PYTHON-模拟练习题目集合

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1. Which of the following expression is Illegal?

O A.['12.56'] * 7
O B. int(7.4)+7.4
O C.['a','b','c']-['a']
O D.str(1.32)*5

考查:列表,整数,字符串的运算规则

---

在Python中，列表的加法运算是将两个列表拼接成一个新的列表，例如：

a = [1, 2, 3]
b = [4, 5, 6]
c = a + b # [1, 2, 3, 4, 5, 6]

列表的乘法运算是重复一个列表多次得到一个新的列表，例如：

a = [1, 2, 3]
b = a * 3 # [1, 2, 3, 1, 2, 3, 1, 2, 3]

列表不存在减法和除法运算。->选c

2.Function defining

Please define a function which can find out the all multiples of 3 (3的倍数) from 0 to nand return the sum of those mutiples.

Header of the function:

在这里描述函数接口。

例如:def acc_three(n):
n is larger than 3 and smaller than 1000

def acc_three(n):
    sum = 0
    for i in range (3,n+1):
        if i % 3 == 0:
            sum += i
    return sum
 
#👇为后台系统调用检查结果
print(acc_three(1000))->166833

考查函数相关的知识点,---> http://t.csdn.cn/TpC92

3.Using Function Recursion(函数递归) to Find the Sum of 1-100

def f(n):
    if n == 1: #递归结束条件
        return 1
    else:
        return n+f(n-1) #递归公式
 
print(f(100))#5050

考查函数递归知识点,-->http://t.csdn.cn/6uEc5

4.Please read through the block carefully and finish the following tasks:Q1. output the average Python score of three students.Q2. Define a List to store the First name of all students.
Remember to copy this code to your answer. All you need to do is to write your
code under each condition below.

student_dict = {
    "Kendrick Ray": {
        "Math": 60,
        "English": 45,
        "Python": 60
    },
    "Jhon Rick":{
          "Math": 90,
        "English": 95,
        "Python": 70
    },
    "Stephen Curry":{
          "Math": 80,
        "English": 90,
        "Python": 50
    }
}
sign = input()
 
if sign == "avg":
#
    py_sum = 0
    for stu in student_dict:
        py_sum = py_sum + student_dict[stu]["Python"]
    print(py_sum / 3)
#
 
elif sign == "name":
#
    name_list =[]    #创建一个空列表接收name
    for stu in student_dict:    #遍历字典里面的字典
        stu_list = stu.split(" ")    
#使用split内置函数切割每个小字典的名字,并保存在新的列表stu_list中
 
        name_list.append(stu_list[0])    
#将新列表对应的name取出来储存在一个刚刚创建的空列表里
    print(name_list)
#

# Output prediction
if 0:
    print("1")
elif 1-1.0:
    print("2")
else:
    print("3")

结果:3

z = -87.7e100
print(type(z))
#结果:float

for x in range(28,31,3):
    print(x)
# 28

fruits = ["apple","banana","cherry"]
for x in  fruits:
    for y in  x:
        if x == "banana":
            break
        print(y)
 
#a
#p
#p
#l
#e
#c
#h
#e
#r
#r
#y

上面两个for循环,x是取到列表里面的元素,y是取得元素的每个字母,虽然它break跳出了离他最近的内循环,但由于print的是y,所以能够取到 apple 和 cherry. 

dict1 ={
"brand": "Ford",
"model": "Mustang",
"year":  1964,
"price": 20000
}
print(dict1["model"][4])
print(len(dict1))
#a
#4

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