Example 1: deque python
>>> from collections import deque
>>> d = deque('ghi')
>>> for elem in d:
... print elem.upper()
G
H
I
>>> d.append('j')
>>> d.appendleft('f')
>>> d
deque(['f', 'g', 'h', 'i', 'j'])
>>> d.pop()
'j'
>>> d.popleft()
'f'
>>> list(d)
['g', 'h', 'i']
>>> d[0]
'g'
>>> d[-1]
'i'
>>> list(reversed(d))
['i', 'h', 'g']
>>> 'h' in d
True
>>> d.extend('jkl')
>>> d
deque(['g', 'h', 'i', 'j', 'k', 'l'])
>>> d.rotate(1)
>>> d
deque(['l', 'g', 'h', 'i', 'j', 'k'])
>>> d.rotate(-1)
>>> d
deque(['g', 'h', 'i', 'j', 'k', 'l'])
>>> deque(reversed(d))
deque(['l', 'k', 'j', 'i', 'h', 'g'])
>>> d.clear()
>>> d.pop()
Traceback (most recent call last):
File "<pyshell#6>", line 1, in -toplevel-
d.pop()
IndexError: pop from an empty deque
>>> d.extendleft('abc')
>>> d
deque(['c', 'b', 'a'])
Example 2: python ordereddict
>>>
>>> d = {'banana': 3, 'apple': 4, 'pear': 1, 'orange': 2}
>>>
>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
>>>
>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
>>>
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])
Example 3: python counter
>>>
>>> cnt = Counter()
>>> for word in ['red', 'blue', 'red', 'green', 'blue', 'blue']:
... cnt[word] += 1
>>> cnt
Counter({'blue': 3, 'red': 2, 'green': 1})
>>>
>>> import re
>>> words = re.findall(r'\w+', open('hamlet.txt').read().lower())
>>> Counter(words).most_common(10)
[('the', 1143), ('and', 966), ('to', 762), ('of', 669), ('i', 631),
('you', 554), ('a', 546), ('my', 514), ('hamlet', 471), ('in', 451)]
Example 4: python counter
>>> Counter('abracadabra').most_common(3)
[('a', 5), ('r', 2), ('b', 2)]
Example 5: 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)]
Example 6: python counter
sum(c.values())
c.clear()
list(c)
set(c)
dict(c)
c.items()
Counter(dict(list_of_pairs))
c.most_common()[:-n-1:-1]
c += Counter()