Python Ordered Dictionary

from collections import OrderedDict

# Remembers the order the keys are added!
x = OrderedDict(a=1, b=2, c=3)

Python ordereddict

>>> # regular unsorted dictionary
>>> d = {'banana': 3, 'apple': 4, 'pear': 1, 'orange': 2}

>>> # dictionary sorted by key
>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])

>>> # dictionary sorted by value
>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])

>>> # dictionary sorted by length of the key string
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])

Source: docs.python.org

Counter most_common

most_common([n])¶
Return a list of the n most common elements and their counts from the most common to the least. If n is omitted or None, most_common() returns all elements in the counter.
Elements with equal counts are ordered arbitrarily:

>>> Counter('abracadabra').most_common(3)
[('a', 5), ('r', 2), ('b', 2)]

Source: docs.python.org

Python counter

sum(c.values())                 # total of all counts
c.clear()                       # reset all counts
list(c)                         # list unique elements
set(c)                          # convert to a set
dict(c)                         # convert to a regular dictionary
c.items()                       # convert to a list of (elem, cnt) pairs
Counter(dict(list_of_pairs))    # convert from a list of (elem, cnt) pairs
c.most_common()[:-n-1:-1]       # n least common elements
c += Counter()                  # remove zero and negative counts

Source: docs.python.org

Ordered dict

#Update: As of Python 3.7 (and CPython 3.6), standard dict is 
	#guaranteed to preserve order and is more performant than OrderedDict. 
#(For backward compatibility and especially readability, 
	#however, you may wish to continue using OrderedDict.)

>>> keywords = ['foo', 'bar', 'bar', 'foo', 'baz', 'foo']

>>> list(dict.fromkeys(keywords))
['foo', 'bar', 'baz']

Source: stackoverflow.com

Similar codes

0 vote274 views2 answers
How to query DNS records using python
Python
-2 votes271 views1 answer
Ceil in python3
Python
0 vote306 views1 answer
Tkinter gui grid and frame
Python
0 vote300 views1 answer
Next function with inherited list python
Python
0 vote260 views2 answers
Python starting multiple processes in a loop
Python
-1 vote286 views2 answers
How to convert 2 dimensional in 1 dimensional array
Python
1 vote295 views2 answers
Pytorch view -1 meaning
Python
0 vote274 views1 answer
Django strptime
Python
0 vote260 views1 answer
How to make a lis in python
Python
2 votes334 views1 answer
Flask if statement
Python Flask