Question

我的问题是当两个模块中有两个相同的命名函数时，计算机如何选择运行函数。其中一个模块已导入。

这是Pythonschool的一个例子。

我有一个名为crops.py的文件：

from wheat_class import *
from potato_class import *

#test program to select a crop and manage the crop

def display_menu():
    print()
    print("Which crop would you like to create?")
    print()
    print("1. Potato")
    print("2. Wheat")
    print()
    print("Please select an option from the above menu")

def select_option():
    valid_option = False
    while not valid_option:
        try:
            choice = int(input("Option selected: "))
            if choice in (1,2):
                valid_option = True
            else:
                print("Please enter a valid option")
        except ValueError:
            print("Please enter a valid option")
    return choice

def create_crop():
    display_menu()
    choice = select_option()
    if choice == 1:
        new_crop = Potato()
    elif choice == 2:
        new_crop = Wheat()
    return new_crop

def main():
    new_crop = create_crop()
    manage_crop(new_crop)

if __name__ == "__main__":
    main()

wheat_class和potato_class是名为Crop的类的子类。 Crop类在crop_class.py中定义：

class Crop:
    """A generic food crop"""

    #constructor = runs automatically when instantiating
    def __init__(self,growth_rate,light_need,water_need):
        #set the attributes
        #if underscore in front of name, private attributes

        self._growth = 0
        self._days_growing = 0
        self._growth_rate = growth_rate
        self._light_need = light_need
        self._water_need = water_need
        self._status = "Seed"
        self._type = "Generic"

    def needs(self):
        some code

    def report(self):
        some code

    def _update_status(self):
        #code for updating status of crop

    def grow(self,light,water):
        #code increasing growth value

def auto_grow(crop,days):
    some code

def manual_grow(crop):
    some code

def display_menu():
    print("1. Grow manually over 1 day")
    print("2. Grow automatically over 30 days")
    print("3. Report status")
    print("0. Exit test program")
    print()
    print("Please select an option from the above menu")

def get_menu_choice():
    option_valid = False
    while not option_valid:
        try:
            choice = int(input("Option Selected: "))
            if 0 <= choice <= 3:
                option_valid = Tsame furue
            else:
                print("Value entered not valid - please enter a value between 0 and 3")
        except ValueError:
            print("Value entered not valid - please enter a value between 0 and 3")
    return choice

def manage_crop(crop):
    print("This is the crop management program")
    print()
    noexit = True
    while noexit:
        display_menu()
        option = get_menu_choice()
        if option == 1:
            manual_grow(crop)
        elif option == 2:
            auto_grow(crop,30)
        elif option == 3:
            print(crop.report())
            print()
        elif option == 0:
            noexit = False
            print()

我的问题是关于函数display_menu（）。如图所示，该函数存在于crops.py和crop_class.py中。运行crops.py中的main函数时，将运行crops.py中的display_menu（） new_crop = create_crop() 同时运行crop_class.py中的display_menu（） manage_crop(new_crop)。

我很困惑，因为这两个函数都没有归因于特定的类。 crop_class.py中的display_menu（）以某种方式缩进，使其不属于Crop类。因此，我对计算机如何选择运行代码感到困惑。对此规则的破坏将非常有帮助。

Answer 1

使用import在Python中*模块时，您可以访问模块中的所有类和方法。

对于manage_crop和create_crop，crops.py文件可以在crop_class.py导入时（通过wheat_class间接访问方法）或potato_class）。

希望有所帮助！

有关详细信息，请查看import和modules文档。

具有相同函数名称Python

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