack on python3.5 and it works:

pprint.PrettyPrinter._dispatch[pprint._collections.OrderedDict.__repr__] = pprint.PrettyPrinter._pprint_dict


def unsorted_pprint(data):
    def fake_sort(*args, **kwargs):
        return args[0]
    orig_sorted = __builtins__.sorted
    try:
        __builtins__.sorted = fake_sort
        pprint.pprint(data)
    finally:
        __builtins__.sorted = orig_sorted

You make pprint use the usual dict based summary and also disable sorting for the duration of the call so that no keys are actually sorted for printing.

Question 19

As of Python 3.8 : pprint.PrettyPrinter exposes the sort_dicts keyword parameter.

True by default, setting it to False will leave the dictionary unsorted.

>>> from pprint import PrettyPrinter

>>> x = {'John': 1,
>>>      'Mary': 2,
>>>      'Paul': 3,
>>>      'Lisa': 4,
>>>      }

>>> PrettyPrinter(sort_dicts=False).pprint(x)

Will output :

{'John': 1, 
 'Mary': 2, 
 'Paul': 3,
 'Lisa': 4}

Reference : https://docs.python.org/3/library/pprint.html

Question 20

The pprint() method is just invoking the __repr__() method of things in it, and OrderedDict doesn’t appear to do much in it’s method (or doesn’t have one or something).

Here’s a cheap solution that should work IF YOU DON’T CARE ABOUT THE ORDER BEING VISIBLE IN THE PPRINT OUTPUT, which may be a big if:

class PrintableOrderedDict(OrderedDict):
    def __repr__(self):
        return dict.__repr__(self)

I’m actually surprised that the order isn’t preserved… ah well.

Question 21

You can also use this simplification of the kzh answer:

pprint(data.items(), indent=4)

It preserves the order and will output almost the same than the webwurst answer (print through json dump).

Question 22

For python < 3.8 (e.g. 3.6):

Monkey patch pprint‘s sorted in order to prevent it from sorting. This will have the benefit of everything working recursively as well, and is more suitable than the json option for whoever needs to use e.g. width parameter:

import pprint
pprint.sorted = lambda arg, *a, **kw: arg

>>> pprint.pprint({'z': 1, 'a': 2, 'c': {'z': 0, 'a': 1}}, width=20)
{'z': 1,
 'a': 2,
 'c': {'z': 0,
       'a': 1}}

Edit: cleaning up

To clean up after this dirty business just run: pprint.sorted = sorted

For a really clean solution can even use a contextmanager:

import pprint
import contextlib

@contextlib.contextmanager
def pprint_ordered():
    pprint.sorted = lambda arg, *args, **kwargs: arg
    yield
    pprint.sorted = sorted

# usage:

with pprint_ordered():
    pprint.pprint({'z': 1, 'a': 2, 'c': {'z': 0, 'a': 1}}, width=20)

# without it    
pprint.pprint({'z': 1, 'a': 2, 'c': {'z': 0, 'a': 1}}, width=20)

# prints: 
#    
# {'z': 1,
#  'a': 2,
#  'c': {'z': 0,
#        'a': 1}}
#
# {'a': 2,
#  'c': {'a': 1,
#        'z': 0},
#  'z': 1}

Question 23

You could redefine pprint() and intercept calls for OrderedDict‘s. Here’s a simple illustration. As written, the OrderedDict override code ignores any optional stream, indent, width, or depth keywords that may have been passed, but could be enhanced to implement them. Unfortunately this technique doesn’t handle them inside another container, such as a list of OrderDict‘s

from collections import OrderedDict
from pprint import pprint as pp_pprint

def pprint(obj, *args, **kwrds):
    if not isinstance(obj, OrderedDict):
        # use stock function
        return pp_pprint(obj, *args, **kwrds)
    else:
        # very simple sample custom implementation...
        print "{"
        for key in obj:
            print "    %r:%r" % (key, obj[key])
        print "}"

l = [10, 2, 4]
d = dict((('john',1), ('paul',2), ('mary',3)))
od = OrderedDict((('john',1), ('paul',2), ('mary',3)))
pprint(l, width=4)
# [10,
#  2,
#  4]
pprint(d)
# {'john': 1, 'mary': 3, 'paul': 2}

pprint(od)
# {
#     'john':1
#     'paul':2
#     'mary':3
# }

Question 24

If the dictionary items are all of one type, you could use the amazing data-handling library pandas:

>>> import pandas as pd
>>> x = {'foo':1, 'bar':2}
>>> pd.Series(x)
bar    2
foo    1
dtype: int64

or

>>> import pandas as pd
>>> x = {'foo':'bar', 'baz':'bam'}
>>> pd.Series(x)
baz    bam
foo    bar
dtype: object

Question 25

I am struggling with the following problem: I want to convert an OrderedDict like this:

OrderedDict([('method', 'constant'), ('data', '1.225')])

into a regular dict like this:

{'method': 'constant', 'data':1.225}

because I have to store it as string in a database. After the conversion the order is not important anymore, so I can spare the ordered feature anyway.

Thanks for any hint or solutions,

Ben

Question 26

>>> from collections import OrderedDict
>>> OrderedDict([('method', 'constant'), ('data', '1.225')])
OrderedDict([('method', 'constant'), ('data', '1.225')])
>>> dict(OrderedDict([('method', 'constant'), ('data', '1.225')]))
{'data': '1.225', 'method': 'constant'}
>>>

However, to store it in a database it’d be much better to convert it to a format such as JSON or Pickle. With Pickle you even preserve the order!

Question 27

Even though this is a year old question, I would like to say that using dict will not help if you have an ordered dict within the ordered dict. The simplest way that could convert those recursive ordered dict will be

import json
from collections import OrderedDict
input_dict = OrderedDict([('method', 'constant'), ('recursive', OrderedDict([('m', 'c')]))])
output_dict = json.loads(json.dumps(input_dict))
print output_dict

Question 28

It is easy to convert your OrderedDict to a regular Dict like this:

dict(OrderedDict([('method', 'constant'), ('data', '1.225')]))

If you have to store it as a string in your database, using JSON is the way to go. That is also quite simple, and you don’t even have to worry about converting to a regular dict:

import json
d = OrderedDict([('method', 'constant'), ('data', '1.225')])
dString = json.dumps(d)

Or dump the data directly to a file:

with open('outFile.txt','w') as o:
    json.dump(d, o)

Question 29

If you are looking for a recursive version without using the json module:

def ordereddict_to_dict(value):
    for k, v in value.items():
        if isinstance(v, dict):
            value[k] = ordereddict_to_dict(v)
    return dict(value)

Question 30

Here is what seems simplest and works in python 3.7

from collections import OrderedDict

d = OrderedDict([('method', 'constant'), ('data', '1.225')])
d2 = dict(d)  # Now a normal dict

Now to check this:

>>> type(d2)
<class 'dict'>
>>> isinstance(d2, OrderedDict)
False
>>> isinstance(d2, dict)
True

NOTE: This also works, and gives same result –

>>> {**d}
{'method': 'constant', 'data': '1.225'}
>>> {**d} == d2
True

As well as this –

>>> dict(d)
{'method': 'constant', 'data': '1.225'}
>>> dict(d) == {**d}
True

Cheers

Question 31

If somehow you want a simple, yet different solution, you can use the {**dict} syntax:

from collections import OrderedDict

ordered = OrderedDict([('method', 'constant'), ('data', '1.225')])
regular = {**ordered}

Question 32

Its simple way

>>import json 
>>from collection import OrderedDict

>>json.dumps(dict(OrderedDict([('method', 'constant'), ('data', '1.225')])))

Question 33

Lets say I have the following code:

import collections
d = collections.OrderedDict()
d['foo'] = 'python'
d['bar'] = 'spam'

Is there a way I can access the items in a numbered manner, like:

d(0) #foo's Output
d(1) #bar's Output

Question 34

If its an OrderedDict() you can easily access the elements by indexing by getting the tuples of (key,value) pairs as follows

>>> import collections
>>> d = collections.OrderedDict()
>>> d['foo'] = 'python'
>>> d['bar'] = 'spam'
>>> d.items()
[('foo', 'python'), ('bar', 'spam')]
>>> d.items()[0]
('foo', 'python')
>>> d.items()[1]
('bar', 'spam')

Note for Python 3.X

dict.items would return an iterable dict view object rather than a list. We need to wrap the call onto a list in order to make the indexing possible

>>> items = list(d.items())
>>> items
[('foo', 'python'), ('bar', 'spam')]
>>> items[0]
('foo', 'python')
>>> items[1]
('bar', 'spam')

Question 35

Do you have to use an OrderedDict or do you specifically want a map-like type that’s ordered in some way with fast positional indexing? If the latter, then consider one of Python’s many sorted dict types (which orders key-value pairs based on key sort order). Some implementations also support fast indexing. For example, the sortedcontainers project has a SortedDict type for just this purpose.

>>> from sortedcontainers import SortedDict
>>> sd = SortedDict()
>>> sd['foo'] = 'python'
>>> sd['bar'] = 'spam'
>>> print sd.iloc[0] # Note that 'bar' comes before 'foo' in sort order.
'bar'
>>> # If you want the value, then simple do a key lookup:
>>> print sd[sd.iloc[1]]
'python'

Question 36

Here is a special case if you want the first entry (or close to it) in an OrderedDict, without creating a list. (This has been updated to Python 3):

>>> from collections import OrderedDict
>>> 
>>> d = OrderedDict()
>>> d["foo"] = "one"
>>> d["bar"] = "two"
>>> d["baz"] = "three"
>>> next(iter(d.items()))
('foo', 'one')
>>> next(iter(d.values()))
'one'

(The first time you say “next()”, it really means “first.”)

In my informal test, next(iter(d.items())) with a small OrderedDict is only a tiny bit faster than items()[0]. With an OrderedDict of 10,000 entries, next(iter(d.items())) was about 200 times faster than items()[0].

BUT if you save the items() list once and then use the list a lot, that could be faster. Or if you repeatedly { create an items() iterator and step through it to to the position you want }, that could be slower.

Question 37

It is dramatically more efficient to use IndexedOrderedDict from the indexed package.

Following Niklas’s comment, I have done a benchmark on OrderedDict and IndexedOrderedDict with 1000 entries.

In [1]: from numpy import *
In [2]: from indexed import IndexedOrderedDict
In [3]: id=IndexedOrderedDict(zip(arange(1000),random.random(1000)))
In [4]: timeit id.keys()[56]
1000000 loops, best of 3: 969 ns per loop

In [8]: from collections import OrderedDict
In [9]: od=OrderedDict(zip(arange(1000),random.random(1000)))
In [10]: timeit od.keys()[56]
10000 loops, best of 3: 104 µs per loop

IndexedOrderedDict is ~100 times faster in indexing elements at specific position in this specific case.

Question 38

This community wiki attempts to collect existing answers.

Python 2.7

In python 2, the keys(), values(), and items() functions of OrderedDict return lists. Using values as an example, the simplest way is

d.values()[0]  # "python"
d.values()[1]  # "spam"

For large collections where you only care about a single index, you can avoid creating the full list using the generator versions, iterkeys, itervalues and iteritems:

import itertools
next(itertools.islice(d.itervalues(), 0, 1))  # "python"
next(itertools.islice(d.itervalues(), 1, 2))  # "spam"

The indexed.py package provides IndexedOrderedDict, which is designed for this use case and will be the fastest option.

from indexed import IndexedOrderedDict
d = IndexedOrderedDict({'foo':'python','bar':'spam'})
d.values()[0]  # "python"
d.values()[1]  # "spam"

Using itervalues can be considerably faster for large dictionaries with random access:

$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 1000;   d = OrderedDict({i:i for i in range(size)})'  'i = randint(0, size-1); d.values()[i:i+1]'
1000 loops, best of 3: 259 usec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 10000;  d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i:i+1]'
100 loops, best of 3: 2.3 msec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 100000; d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i:i+1]'
10 loops, best of 3: 24.5 msec per loop

$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 1000;   d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); next(itertools.islice(d.itervalues(), i, i+1))'
10000 loops, best of 3: 118 usec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 10000;  d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); next(itertools.islice(d.itervalues(), i, i+1))'
1000 loops, best of 3: 1.26 msec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 100000; d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); next(itertools.islice(d.itervalues(), i, i+1))'
100 loops, best of 3: 10.9 msec per loop

$ python2 -m timeit -s 'from indexed import IndexedOrderedDict; from random import randint; size = 1000;   d = IndexedOrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i]'
100000 loops, best of 3: 2.19 usec per loop
$ python2 -m timeit -s 'from indexed import IndexedOrderedDict; from random import randint; size = 10000;  d = IndexedOrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i]'
100000 loops, best of 3: 2.24 usec per loop
$ python2 -m timeit -s 'from indexed import IndexedOrderedDict; from random import randint; size = 100000; d = IndexedOrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i]'
100000 loops, best of 3: 2.61 usec per loop

+--------+-----------+----------------+---------+
|  size  | list (ms) | generator (ms) | indexed |
+--------+-----------+----------------+---------+
|   1000 | .259      | .118           | .00219  |
|  10000 | 2.3       | 1.26           | .00224  |
| 100000 | 24.5      | 10.9           | .00261  |
+--------+-----------+----------------+---------+

Python 3.6

Python 3 has the same two basic options (list vs generator), but the dict methods return generators by default.

List method:

list(d.values())[0]  # "python"
list(d.values())[1]  # "spam"

Generator method:

import itertools
next(itertools.islice(d.values(), 0, 1))  # "python"
next(itertools.islice(d.values(), 1, 2))  # "spam"

Python 3 dictionaries are an order of magnitude faster than python 2 and have similar speedups for using generators.

+--------+-----------+----------------+---------+
|  size  | list (ms) | generator (ms) | indexed |
+--------+-----------+----------------+---------+
|   1000 | .0316     | .0165          | .00262  |
|  10000 | .288      | .166           | .00294  |
| 100000 | 3.53      | 1.48           | .00332  |
+--------+-----------+----------------+---------+

Question 39

It’s a new era and with Python 3.6.1 dictionaries now retain their order. These semantics aren’t explicit because that would require BDFL approval. But Raymond Hettinger is the next best thing (and funnier) and he makes a pretty strong case that dictionaries will be ordered for a very long time.

So now it’s easy to create slices of a dictionary:

test_dict = {
                'first':  1,
                'second': 2,
                'third':  3,
                'fourth': 4
            }

list(test_dict.items())[:2]

Note: Dictonary insertion-order preservation is now official in Python 3.7.

Question 40

for OrderedDict() you can access the elements by indexing by getting the tuples of (key,value) pairs as follows or using ‘.values()’

>>> import collections
>>> d = collections.OrderedDict()
>>> d['foo'] = 'python'
>>> d['bar'] = 'spam'
>>> d.items()
[('foo', 'python'), ('bar', 'spam')]
>>>d.values()
odict_values(['python','spam'])
>>>list(d.values())
['python','spam']

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