Python Ordered Dictionary
from collections import OrderedDict
# Remembers the order the keys are added!
x = OrderedDict(a=1, b=2, c=3)Python Ordered Dictionary
from collections import OrderedDict
# Remembers the order the keys are added!
x = OrderedDict(a=1, b=2, c=3)collections
LIST: Can store duplicate values,Keeps the insertion order.
- ArrayList not syncronized, array based class
- LinkedList not synchronized, doubly linked
- Vector is synchronized, thread safe
SET: Can only store unique values,And does not maintain order
- HashSet can have null, order is not guaranteed
- LinkedHashSet can have null and keeps the order
- TreeSet sorts the order and don't accept null
QUQUE :Accepts duplicates, Doesn't have index num,
First in first our order.
MAP:is a (key-value format)
and keys are always unique,
and value can be duplicated.
- HashTable don't have null key, sychronized(thread-safe)
- LinkedHashMap can have null key, keeps order
- HasHMap can have null key, order is not guaranteed
- TreeMap doesn't have null key and keys are sorted
Examples
■ I have used TreeSet to print dropdown list in for
non duplicate values and ascending order
■ I have used HashMaps to compare values from
a database with expected valuespython 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)])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)]what type of collections used
Depending on the data that I am working with, I use
Arrays, Lists, Sets, Maps.collections
import collections
numShoes = int(raw_input())
shoes = collections.Counter(map(int, raw_input().split()))
numCust = int(raw_input())
income = 0
for i in range(numCust):
size, price = map(int, raw_input().split())
if shoes[size]:
income += price
shoes[size] -= 1
print incomeCopyright © 2021 Codeinu
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