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Gaidy's

8.python3实用编程技巧进阶(三)

3.1.如何实现可迭代对象和迭代器对象

#3.1 如何实现可迭代对象和迭代器对象

import requests
from collections.abc import Iterable,Iterator

class WeatherIterator(Iterator):
    def __init__(self,cities):
        self.cities = cities
        #从列表中迭代一个city,index就+1
        self.index = 0

    def __next__(self):
        #如果所有的城市都迭代完了,就抛出异常
        if self.index == len(self.cities):
            raise StopIteration
        #当前迭代的city
        city = self.cities[self.index]
        #迭代完当前city,index就+1
        self.index += 1
        return self.get_weather(city)

    def get_weather(self,city):
        url = 'http://wthrcdn.etouch.cn/weather_mini?city=' + city
        r = requests.get(url)
        #获取当天的天气信息
        data = r.json()['data']['forecast'][0]
        #返回城市名字、最高和最低气温
        return city, data['high'], data['low']


class WeatherIterable(Iterable):
    def __init__(self,cities):
        self.cities = cities

    def __iter__(self):
        return WeatherIterator(self.cities)


def show(w):
    for x in w:
        print(x)

weather = WeatherIterable(['北京','上海','广州','深圳','东莞'])
show(weather)

结果

3.2如何使用生成器函数实现可迭代对象

#3.2如何使用生成器函数实现可迭代对象

from collections.abc import Iterable

class PrimeNumbers(Iterable):
    def __init__(self,a,b):
        self.a = a
        self.b = b

    def __iter__(self):
        for k in range(self.a,self.b):
            if self.is_prime(k):
                yield k

    def is_prime(self,k):
        return False if k < 2 else all(map(lambda x : k % x, range(2, k)))

#打印1到30直接的素数
pn = PrimeNumbers(1, 30)
for n in pn:
    print(n)

3.3.如何进行反向迭代以及如何实现反向迭代

反向迭代

In [75]: l = [1,2,3,4,5]

In [76]: for x in l:
    ...:     print(x)
    ...:
1
2
3
4
5

In [77]: for x in reversed(l):
    ...:     print(x)
    ...:
5
4
3
2
1

要想实现反向迭代必须实现__reversed__方法

#3.3.如何进行反向迭代以及如何实现反向迭代

class IntRange:
    def __init__(self,a,b,step):
        self.a = a
        self.b = b
        self.step = step

    def __iter__(self):
        t = self.a
        while t <= self.b:
            yield t
            t += self.step
    
    def __reversed__(self):
        t = self.b
        while t >= self.a:
            yield t
            t -= self.step

fr = IntRange(1, 10, 2)

for x in fr:
    print(x)

print('=' * 30)

#反向迭代
for y in reversed(fr):
    print(y)

 3.4.如何对迭代器做切片操作

(1)切片的实质是__getitem__方法

In [9]: l = list(range(10))

In [10]: l[3]
Out[10]: 3

In [11]: l.__getitem__(3)
Out[11]: 3

In [12]: l[2:6]
Out[12]: [2, 3, 4, 5]

In [13]: l.__getitem__(slice(2,6))
Out[13]: [2, 3, 4, 5]

(2)打印文件第2~5行

islice能返回一个迭代对象切片的生成器

#3.4.如何对迭代器做切片操作

from itertools import islice

f= open('iter_islice')

#打印文件的第2~5行内容
for line in islice(f, 1, 5):
    print(line)

(3)自己实现islice功能

#自己实现一个类似islice的功能
def my_slice(iterable, start, end, step=1):
    for i, x in enumerate(iterable):
        if i >= end:
            break
        if i >= start:
            yield x

print(list(my_slice(range(1,20), 4, 10)))   #[5, 6, 7, 8, 9, 10]

from itertools import islice

print(list(islice(range(1,20),4, 10)))      #[5, 6, 7, 8, 9, 10]

(4)加step

#3.4.如何对迭代器做切片操作

from itertools import islice

f= open('iter_islice')

#打印文件的第2~5行内容
for line in islice(f, 1, 5):
    print(line)


#自己实现一个类似islice的功能
def my_slice(iterable, start, end, step=1):
    tmp = 0
    for i, x in enumerate(iterable):
        if i >= end:
            break
        if i >= start:
            if tmp == 0:
                tmp = step
                yield x
            tmp -= 1

print(list(my_slice(range(1,20), 4, 10)))   #[5, 6, 7, 8, 9, 10]
print(list(my_slice(range(1,20), 4, 10,2)))   #[5, 7, 9]

from itertools import islice

print(list(islice(range(1,20),4, 10)))      #[5, 6, 7, 8, 9, 10]
print(list(islice(range(1,20),4, 10,2)))    #[5, 7, 9]

3.5.如何在一个for语句中迭代多个可迭代对象

计算学生的三科成绩总分,用zip()函数

In [25]: from random import randint

In [26]: chinese = [randint(60,100) for _ in range(10)]

In [27]: math = [randint(60,100) for _ in range(10)]

In [28]: english = [randint(60,100) for _ in range(10)]

In [29]: chinese
Out[29]: [70, 63, 85, 74, 70, 96, 60, 69, 62, 83]

In [30]: math
Out[30]: [76, 81, 86, 93, 74, 83, 69, 63, 60, 80]

In [31]: english
Out[31]: [100, 96, 83, 89, 71, 79, 82, 87, 81, 71]

In [32]: t = []

In [33]: for s1, s2, s3 in zip(chinese, math, english):
    ...:     t.append(s1 + s2 +s3)
    ...:

In [34]: t
Out[34]: [246, 240, 254, 256, 215, 258, 211, 219, 203, 234]

求三个班级中分数高于90分的总人数,用chain

In [53]: c1 = [randint(60,100) for _ in range(1,10)]

In [54]: c2 = [randint(60,100) for _ in range(1,10)]

In [55]: c3 = [randint(60,100) for _ in range(1,10)]

In [56]: c1
Out[56]: [60, 79, 89, 84, 68, 68, 89, 68, 82]

In [57]: c2
Out[57]: [69, 64, 87, 89, 60, 77, 89, 81, 90]

In [58]: c3
Out[58]: [80, 92, 64, 73, 68, 84, 97, 71, 65]

In [59]: from itertools import chain

In [60]: len([ x for x in chain(c1, c2, c3) if x > 90])
Out[60]: 2

 

 
 
 

posted on 2019-12-23 17:49  Gaidy  阅读(246)  评论(0编辑  收藏  举报

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