python--collections容器

Counter:

from collections import *

cnt = Counter()  #创建对象--字典的之类
wordList = ["a","b","c","c","a","a"]
for word in wordList:
    cnt[word] += 1  #【说明：针对相应key的值进行计算】
print(cnt)   #Counter({‘a‘: 3, ‘c‘: 2, ‘b‘: 1})

from collections import *

c = Counter()#一个新的，空的counter
print(c)  #Counter()

c = Counter("gallahad")#从可迭代的字符串初始化counter
print(c)   #Counter({‘a‘: 3, ‘l‘: 2, ‘g‘: 1, ‘h‘: 1, ‘d‘: 1})

c = Counter({‘red‘:4,‘blue‘:2})  #从映射初始化counter
print(c)   #Counter({‘red‘: 4, ‘blue‘: 2})

c = Counter(cats = 4,dogs = 8)  #从args初始化counter
print(c)   #Counter({‘dogs‘: 8, ‘cats‘: 4})

from collections import *#Counter对象类似于字典，如果某个项缺失，会返回0，而不是报出KeyError；c = Counter([‘eggs‘,‘ham‘])print(c[‘bacon‘])    #没有‘bacon‘,返回0del c[‘eggs‘]  #删除eggs元素print(c)   #Counter({‘ham‘: 1})

Counter对象支持以下三个字典不支持的方法，elements()，most_common()，subtract()；

from collections import *

c = Counter(a=2,b=4,c=0,d=-2,e = 1)
print(c)  #Counter({‘b‘: 4, ‘a‘: 2, ‘e‘: 1, ‘c‘: 0, ‘d‘: -2})
print(c.elements())  #c.elements()是一个迭代器--返回迭代器的地址
L=list(c.elements())   
#element()，返回一个迭代器，每个元素重复的次数为它的数目，顺序是任意的顺序，如果一个元素的数目少于1，那么elements()就会忽略它；
print(L)   #[‘a‘, ‘a‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘e‘]

from collections import *

c = Counter(‘abracadabra‘)
print(c)  #Counter({‘a‘: 5, ‘b‘: 2, ‘r‘: 2, ‘c‘: 1, ‘d‘: 1})
a=c.most_common(3)
#most_common()，返回一个列表，包含counter中n个最大数目的元素，如果忽略n或者为None，most_common()将会返回counter中的所有元素，元素有着相同数目的将会以任意顺序排列；
print(a)   #[(‘a‘, 5), (‘b‘, 2), (‘r‘, 2)]
b=c.most_common()
print(b)
d=c.most_common(None)
print(d)

from collections import *

c = Counter(a=4,b=2,c=0,d=-2,e=9)
d = Counter(a=1,b=2,c=-3,d=4,f=77)
print(c,d)
c.subtract(d)  #c中相应的值减去d中相应的值，结果保存到c中
print(c)

c = Counter(a=4,b=2,c=0,d=-2,e=9)
c.update(d)   #c中相应的值加上d中相应的值，结果保存到c中
print(c)