Python技巧之--collections（OrderedDict，Counter，deque）

目录

OrderedDict

import collections
print "Regular dictionary"
d={}
d['a']='A'
d['b']='B'
d['c']='C'
for k,v in d.items():
    print k,v

print "nOrder dictionary"
d1 = collections.OrderedDict()
d1['a'] = 'A'
d1['b'] = 'B'
d1['c'] = 'C'
d1['1'] = '1'
d1['2'] = '2'
for k,v in d1.items():
    print k,v

输出：
Regular dictionary
a A
c C
b B

Order dictionary
a A
b B
c C
1 1
2 2

print 'Regular dictionary:'
d2={}
d2['a']='A'
d2['b']='B'
d2['c']='C'

d3={}
d3['c']='C'
d3['a']='A'
d3['b']='B'

print d2 == d3

print 'nOrderedDict:'
d4=collections.OrderedDict()
d4['a']='A'
d4['b']='B'
d4['c']='C'

d5=collections.OrderedDict()
d5['c']='C'
d5['a']='A'
d5['b']='B'

print  d1==d2

输出：
Regular dictionary:
True

OrderedDict:
False

dd = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
#按key排序
kd = collections.OrderedDict(sorted(dd.items(), key=lambda t: t[0]))
print kd
#按照value排序
vd = collections.OrderedDict(sorted(dd.items(),key=lambda t:t[1]))
print vd

#输出
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])

Counter

    >>> c = Counter('abcdeabcdabcaba')  # count elements from a string

    >>> c.most_common(3)                # three most common elements
    [('a', 5), ('b', 4), ('c', 3)]
    >>> sorted(c)                       # list all unique elements
    ['a', 'b', 'c', 'd', 'e']
    >>> ''.join(sorted(c.elements()))   # list elements with repetitions
    'aaaaabbbbcccdde'
    >>> sum(c.values())                 # total of all counts
    15

    >>> c['a']                          # count of letter 'a'
    5
    >>> for elem in 'shazam':           # update counts from an iterable
    ...     c[elem] += 1                # by adding 1 to each element's count
    >>> c['a']                          # now there are seven 'a'
    7
    >>> del c['b']                      # remove all 'b'
    >>> c['b']                          # now there are zero 'b'
    0

    >>> d = Counter('simsalabim')       # make another counter
    >>> c.update(d)                     # add in the second counter
    >>> c['a']                          # now there are nine 'a'
    9

    >>> c.clear()                       # empty the counter
    >>> c
    Counter()

    Note:  If a count is set to zero or reduced to zero, it will remain
    in the counter until the entry is deleted or the counter is cleared:

    >>> c = Counter('aaabbc')
    >>> c['b'] -= 2                     # reduce the count of 'b' by two
    >>> c.most_common()                 # 'b' is still in, but its count is zero
    [('a', 3), ('c', 1), ('b', 0)]

deque

最后

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