Answers for "Python Ordered Dictionary"

3

Python Ordered Dictionary

from collections import OrderedDict

# Remembers the order the keys are added!
x = OrderedDict(a=1, b=2, c=3)
Posted by: Guest on September-19-2020
1

ordered dictionary python

import collections

print 'Regular dictionary:'
d = {}
d['a'] = 'A'
d['b'] = 'B'
d['c'] = 'C'
d['d'] = 'D'
d['e'] = 'E'

for k, v in d.items():
    print k, v

print '\nOrderedDict:'
d = collections.OrderedDict()
d['a'] = 'A'
d['b'] = 'B'
d['c'] = 'C'
d['d'] = 'D'
d['e'] = 'E'

for k, v in d.items():
    print k, v
Posted by: Guest on October-21-2020
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)])
Posted by: Guest on April-01-2020
0

how to make an ordered dictionary in python

# A Python program to demonstrate working of key  
# value change in OrderedDict 
from collections import OrderedDict 
  
print("Before:\n") 
od = OrderedDict() 
od['a'] = 1
od['b'] = 2
od['c'] = 3
od['d'] = 4
for key, value in od.items(): 
    print(key, value) 
  
print("\nAfter:\n") 
od['c'] = 5
for key, value in od.items(): 
    print(key, value) 
""" 
Ouptut:
Before:

('a', 1)
('b', 2)
('c', 3)
('d', 4)

After:

('a', 1)
('b', 2)
('c', 5)
('d', 4)
"""
Posted by: Guest on November-05-2020
1

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)]
Posted by: Guest on May-18-2020
1

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
Posted by: Guest on March-11-2020

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