# Dictonary which stores order of data insertion like a queue
from collections import OrderedDict
#initialization
od = OrderedDict()
# or
od = OrderedDict({'a':1})
#Storing data from small to big values (not keys)
# Sorted with value, then key
#Adding Data
od['a'] = 1
od['d'] = 3
od['e'] = 7
od['f'] = 1
od['c'] = 7
od['b'] = 5
#Get all keys
od.keys()
#Get all key & value as list
od.items()
# Move an item to end
od.move_to_end('a', last=True) #Default last value = True
# Move an item to Start
od.move_to_end('a', last=False)
#Removing Data with key
del od['c']
od.pop('c') #Also returns the value
#Removing Top Element
od.popitem(last = False) #also return key/value pair
#Removing Bottom Element
od.popitem(last = True) #also return key/value pair>>> # 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)])>>> # 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)])
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)]