python collections

1from collections import OrderedDict
2
3# Remembers the order the keys are added!
4x = OrderedDict(a=1, b=2, c=3)

1>>> # regular unsorted dictionary
2>>> d = {'banana': 3, 'apple': 4, 'pear': 1, 'orange': 2}
3
4>>> # dictionary sorted by key
5>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))
6OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
7
8>>> # dictionary sorted by value
9>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))
10OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
11
12>>> # dictionary sorted by length of the key string
13>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
14OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])
15

1>>> # Tally occurrences of words in a list
2>>> cnt = Counter()
3>>> for word in ['red', 'blue', 'red', 'green', 'blue', 'blue']:
4...     cnt[word] += 1
5>>> cnt
6Counter({'blue': 3, 'red': 2, 'green': 1})
7
8>>> # Find the ten most common words in Hamlet
9>>> import re
10>>> words = re.findall(r'\w+', open('hamlet.txt').read().lower())
11>>> Counter(words).most_common(10)
12[('the', 1143), ('and', 966), ('to', 762), ('of', 669), ('i', 631),
13 ('you', 554),  ('a', 546), ('my', 514), ('hamlet', 471), ('in', 451)]
14

1
2# Dictionary where key values default to a list:
3	a = collections.defaultdict(list)   
4# Dictionary where key values default to 0:
5	a = collections.defaultdict(int)
6# Dictionary where key values default to another dictionary:
7	a = collections.defaultdict(dict)
8
9'''
10Example:
11  In a regular dictionary, if you did the following:
12  
13    a = {}
14    if a['apple'] == 1:
15        return True    
16    else:
17        return False
18        
19  You'd get an error, because 'apple' does not exist in dictionary. 
20  However, if you use collections:
21    
22    a = collections.defaultdict(int)
23    if a['apple'] == 1:
24		return True
25    else:
26        return False
27    
28  This would not give an error, and False would be returned.
29  Also, the dictionary would now have the key 'apple' with a
30  default integer value of 0 inside it.
31  
32  If you used say, collections.defaultdict(list), the default value
33  would be an empty list instead of 0.
34  
35'''  
36

1most_common([n])¶
2Return 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.
3Elements with equal counts are ordered arbitrarily:
4
5>>> Counter('abracadabra').most_common(3)
6[('a', 5), ('r', 2), ('b', 2)]
7

1list = [1,2,3,4,1,2,6,7,3,8,1]
2Counter(list)
3