Python中collections模块的基本使用教程

#!/usr/bin/env python
# -*- coding:utf-8 -*-
# __author__ = 'liao gao xiang'

"""
保留最后n个元素
"""
from collections import deque


def search(file, pattern, history=5):
 previous_lines = deque(maxlen=history)
 for l in file:
 if pattern in l:
  yield l, previous_lines # 使用yield表达式的生成器函数，将搜索过程的代码和搜索结果的代码解耦
 previous_lines.append(l)


with open(b'file.txt', mode='r', encoding='utf-8') as f:
 for line, prevlines in search(f, 'Python', 5):
 for pline in prevlines:
  print(pline, end='')
 print(line, end='')

d = deque()
d.append(1)
d.append("2")
print(len(d))
print(d[0], d[1])
d.extendleft([0])
print(d)
d.extend([6, 7, 8])
print(d)

d2 = deque('12345')
print(len(d2))
d2.popleft()
print(d2)
d2.pop()
print(d2)

# 在队列两端插入或删除元素时间复杂度都是 O(1) ，区别于列表，在列表的开头插入或删除元素的时间复杂度为 O(N)
d3 = deque(maxlen=2)
d3.append(1)
d3.append(2)
print(d3)
d3.append(3)
print(d3)

class deque(object):
 """
 deque([iterable[, maxlen]]) --> deque object
 
 A list-like sequence optimized for data accesses near its endpoints.
 """
 def append(self, *args, **kwargs): # real signature unknown
 """ Add an element to the right side of the deque. """
 pass

 def appendleft(self, *args, **kwargs): # real signature unknown
 """ Add an element to the left side of the deque. """
 pass

 def clear(self, *args, **kwargs): # real signature unknown
 """ Remove all elements from the deque. """
 pass

 def copy(self, *args, **kwargs): # real signature unknown
 """ Return a shallow copy of a deque. """
 pass

 def count(self, value): # real signature unknown; restored from __doc__
 """ D.count(value) -> integer -- return number of occurrences of value """
 return 0

 def extend(self, *args, **kwargs): # real signature unknown
 """ Extend the right side of the deque with elements from the iterable """
 pass

 def extendleft(self, *args, **kwargs): # real signature unknown
 """ Extend the left side of the deque with elements from the iterable """
 pass

 def index(self, value, start=None, stop=None): # real signature unknown; restored from __doc__
 """
 D.index(value, [start, [stop]]) -> integer -- return first index of value.
 Raises ValueError if the value is not present.
 """
 return 0

 def insert(self, index, p_object): # real signature unknown; restored from __doc__
 """ D.insert(index, object) -- insert object before index """
 pass

 def pop(self, *args, **kwargs): # real signature unknown
 """ Remove and return the rightmost element. """
 pass

 def popleft(self, *args, **kwargs): # real signature unknown
 """ Remove and return the leftmost element. """
 pass

 def remove(self, value): # real signature unknown; restored from __doc__
 """ D.remove(value) -- remove first occurrence of value. """
 pass

 def reverse(self): # real signature unknown; restored from __doc__
 """ D.reverse() -- reverse *IN PLACE* """
 pass

 def rotate(self, *args, **kwargs): # real signature unknown
 """ Rotate the deque n steps to the right (default n=1). If n is negative, rotates left. """
 pass

d = {}
for key, value in pairs:
 if key not in d:
 d[key] = []
 d[key].append(value)

d = defaultdict(list)
for key, value in pairs:
 d[key].append(value)

#!/usr/bin/env python
# -*- coding:utf-8 -*-
# __author__ = 'liao gao xiang'

# 字典中的键映射多个值
from collections import defaultdict

d = defaultdict(list)
print(d)
d['a'].append([1, 2, 3])
d['b'].append(2)
d['c'].append(3)

print(d)

d = defaultdict(set)
print(d)
d['a'].add(1)
d['a'].add(2)
d['b'].add(4)

print(d)

from collections import namedtuple

websites = [
 ('Sohu', 'http://www.sohu.com/', u'张朝阳'),
 ('Sina', 'http://www.sina.com.cn/', u'王志东'),
 ('163', 'http://www.163.com/', u'丁磊')
]

Website = namedtuple('Website', ['name', 'url', 'founder'])

for website in websites:
 website = Website._make(website)
 print website


# 输出结果:
Website(name='Sohu', url='http://www.sohu.com/', founder=u'\u5f20\u671d\u9633')
Website(name='Sina', url='http://www.sina.com.cn/', founder=u'\u738b\u5fd7\u4e1c')
Website(name='163', url='http://www.163.com/', founder=u'\u4e01\u78ca')

#!/usr/bin/env python
# -*- coding:utf-8 -*-
# __author__ = 'liao gao xiang'

from collections import Counter

words = [
 'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
 'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
 'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
 'my', 'eyes', "you're", 'under'
]

word_counts = Counter(words)

# 出现频率最高的三个单词
top_three = word_counts.most_common(3)
print(top_three)
# Outputs [('eyes', 8), ('the', 5), ('look', 4)]
print(word_counts['eyes'])

morewords = ['why', 'are', 'you', 'not', 'looking', 'in', 'my', 'eyes']

# 如果你想手动增加计数，可以简单的用加法：
for word in morewords:
 print(word)
 word_counts[word] += 1
print(word_counts['eyes'])

#!/usr/bin/env python
# -*- coding:utf-8 -*-
# __author__ = 'liao gao xiang'

from collections import OrderedDict

d = OrderedDict()
d['foo'] = 1
d['bar'] = 2
d['spam'] = 3
d['grok'] = 4
# d['bar'] = 22 #对于一个已经存在的键，重复赋值不会改变键的顺序
for key in d:
 print(key, d[key])

print(d)

import json

print(json.dumps(d))

#!/usr/bin/env python
# -*- coding:utf-8 -*-
# __author__ = 'liao gao xiang'

from collections import ChainMap
from itertools import chain

# 不同集合上元素的迭代
a = [1, 2, 3, 4]
b = ('x', 'y', 'z')
c = {1, 'a'}

# 方法一，使用chain
for i in chain(a, b, c):
 print(i)
print('--------------')
# 方法二，使用chainmap
for j in ChainMap(a, b, c):
 print(j)

# 这两种均为节省内存，效率更高的迭代方式

#!/usr/bin/env python
# -*- coding:utf-8 -*-
# __author__ = 'liao gao xiang'

# 合并多个字典和映射
a = {'x': 1, 'z': 3}
b = {'y': 2, 'z': 4}
# 现在假设你必须在两个字典中执行查找操作
# (比如先从 a 中找，如果找不到再在 b 中找)。
# 一个非常简单的解决方案就是使用collections模块中的ChainMap类
from collections import ChainMap

c = ChainMap(a, b)

print(c)
a['x'] = 11 # 使用ChainMap时，原字典做了更新，这种更新会合并到新的字典中去

print(c) # 按顺序合并两个字典
print(c['x'])
print(c['y'])
print(c['z'])

# 对于字典的更新或删除操作影响的总是列中的第一个字典。
c['z'] = 10
c['w'] = 40
del c['x']
print(a)
# del c['y']将出现报错

# ChainMap对于编程语言中的作用范围变量（比如globals,locals等）
# 是非常有用的。事实上，有一些方法可以使它变得简单：
values = ChainMap() # 默认会创建一个空字典
print('\t', values)
values['x'] = 1
values = values.new_child() # 添加一个空字典
values['x'] = 2
values = values.new_child()
values['x'] = 30
# values = values.new_child()
print(values, values['x']) # values['x']输出最后一次添加的值
values = values.parents # 删除上一次添加的字典
print(values['x'])
values = values.parents
print(values)

a = {'x': 1, 'y': 2}
b = {'y': 2, 'z': 3}
merge = dict(b)
merge.update(a)
print(merge['x'], merge['y'], merge['z'])
a['x'] = 11
print(merge['x'])