python内置函数代码练习

Posted 2022-07-18 jiangzuofenghua

练习1：

# 输入两个数，比较大小后，从小到大升序打印

a = input(‘fisrt:‘)
b = input(‘secend:‘)
if a > b:
    print(b,a)
else:
    print(a,b)

 

 

练习2：

#九九乘法表
for i in range(1,10):
    for j in range(1,i+1):
        print(str(j)+‘*‘+str(i)+"="+str(i*j),end=‘ ‘)
    print()

:

练习3：

#打印倒着的九九乘法表

for i in range(1,10):
    print(‘ ‘*7*(i-1), end=‘‘)
    for j in range(i,10):
        product = i*j
        if product<10:
            end = ‘  ‘
        else:
            end = ‘ ‘
            
        print(str(i)+‘*‘+str(j)+‘=‘+str(i*j),end=end)
    print()

练习4：

#依次接手用户输入的3个数，排序后打印
#方法1：转换成int后判断大小排序
nums = []

for i in range(3):
    nums.append(int(input(‘number:‘.format(i))))

if nums[0] > nums[1]:
    if nums[0] > nums[2]:
        i3 = nums[0]
        if nums[1] > nums[2]:
            i2 = nums[1]
            i1 = nums[2]
        else:
            i2 = nums[2]
            i1 = nums[1]
    else:
        i2 = nums[0]
        i3 = nums[2]
        i1 = nums[1]

else:  # nums[0] < nums[1]
    if nums[0] > nums[2]:
        i3 = nums[1]
        i2 = nums[0]
        i1 = nums[2]
    else:
        if nums[1] < nums[2]:  #1<2
            i1 = nums[0]
            i2 = nums[1]
            i3 = nums[2]
        else:    #1>2
            i1 = nums[0]
            i2 = nums[2]
            i3 = nums[1]
print(i1,i2,i3)        
            

 

练习5：

#依次接手用户输入的3个数，排序后打印
#方法2：max min 方法实现
nums = []
out = None
for i in range(3):
    nums.append(int(input(‘number:‘.format(i))))

while True:
    cur = min(nums)
    print(cur)
    nums.remove(cur)
    
    if len(nums) == 1:
        print(nums[0])
        break
        
# sort方法实现
nums = []
out = None
for i in range(3):
    nums.append(int(input(‘number:‘.format(i))))
nums.sort()
print(nums)

 

练习6：

# 冒泡法实现
num_list = [
    [1,9,8,5,6,7,4,3,2],
    [1,2,3,4,5,6,7,8,9],
    [1,2,3,4,5,6,7,9,8]
]

nums = num_list[2]
print(nums)
length = len(nums)
count_swap = 0
count = 0

for i in range(length):
    flag = False
    for j in range(length-i-1):
        count += 1
        if nums[j] > nums[j+1]:
            tmp = nums[j]
            nums[j] = nums[j+1]
            nums[j+1] = tmp
            flag = True
            count_swap += 1
    
    if not flag:
        break
print(nums,count_swap,count)

 

练习7：

# 用户输入一个数字
‘‘‘
1、判断是几位数，去掉前面空格和前面的0
2、打印每一位数字及其重复的次数
3、依次打印每一位数字，顺序个十百千万
‘‘‘
num = ‘‘


while True:
    num = input(‘Input a positive number >>>‘).strip().lstrip(‘0‘)   # strip删除前面空格换行0等
    
    if num.isdigit():   #判断是否全部数字
        break
    
print("The length of  is .".format(num, len(num)))  

#倒序打印
for i in range(len(num),0,-1):
    print(num[i-1],end=‘‘)
print()

#判断0-9的数字在字符串中出现的次数，每一次迭代都是用count，都是O（n）问题

counter = [0]*10
for i in range(10):
    counter[i] = num.count(str(i))  # 统计num字符串中数字i的个数
    if counter[i]:
        print("The count of  is ".format(i, counter[i]))
    
# 迭代字符串本身的字符
counter = [0]*10
for x in num:
    i = int(x)
    counter[i] += 1
    
for i in range(len(counter)):
    if counter[i]:
        print("The count of  is ".format(i, counter[i]))
        

 

练习8：

#输入5个数字，打印每个数字的位数，将这些数字排序打印，要求升序打印

nums = []
while len(nums) < 5:
    num = input(‘Input a positive number >>>‘).strip().lstrip(‘0‘)
    if not num.isdigit():
        continue
    print(‘The length of  is ‘.format(num, len(num)))
    
    nums.append(int(num))
print(nums)


# 冒泡法排序
for i in range(len(nums)):
    flag = False
    for j in range(len(nums)-i-1):
        if nums[j] > nums[j+1]:
            tmp = nums[j]
            nums[j] = nums[j+1]
            nums[j+1] = tmp
            flag = True
    if not flag:
        break
print(nums)

 

练习9：

# 随机产生10个数字
‘‘‘
要求每个数字取值范围【1,20】
统计重复的数字有几个，分别是什么
统计不重复的数字有几个，分别是什么

‘‘‘
import random

nums = []
for _ in range(10):
    nums.append(random.randrange(21))
print("Origin numbers = ".format(nums))
print()

length = len(nums)

samenums = []   # 记录相同的数字
diffnums = []   # 记录不同的数字
states = [0]*length  # 记录不同的索引异同状态
print(states)

for i in range(length):
    flag = False  # 假定没有重复
    if states[i] == 1:
        continue
    for j in range(i+1,length):
        if states[j] == 1:
            continue
        if nums[i] == nums[j]:
            flag = True
            states[j] = 1
    if flag:
        samenums.append(nums[i])
        states[i] = 1
    else:
        diffnums.append(nums[i])
print("Same numbers = 1, Counter=0".format(len(samenums), samenums))
print("Diff numbers = 1, Counter=0".format(len(diffnums), diffnums))