YanYuSheng45的博客

import time def sum12n(n): s = 0 start = time.time() for i in range(1, n + 1): s += (i + 1) end = time.time() return s, end - start for i in range(6): sum1 = sum12...

# 列表推导式 list1 = [] # 用while循环实现 i = 0 while i < 10: list1.append(i) i += 1 print(list1) # 用for循环实现 for j in range(10): list1.append(j) print(list1) # 利用列表推导式实现 list2 = [i for i in r...

list1 = [1, 2, 3, 4, 5, 6, 6, 6] tuple1 = (11, 22, 33, 44, 55, 66) set1 = {'a', 'b', 'c', 'd', 'e'} # 将列表以及集合转换成元组 list1_2_tuple = tuple(list1) set1_2_tuple = tuple(set1) print(list1_2_tuple) print(...

yy = "c:\news" xx = "c:\\news" print(yy) print(xx)

# 定义字符串时，绝大多数情况下使用的是双引号，但是当字符串中需要使用双引号时，可以使用单引号来定义字符串 str1 = "我们都是追梦人！ 我们都是中国人！ 我们每个人都不一样！ " # print(str1[2]) # for item in str1: # print(item) k = len(str1) print(k) kk = str1.count("123", 0,...

str1 = "Hello Word!" print(str1.startswith("H", 6, 10)) # 判断字符串是否以哪个开头，也可以指定范围 print(str1.endswith("!")) # 判断字符串是否以哪个结尾，也可以指定范围 # index查找与find查找的异同 find_num = str1.find("Word!") # 查找要查找的字符串在字符串...

address = 'wo shi zhong bei da xue de xue sheng!' print(address.rfind('bei')) # 输出'bei'开始出现位置的下标13 print(address.rfind('bei', 0, 12)) # 输出-1,表示在指定范围内没有找到要查找的字符串 print(address.rfind('bei', 10)) # 输...

address = 'wo shi zhong bei da xue de xue sheng!' print(address.find('bei')) # 输出'bei'开始出现位置的小标13 print(address.find('bei', 0, 12)) # 输出-1,表示在指定范围内没有找到要查找的字符串 print(address.find('bei', 10)) # 输出13...

num = '012345678' print(num[0:5:1]) # 输出01234 print(num[0:5]) # 输出01234 print(num[0:]) # 输出012345678 print(num[:]) # 输出012345678 print(num[:6]) # 输出012345 print(num[::2]) # 输出02468 print(num[::...

# 九九乘法表输出 i = 1 while i <= 9: j = 1 while j <= i: print(f"{j}*{i}={i*j}", end="\t") j += 1 print() i += 1

# 三目运算符 # 生成随机数，取最大值 import random as r a = r.random() b = r.random() c = a if a > b else b print(f"a = {a}, b = {b}, 最大的数是c = {c} ")

import random player = int(input("请出拳： 0-剪刀； 1-石头； 2-布 ")) computer = random.randint(0, 2) if ((player == 0) and (computer == 2)) or ((player == 1) and (computer == 0)) or ((player == 2) and (comp...