多次运行一个函数，然后返回所有不同的结果？答案

【问题标题】：Run a function multiple times and then return all the different results?多次运行一个函数，然后返回所有不同的结果？
【发布时间】：2019-05-03 12:14:10
【问题描述】：

这是我的代码：

import random
def name_generator():
    for x in range (0, 11):
        color = ["Red", "Green", "Blue", "White", "Black", "Yellow", "Purple", "Orange", "Pink"]
        animal = ["Cat", "Dog", "Snake", "Mouse", "Tiger", "Leopard", "Moose", "Wolf", "Bear"]
        number = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"]

        randomColor = random.randrange(0, len(color))
        randomAnimal = random.randrange(0, len(animal))
        randomNumber = random.randrange(0, len(number))

        name = "Username: " + color[randomColor] + animal[randomAnimal] + number[randomNumber]
        return name

我正在尝试生成十个不同的名称，然后打印所有不同的名称。现在，当我运行该函数时，我只会返回一个名称，而不是十个不同的名称。

for x in range (0, 11)

据我所知，这应该执行十次？

这是我现在得到的结果：

name_generator()
'Username: PurpleLeopard13'

但我想得到：

name_generator()
'Username: PurpleLeopard13'
'Username: GreenDog10'
'Username: PurpleCat1'...(10 times)

谢谢！

【问题讨论】：

你调用你的函数name_generator...你真的想创建一个可以循环的生成器函数吗？如果是这样，您可能想要yield 而不是return - 或者作为答案意义并调用它10次......
附带说明 - 用于选择单个元素 - 而不是 random.randrange(0, len(whatever)) 并使用它来索引到 whatever... 你可以只使用 random_color = random.choice(color) 然后 random_color 将是列表中的一种颜色...更清晰、更简单。
我同意@JonClements。要么把它变成一个实际的生成器并产生结果，要么制作一个内部列表，将结果附加到循环中的列表中，然后返回最终列表。这取决于您的需求。

标签： python function loops names

【解决方案1】：

问题是目前您已将 return 语句放在 for 循环中。所以一旦你生成了名字，你就返回并离开这个函数。

您需要稍作修改：将函数调用移到 for 循环中。或者，请遵循以下 cmets 中 @Jon Clements 的建议。

import random
def name_generator():
        color = ["Red", "Green", "Blue", "White", "Black", "Yellow", "Purple", "Orange", "Pink"]
        animal = ["Cat", "Dog", "Snake", "Mouse", "Tiger", "Leopard", "Moose", "Wolf", "Bear"]
        number = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"]

        randomColor = random.randrange(0, len(color))
        randomAnimal = random.randrange(0, len(animal))
        randomNumber = random.randrange(0, len(number))

        name = "Username: " + color[randomColor] + animal[randomAnimal] + number[randomNumber]
        return name

for x in range (0, 11):    
    print (name_generator())

# Username: PinkMouse3
# Username: GreenCat8
# Username: GreenMoose2
# Username: BlackTiger14
# Username: BlackSnake11
# Username: PurpleMoose13
# Username: GreenMouse2
# Username: RedBear5
# Username: BlueBear2
# Username: WhiteDog2
# Username: WhiteTiger3

【讨论】：

那个或者他们想yield name 然后for name in name_generator(): print(name)... 或者print(*name_generator(), sep='\n') 或者其他什么......
@JonClements：这里的函数名称不明确。但我想我试图解释他/她的代码出了什么问题。我希望 OP 从这个答案中学到一些东西

【解决方案2】：

试试下面的代码。而不是在 for 循环中返回尝试使用打印。您的返回会阻止循环进一步交互。

import random
def name_generator():
    for x in range (0, 11):
        color = ["Red", "Green", "Blue", "White", "Black", "Yellow", "Purple", "Orange", "Pink"]
        animal = ["Cat", "Dog", "Snake", "Mouse", "Tiger", "Leopard", "Moose", "Wolf", "Bear"]
        number = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"]

        randomColor = random.randrange(0, len(color))
        randomAnimal = random.randrange(0, len(animal))
        randomNumber = random.randrange(0, len(number))

        name = "Username: " + color[randomColor] + animal[randomAnimal] + number[randomNumber]

        print(name)

name_generator()

【讨论】：

【解决方案3】：

作为使用生成器的附加 riff，使您的生成器能够提供任意个名称。用户可以决定要检索多少个名称。

from random import choice
from itertools import islice

def name_generator():
    colors = ["Red", "Green", "Blue", "White", "Black", "Yellow", "Purple", "Orange", "Pink"]
    animals = ["Cat", "Dog", "Snake", "Mouse", "Tiger", "Leopard", "Moose", "Wolf", "Bear"]
    numbers = ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15"]

    while True:
        yield "Username: {}{}{}".format(
           choice(colors),
           choice(animals),
           choice(numbers)
        )


names = list(islice(name_generator(), 10))

【讨论】：