Python 列表未附加到函数中

Question

由于变量的值在函数内部不会改变，我首先运行一个列表测试器。事实证明，它们确实附加了函数中提供的新值。这是我试过的


def test():
    testlist.append(6)


testlist=[3,4]

print(testlist)

test()

print(testlist)

输出：-

[3, 4]

[3, 4, 6]

但是当我在我的实际项目中尝试这个时，它没有用。这是代码的一部分：-

student_name=[]
roll_number=[]

def student_info()
            
            sn=input("Student Name : ")
            student_name.append(sn)

            rn=input("Roll no. : ")
            roll_number.append(rn)

有多少学生，student_info()就运行多少次，这也是一个变量。

每次，列表都会为每个学生附加一个值。所有这些值稍后将在需要时使用索引号提取。但是，当我尝试调用列表值时，它始终是出现的列表的第一个值。所以没有附加列表？


def print_card():
        for p in range(0,no_of_classes,1):
            for l in range(0,class_strength,1):
           
                print("Student Name:",student_name[l],x*50,"Roll No:",roll_number[l])

（请忽略不必要的复杂格式部分）


#This is the main part of the code.

for j in range (no_of_classes):
    
    class_=input("Class name")
    print("Class strength of ",class_)
    class_strength=int(input())
    
    for i in range (class_strength):
        student_info()
print_card()

这是最小可重现程序


student_name=[]
roll_number=[]

def student_info():
            
            sn=input("Student Name : ")
            student_name.append(sn)

            rn=input("Roll no. : ")
            roll_number.append(rn)
            
def print_card():
        for p in range(0,no_of_classes,1):
            for l in range(0,class_strength,1):
                x=' '
                print('\n')
                print("Student Name:",student_name[l],x*50,"Roll No:",roll_number[l])

no_of_classes=2
class_strength=1

for j in range (0,no_of_classes):
    
    for i in range (0,class_strength):
        student_info()
        
        
print_card()

卡片是为第一个 class 创建的，但对于下一个 class，输出相同。为什么会这样？

预期产出与实际产出

（假设 2 个学生）

                                      kode skool                                     
Student Name: 1                                                    Roll No: 1
Class: 1                                                           Section: 1
Address    1
           1


City:  1                                                        Pin Code: 1
Guardian's Phone Number 1


                                      kode skool                                        
Student Name: 2                                                    Roll No: 2
Class: 2                                                           Section: 2
Address    2
           2


City: 2                                                     Pin Code: 2
Guardian's Phone Number 2



-------------------------------------------------------------------



                                      kode skool  
Student Name: 1                                                    Roll No: 1
Class: 1                                                           Section: 1
Address    1
           1


City:  1                                                        Pin Code: 1
Guardian's Phone Number 1





                                      kode skool  
Student Name: 1                                                    Roll No: 1
Class: 1                                                           Section: 1
Address    1
           1


City:  1                                                        Pin Code: 1
Guardian's Phone Number 1

Answer 1

你所有的问题是你让所有学生都在一个单一的名单上

students = [name1, name2, name3, ...]

所以稍后它会导致第二个 class.

的正确数据出现问题

您使用索引 [0]、[1] 等来获取第二名 class 的学生，但它需要

[class_1_strength + 0], [class_1_strength + 1], etc.

要让第三名的学生 class 你需要

[class_1_strength + class_2_strength + 0], [class_1_strength + class_2_strength + 1], etc.

所以这是个大问题。

如果您将每个 class 保留为子列表

，将会简单得多

school = [
    [name1, name2, name3, ...], # class 1
    [name1, name2, name3, ...], # class 2
    # etc.
]

或作为字典：

school = {
    'class_1_name': [name1, name2, name3, ...], # class 1
    'class_2_name': [name1, name2, name3, ...], # class 2
    # etc.
}

坦率地说，最好将所有值放在一起：

school = [
    [ (name1, roll_number, ...),  (name2, roll_number, ...), ...], # class 1
    [ (name1, roll_number, ...),  (name2, roll_number, ...), ...], # class 2
    # etc.
]

最简单的工作示例。

我使用 random、string 只是为了创建一些测试数据 - 所以我不必使用 input().

我使用 random.seed(0) 来获取始终相同的数据 - 因此我可以将它们与之前的执行进行比较。

import random
import string

random.seed(0)

def generate_random_name(lenght=5):
    return "".join(random.choices(string.ascii_uppercase, k=lenght))

def get_student_info():
    
    #student_name = input("Student Name : ")
    #roll_number  = input("Roll no. : ")

    student_name = generate_random_name()
    roll_number  = 1

    return student_name, roll_number
    
def print_card():
    for class_number, class_students in enumerate(school, 1):
        print('--- class:', class_number, '---\n')
        for item in class_students:
            student_name, roll_number, class_number = item

            print(f"Student Name: {student_name:}")
            print(f"Roll No: {roll_number}")
            print(f"Class: {class_number}")
            print()
        
# --- main ---

school = []

no_of_classes = 2
class_strength = 3

for class_number in range(1, no_of_classes+1):

    class_students = [] # list for all students in one clas

    for student_number in range(1, class_strength+1):

        student_name, roll_number = get_student_info()

        # all information about one student        
        item = [student_name, student_number, class_number]

        class_students.append(item)

    school.append(class_students)
        
print_card()

结果：

--- class: 1 ---

Student Name: VTKGN
Roll No: 1
Class: 1

Student Name: KUHMP
Roll No: 2
Class: 1

Student Name: XNHTQ
Roll No: 3
Class: 1

--- class: 2 ---

Student Name: GXZVX
Roll No: 1
Class: 2

Student Name: ISXRM
Roll No: 2
Class: 2

Student Name: CLPXZ
Roll No: 3
Class: 2

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