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from collections import OrderedDict
# Remembers the order the keys are added!
x = OrderedDict(a=1, b=2, c=3)
Delta Sierra
from collections import OrderedDict
# Remembers the order the keys are added!
x = OrderedDict(a=1, b=2, c=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)])
# A Python program to demonstrate working of OrderedDict
from collections import OrderedDict
print("This is a Dict:\n")
d = {}
d['a'] = 1
d['b'] = 2
d['c'] = 3
d['d'] = 4
for key, value in d.items():
print(key, value)
print("\nThis is an Ordered Dict:\n")
od = OrderedDict()
od['a'] = 1
od['b'] = 2
od['c'] = 3
od['d'] = 4
for key, value in od.items():
print(key, value)# ordered dictionary
#Regular Dictionary
d = {'apple':1, 'oranges':2, 'bananas':3}
d['grapes'] = 4
print(d)
#Output:
# {'apple': 1, 'bananas': 3, 'grapes': 4, 'oranges': 2}
#Ordered Dictionary
from collections import OrderedDict
d = OrderedDict({'apple':1, 'oranges':2, 'bananas':3})
d['grapes'] = 4
print(d)
# OrderedDict([('apple', 1), ('oranges', 2), ('bananas', 3), ('grapes', 4)])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)]
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