zip

• 把两个可迭代内容生成一个可迭代的tuple元素类型组成的内容

# zip 案例
l1 = [ 1,2,3,4,5]
l2 = [11,22,33,44,55]

z = zip(l1, l2)

print(type(z))
print(z)

for i in z:
print(i)

enumerate

• 跟zip功能比较像
• 对可迭代对象里的每一元素，配上一个索引，然后索引和内容构成tuple类型

# enumerate案例1
l1 = [11,22,33,44,55]

em = enumerate(l1)

l2 = [i for i in em]
print(l2)

• 自定义索引开始位置

em = enumerate(l1, start=100)

l2 = [ i for i in em]
print(l2)

collections模块

• namedtuple
• deque

namedtuple

• tuple类型
• 是一个可命名的tuple

import collections
Point = collections.namedtuple("Point", ['x', 'y'])
p = Point(11, 22)
print(p.x)
print(p[0])

Circle = collections.namedtuple("Circle", ['x', 'y', 'r'])

c = Circle(100, 150, 50)
print(c)
print(type(c))

# 想检测以下namedtuple到底属于谁的子类
isinstance(c, tuple)

deque

• 比较方便的解决了频繁删除插入带来的效率问题

from collections import deque

q = deque(['a', 'b', 'c'])
print(q)

q.append("d")
print(q)

q.appendleft('x')
print(q)

defaultdict

• 当直接读取dict不存在的属性时，直接返回默认值

d1 = {"one":1, "two":2, "three":3}
print(d1['one'])
print(d1['four'])  #这句会报错

from collections import defaultdict
# lambda表达式，直接返回字符串
func = lambda: "刘大拿"
d2 = defaultdict(func)

d2["one"] = 1
d2["two"] = 2

print(d2['one'])
print(d2['four'])

Counter

• 统计字符串个数

from collections import Counter

# 为什么下面结果不把abcdefgabced.....作为键值，而是以其中每一个字母作为键值
# 需要括号里内容为可迭代
c = Counter("abcdefgabcdeabcdabcaba")
print(c)

s = ["liudana", "love", "love", "love", "love", "wangxiaona"]
c = Counter(s)

print(c)