【Python 基础 2022 最新】练习 2

【Python 基础 2022 最新】练习 2

【Python 基础 2022 最新】练习 2

​​概述​​​​第五题​​​​第六题​​​​第七题​​​​第八题​​​​第五题 (答案)​​​​第六题 (答案)​​​​第七题 (答案)​​​​第八题 (答案)​​

概述

从今天开始, 小白我将带领大家学习一下 Python 零基础入门的内容. 本专栏会以讲解 + 练习的模式, 带领大家熟悉 Python 的语法, 应用, 以及代码的基础逻辑.

第五题

def assignment(): # a. assign x the value 8 # b. assign both x and y the value 9 in one line of code # c. assign x the value 10 and y the value 12 using one line of code # d.# run the programassignment()

预期输出:

89 910 127

第六题

def even_odd(x): # your code# define some sample dataa=45# run the programeven_odd(a)

预期输出:

odd

第七题

def check_max(lst): # your code# define some sample valuesnums=[3, 41, 12, 9, 74, 15]# run the programcheck_max(nums)

预期结果:

74

第八题

# your codecheck_balance(1000,1400)check_balance(1000,800)check_balance(1000,1200)

预期输出:

your account balance has increased by $ 400your account balance has decreased by $ 200No action is required at this time

第五题 (答案)

def assignment(): # a. assign x the value 8 x=8 print(x) # b. assign both x and y the value 9 in one line of code x=y=9 print(x,y) # c. assign x the value 10 and y the value 12 using one line of code x,y=10,12 print(x,y) # d. x=4 x+=1 x+=1 x+=1 print(x)# run the programassignment()

第六题 (答案)

def even_odd(x): # your code if x%2==0: print('even') else: print('odd')# define some sample dataa=45# run the programeven_odd(a)

第七题 (答案)

def check_max(lst): # your code Max_num = None for num in lst: if (Max_num is None or num > Max_num): Max_num = num print(Max_num)# define some sample valuesnums=[3, 41, 12, 9, 74, 15]# run the programcheck_max(nums)

第八题 (答案)

# your codedef check_balance(num1, num2): # 判断 if (num2 > num1 * 1.25): print("your account balance has increased by $", num2 - num1) elif (num2 < num1 * 0.98): print("your account balance has decreased by $", num1 - num2) else: print("No action is required at this time")# 调用check_balance(1000,1400)check_balance(1000,800)check_balance(1000,1200)

输出结果:

your account balance has increased by $ 400your account balance has decreased by $ 200No action is required at this time

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