python學習(8)

來源:互聯網
上載者:User

標籤:範圍   cep   end   false   退出   put   code   pre   習題   

退出雙層迴圈:
方式1:try--except

try:    for i in range(5):        for j in range(5):            if i==3 and j ==3:                raiseexcept:    print(1)pass

方式2:函數實現的return

def fun():    for i in range(5):        for j in range(5):            print(i,j)            if i==3 and j ==3:                return Truefun()

方式3:多層break

for i in range(5):    for j in range(5):        for k in range(5):            if i == j == k == 3:                break            else:                print (i, ‘----‘, j, ‘----‘, k)        else: continue        break    else: continue    break

習題12:輸入3個數字,達到3個數字求和,結束程式

result = 0for i in range(3):    number = input("please input number: ")    result += int(number)print(result)習題13、 使用者輸入不同的資料,當輸入的資料達到3個數位時候,求和結束程式。(數字可以是整數)提示:判斷是否整數的方法,isdigit()遍曆所有的輸入資料,判斷是否在0-9的字串範圍內方式1: #coding=utf-8result = 0count = 0while True:    s = input("please input the number: ")    for v in s:        if v not in "0123456789":#如果不是數字跳出當前迴圈            break    else:        count+=1        result += int(s)    if count ==3:        breakprint(result)

方式2:先定義一個判斷數位函數

#encoding=utf-8def is_int(num):    for n in num:        if n not in "0123456789":            return Falsereturn Trueresult = 0number_count = 0while True:    s = input("please input the number: ")    if is_int(s):        result += int(s)        number_count += 1      if number_count == 3:        break print(result)

方式3:利用isdigit()函數

result1 = 0count1=0while True:    s = input("please input the number: ")    if s.isdigit():        count1+=1        result1 += int(s)    if count1 ==3:        breakprint(result1)

習題14:用嵌套列表的方式,遍曆輸出一個矩陣
方式1:

l = [     [1,2,3],     [4,5,6],     [7,8,9]]for i in l:    for j in i:        print(j,end=" ")    print()

方式2:

for i in range(len(l)):    for j in range(len(l[i])):        print(l[i][j],end = " ")    print()

習題15:嵌套列表的正、反對角線之和
正對角線之和

l = [     [1,2,3],      [4,5,6],      [7,8,9]      ]rusult = 0for i in range(len(l)):    for j in range(len(l[i])):        if i==j:            rusult += l[i][j]print(rusult)

反對角線之和

rusult = 0for i in range(len(l)):    for j in range(len(l[i])):        if (i+j)==2:            rusult += l[i][j]print(rusult)

習題16:求以下矩陣四邊元素之和
l = [
[1,2,3,4,5],
[1,2,3,4,5],
[1,2,3,4,5],
[1,2,3,4,5],
[1,2,3,4,5]
]

方法1:

1、第1行和第5行所有元素求和
2、其他行 只要第1列和第5列求和

rusult = 0for i in range(len(l)):    for j in range(len(l[i])):        if i == 0 or i == 4:            rusult += l[i][j]        else:            if j==0 or j==4:                rusult += l[i][j]print(rusult)

方法2:所有元素之和,減去中間矩陣之和

l = [     [1,2,3,4,5],      [1,2,3,4,5],      [1,2,3,4,5],     [1,2,3,4,5],     [1,2,3,4,5]  ]matrix_element_sum = 0sub_matrix_element_sum = 0for i in range(len(l)):    for j in range(len(l[i])):        matrix_element_sum += l[i][j]result_mid = 0for i in range(len(l)):    for j in range(len(l[i])):        if i == 0 or i ==4:            continue        else:            if j !=0 and j!=4:                sub_matrix_element_sum += l[i][j]print(matrix_element_sum - sub_matrix_element_sum)

python學習(8)

相關文章

聯繫我們

該頁面正文內容均來源於網絡整理,並不代表阿里雲官方的觀點,該頁面所提到的產品和服務也與阿里云無關,如果該頁面內容對您造成了困擾,歡迎寫郵件給我們,收到郵件我們將在5個工作日內處理。

如果您發現本社區中有涉嫌抄襲的內容,歡迎發送郵件至: info-contact@alibabacloud.com 進行舉報並提供相關證據,工作人員會在 5 個工作天內聯絡您,一經查實,本站將立刻刪除涉嫌侵權內容。

A Free Trial That Lets You Build Big!

Start building with 50+ products and up to 12 months usage for Elastic Compute Service

  • Sales Support

    1 on 1 presale consultation

  • After-Sales Support

    24/7 Technical Support 6 Free Tickets per Quarter Faster Response

  • Alibaba Cloud offers highly flexible support services tailored to meet your exact needs.