列表生成式
# 需求1:
# 1.接收变量k,a,b
s = '51 5000 10000'
# li = []
# for item in s.split():
# li.append(int(item))
# k,a,b = li
# print(k,a,b)
# li=[int(item) for item in s.split()]
# print(li)
# print(k,a,b)
# 需求2:生成一个列表,列表元素分别为[1**1,2**2,3**3,4**4...n*n]
# li = []
# for i in range(1,8):
# li.append(i**i)
# print(li)
# li = [i**i for i in range(1,8)]
# print(li)
"""
[experssion for item in 序列 if 判断语句]
"""
# li = [i ** 2 for i in range(1, 8) if i % 2 == 0]
# print(li)
# 找出1~10之间的所有的偶数
print([i for i in range(1, 11) if i % 2 == 0])
# 找出1~1000之间的所有质数
# def isPrime(num):
# pass
# print([i for i in range(1, 1000) if isPrime(i)])
# s1='ABC' s2='123'(for嵌套循环)
# 'A1', 'A2', 'A3', 'B1', 'B2', 'B3', 'C1', 'C2', 'C3'
print([i + j for i in 'ABC' for j in '123'])
需求2:
列表生成式变形之for循环嵌套
"""
需求:讲3x3的矩阵转换成一堆数组
[
[1,2,3],
[4,5,6],
[7,8,9]
]
[1,2,3,4,5,6,7,8,9]
"""
li = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
resLi = []
for item1 in li: # [1,2,3] [4,5,6] [7,8,9]
for item2 in item1:
if item2 % 2 == 0:
resLi.append(item2)
print(resLi)
# print([item2 for item1 in li for item2 in item1])
# from itertools import chain
# print(list(chain(*li)))
列表生成式案例
# 将列表中所有内容都变成小写
# li = ['dasdaFSDFSFDSF','dadDSADSAsdaDASsadas']
# print([i.lower() for i in li])
# 找出/var/log目录中,所有以.log结尾的文件名或者目录名
import os
#print(os.listdir('/var/log'))
print([filename for filename in os.listdir('/var/log') if filename.endswith('.log')])
列表生成式练习
"""
[i**2 for i in range(2,10,2)]
[fun(i) for i in range(2,10,2)]
"""
import math
# 1. 找出1~10之间所有偶数, 并且返回一个列表,(包含以这个偶数为半径的圆的面积)
# 方法一:
li = []
for r in range(2, 11, 2):
square = math.pi * r * r
li.append(square)
print(li)
# 方法二:
print([math.pi * r * r for r in range(2, 11, 2)])
# 方法三:
def square(r):
"""求以r为半径的圆"""
res = math.pi *r*r
return res
print([square(r) for r in range(2,11,2)])
# 2. 找出1~100之间所有的质数.
"""
判断num是否为质数,如果是质数,返回True,否则返回False
什么是质数?
只能被1和本身整除的数,就是质数
方法:
依次判断num能被(2,num)整除
如果能被其中其中一个数整除,不是质数
当循环结束,都没有发现能被整除的数,那么就是质数
"""
def isPrime(num):
for i in range(2,num):
if num % i == 0:
return False
else:
return True
print([i for i in range(2,101) if isPrime(i)])
列表生成式之腾讯笔试题
"""
4. (2018-腾讯-在线编程题)
- 题目描述:
给定一个正整数,编写程序计算有多少对质数的和等于输入的这个正整数,并输出结果。输
入值小于1000。
如,输入为10, 程序应该输出结果为2。(共有两对质数的和为10,分别为(5,5),(3,7))
#[2,3,5,7]
- 输入描述:
输入包括一个整数n,(3 ≤ n < 1000)
- 输出描述:
输出对数
- 示例1 :
输入:
10
输出:
2
"""
num = int(input('N:'))
def isPrime(num):
for i in range(2, num):
if num % i == 0:
return False
else:
return True
primeLi = [i for i in range(2, num) if isPrime(i)]
primePairCount = 0
#[2,3,5,7] # 2====10-2 3===10-3 5===10-5 7==10-7
for item1 in primeLi:
if (num - item1) in primeLi and item1 <= num - item1:
primePairCount += 1
print(primePairCount)
字典生成式
# 需求1:假设有20个学生,学生分数在60~100之间,筛选出成绩在90分以上的学生
import random
# stuInfo={}
# for i in range(20):
# name = 'westos' + str(i)
# score = random.randint(60,100)
# stuInfo[name] = score
# print(stuInfo)
# stuInfo = {'westos'+ str(i):random.randint(60,100) for
# i in range(20)}
# highscore = {}
# for name,score in stuInfo.items():
# if score > 90:
# highscore[name] = score
# print(highscore)
# print({name:score for name,score in stuInfo.items() if score > 90})
# 需求2:将所有的key值变为大写
# d = dict(a=1,b=2)
# new_d = {}
# for i in d:
# new_d[i.upper()] = d[i]
# print('key转化为大写的字典:',new_d)
# print({k.upper():v for k,v in d.items()})
# 需求3:大小写key值合并,统一以小写输出
d = dict(a=2, b=1, c=2, B=9, A=10)
print({k.lower(): d.get(k.lower(), 0) + d.get(k.upper(), 0) for k in d})
# for k, v in d.items():
# low_k = k.lower()
# if low_k not in new_d:
# new_d[low_k] = v
# else:
# new_d[low_k] += v
#
# print(new_d)
集合生成式
print({i ** 2 for i in {1, 2, 3}})
print({i ** 2 for i in {1, 2, 3, 9, 12} if i % 3 == 0})
生成器
"""
1.创建生成器的第一重方式
In [7]: nums = (x*2 for x in range(10))
In [8]: nums
Out[8]: <generator object <genexpr> at 0x7fb2a6137fc0>
In [9]: for num in nums:
...: print(num)
...:
0
2
4
6
8
10
12
14
16
18
2.创建生成器的方法2(定义一个函数,让这个函数变成生成器)
"""
# 1 1 2 3 5 8 13...
# def fib(num):
# a,b,count = 0,1,1
# while count <= num:
# print(b)
# a,b = b,a+b
# count +=1
#
# fib(1000)
"""
如果在调用函数的时候,发现这个函数中有yeild
那么此时,也就不是调用函数了,而是创建了一个生成器对象
"""
def creat_num(all_num):
print('~~~~~~~~~~~1~~~~~~~~~~~~~~~~~')
a,b=0,1
current_num = 0
while current_num < all_num:
print('~~~~~~~~~2~~~~~~~~~~~~~~~~~')
yield a # 相当于暂停了程序
#print(a)
print('~~~~~~~~~~~~~~3~~~~~~~~~~~~')
a,b = b,b+a
current_num += 1
print('~~~~~~~~~~~4~~~~~~~~~~~~~~~~')
obj = creat_num(5)
# 此时.我们发现程序会报错(告诉我们,生成器里面没有东西了)
while True:
try:
ret = next(obj)
print('obj:',ret)
except Exception as red:
print(red.value)
break
# obj = creat_num(5)
# print(obj)
# ret = next(obj)
# print(ret)
# ret1 = next(obj)
# print(ret1)
# ret2 = next(obj)
# print(ret2)
# ret3 = next(obj)
# print(ret3)
# ret4 = next(obj)
# print(ret4)
# ret5 = next(obj)
# print(ret5)
# for num in obj:
# print(num)
# 两个生成器对象之前并没有任何关系
# obj2 = creat_num(100)
# print(obj2)
# ret4 = next(obj2)
# print(ret4)
生成器——send
"""
使用send唤醒程序
使用send()函数来唤醒程序执行,使用send()函数的好处是
可以在唤醒的同时向断点中传入一个附加的数据
"""
def create_num(all_num):
a,b = 0,1
current_num = 0
while current_num < all_num:
ret = yield a
print('>>>>>>>>ret>>>>>>>',ret)
a,b = b,a+b
current_num += 1
obj = create_num(100)
# red = next(obj)
# print(red)
red = obj.send(None)
print(red)
"""
next和send得到的都是yield后面的值
不同的是send传递值而next不传递值
注意:
不能把send放在第一个,因为第一次执行程序是从开始执行
并没有值来接收send
如果你非要把send放在第一个,那么传递的值应该是None
"""