If else 语句追加到 for 循环中的列表
If else statement append to list in for loop
我想将乌龟 'person' 添加到列表 'infected_people'，但出现错误 'TurtleGraphicsError: bad color string: red'。
如何将 'person' 添加到列表 'infected_people' 而不会出现错误。
我是编程新手，如果我的解释不清楚，请见谅。
infected_people = []
i = 0
while i < 30: #number of steps (time)
for person in people:
random_walk(30, 400)
i + 1
for infected_person in infected_people:
if person.distance(infected_person) < 30:
person.color("red")
infected_people.append(person)
这是我的代码的另一部分：
import turtle
import random
wn = turtle.Screen()
def person_characteristics():
for person in people:
person.penup()
person.shape("circle")
person.shapesize(0.2)
x = random.randint(-200, 200)
y = random.randint(-200, 200)
person.setpos(x, y)
person.speed(0)
turtle.update()
return person
def population(population_size):
turtle.update()
people = []
for _ in range(population_size):
people.append(turtle.Turtle())
return people
def person_characteristics():
for person in people:
turtle.update()
person.penup()
person.shape("circle")
person.shapesize(0.2)
person.speed(0)
x = random.randint(-200, 200)
y = random.randint(-200, 200)
person.setpos(x, y)
return person
def random_walk(step_size, area_size):
person.clear()
count = 0
while count < 1:
count += 1
if (-area_size < person.xcor() <area_size) and (-area_size < person.ycor() <area_size):
person.right(random.randint(0,360))
person.forward(step_size)
else:
person.right(180)
person.forward(step_size)
turtle.update()
people = population(50)
person = person_characteristics()
def infect_random(people):
infected = random.choice(people)
infected.color("red")
return infected
infected_people = []
initial_infected = infect_random(people)
infected_people.append(initial_infected)
print(infected_people)
counted_infections = 0
我的目标是每个 infected_person（= 红点）都会感染所有靠近他的人。现在只有最初的感染者会感染其他人。所以我想如果我将每个被感染的人都添加到 infected_people 的列表中，那么它就会起作用。但是，添加行 infected_people.append(person) 时出现错误。
i = 0
while i < 30: #number of steps (time)
for person in people:
random_walk(30, 400)
i += 1
for infected_person in infected_people:
if person.distance(infected_person) < 30:
person.color("red")
infected_people.append(person)
turtle.done()
wn.exitonclick()
这是我收到的错误：
runfile('C:/Users/Noa Hoogeweg/Documents/BMT/PvL/Virus/Noa_Virus_goed.py', wdir='C:/Users/Noa Hoogeweg/Documents/BMT/PvL/Virus')
[<turtle.Turtle object at 0x000001FF4D3D9908>]
Traceback (most recent call last):
File "C:\Users\Noa Hoogeweg\Documents\BMT\PvL\Virus\Noa_Virus_goed.py", line 78, in <module>
person.color("red")
File "C:\Users\Noa Hoogeweg\anaconda3\lib\turtle.py", line 2216, in color
pcolor = self._colorstr(pcolor)
File "C:\Users\Noa Hoogeweg\anaconda3\lib\turtle.py", line 2696, in _colorstr
return self.screen._colorstr(args)
File "C:\Users\Noa Hoogeweg\anaconda3\lib\turtle.py", line 1158, in _colorstr
raise TurtleGraphicsError("bad color string: %s" % str(color))
TurtleGraphicsError: bad color string: red
我不知道你遇到了什么错误，但是 i + 1 行不会增加 i，要增加 i 的值你需要使用 i = i + 1 或 i += 1 .
你对 person.color("red") 的调用看起来完全有效，我 运行 代码时没有收到错误。
根据错误的堆栈跟踪，乌龟只对颜色字符串做了两件事。首先，它测试它是否是 str —— 你的必须通过该测试才能得到你得到的错误。其次，它将颜色字符串传递给 tkinter 的 winfo_rgb() 方法，该方法 returns 一个 RGB 三元组。 Turtle 忽略三元组，它只是想看看该函数是成功还是抛出错误。如果它抛出错误，您会收到显示的消息：
TurtleGraphicsError: bad color string: red
所以球在 tkinter 的球场上，试试下面的小程序，看看 tkinter 是否在做正确的事情：
from tkinter import Button
# Button is an arbitrary widget choice
print(Button().winfo_rgb("red"))
如果有效，您应该会得到如下内容：
> python3 test.py
(65535, 0, 0)
>
在这种情况下，您的 turtle 代码与 tkinter 库连接的方式有些可疑。如果上面的操作失败，你应该得到类似的东西：
> python3 test.py
Traceback (most recent call last):
File "test.py", line 4, in <module>
print(Button().winfo_rgb("red"))
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/tkinter/__init__.py", line 1156, in winfo_rgb
self.tk.call('winfo', 'rgb', self._w, color))
_tkinter.TclError: unknown color name "red"
>
在这种情况下，您可能需要检查 tkinter 安装的有效性。
即使您解决了 "red" 问题，您的代码中还有其他错误会阻止它工作。下面是我的重写，以使其正确地达到 运行：
from turtle import Screen, Turtle
from random import randint, choice
def person_characteristics(people):
for person in people:
person.shape('circle')
person.shapesize(0.2)
person.speed('fastest')
person.penup()
x = randint(-200, 200)
y = randint(-200, 200)
person.setpos(x, y)
person.showturtle()
def population(population_size):
people = []
for _ in range(population_size):
people.append(Turtle(visible=False))
return people
def random_walk(person, step_size, area_size):
if -area_size < person.xcor() < area_size and -area_size < person.ycor() < area_size:
person.right(randint(0, 360))
person.forward(step_size)
else:
person.right(180)
person.forward(step_size)
def infect_random(people):
infected = choice(people)
infected.color('red')
return infected
screen = Screen()
people = population(50)
person_characteristics(people)
infected_people = []
initial_infected = infect_random(people)
infected_people.append(initial_infected)
counted_infections = 1
for _ in range(3000): # number of steps (time)
for person in people:
random_walk(person, 30, 400)
for infected_person in infected_people:
if person.pencolor() != 'red' and person.distance(infected_person) < 30:
person.color('red')
infected_people.append(person)
break
screen.exitonclick()
我想将乌龟 'person' 添加到列表 'infected_people'，但出现错误 'TurtleGraphicsError: bad color string: red'。
如何将 'person' 添加到列表 'infected_people' 而不会出现错误。
我是编程新手，如果我的解释不清楚，请见谅。
infected_people = []
i = 0
while i < 30: #number of steps (time)
for person in people:
random_walk(30, 400)
i + 1
for infected_person in infected_people:
if person.distance(infected_person) < 30:
person.color("red")
infected_people.append(person)
这是我的代码的另一部分：
import turtle
import random
wn = turtle.Screen()
def person_characteristics():
for person in people:
person.penup()
person.shape("circle")
person.shapesize(0.2)
x = random.randint(-200, 200)
y = random.randint(-200, 200)
person.setpos(x, y)
person.speed(0)
turtle.update()
return person
def population(population_size):
turtle.update()
people = []
for _ in range(population_size):
people.append(turtle.Turtle())
return people
def person_characteristics():
for person in people:
turtle.update()
person.penup()
person.shape("circle")
person.shapesize(0.2)
person.speed(0)
x = random.randint(-200, 200)
y = random.randint(-200, 200)
person.setpos(x, y)
return person
def random_walk(step_size, area_size):
person.clear()
count = 0
while count < 1:
count += 1
if (-area_size < person.xcor() <area_size) and (-area_size < person.ycor() <area_size):
person.right(random.randint(0,360))
person.forward(step_size)
else:
person.right(180)
person.forward(step_size)
turtle.update()
people = population(50)
person = person_characteristics()
def infect_random(people):
infected = random.choice(people)
infected.color("red")
return infected
infected_people = []
initial_infected = infect_random(people)
infected_people.append(initial_infected)
print(infected_people)
counted_infections = 0
我的目标是每个 infected_person（= 红点）都会感染所有靠近他的人。现在只有最初的感染者会感染其他人。所以我想如果我将每个被感染的人都添加到 infected_people 的列表中，那么它就会起作用。但是，添加行 infected_people.append(person) 时出现错误。
i = 0
while i < 30: #number of steps (time)
for person in people:
random_walk(30, 400)
i += 1
for infected_person in infected_people:
if person.distance(infected_person) < 30:
person.color("red")
infected_people.append(person)
turtle.done()
wn.exitonclick()
这是我收到的错误：
runfile('C:/Users/Noa Hoogeweg/Documents/BMT/PvL/Virus/Noa_Virus_goed.py', wdir='C:/Users/Noa Hoogeweg/Documents/BMT/PvL/Virus')
[<turtle.Turtle object at 0x000001FF4D3D9908>]
Traceback (most recent call last):
File "C:\Users\Noa Hoogeweg\Documents\BMT\PvL\Virus\Noa_Virus_goed.py", line 78, in <module>
person.color("red")
File "C:\Users\Noa Hoogeweg\anaconda3\lib\turtle.py", line 2216, in color
pcolor = self._colorstr(pcolor)
File "C:\Users\Noa Hoogeweg\anaconda3\lib\turtle.py", line 2696, in _colorstr
return self.screen._colorstr(args)
File "C:\Users\Noa Hoogeweg\anaconda3\lib\turtle.py", line 1158, in _colorstr
raise TurtleGraphicsError("bad color string: %s" % str(color))
TurtleGraphicsError: bad color string: red
我不知道你遇到了什么错误，但是 i + 1 行不会增加 i，要增加 i 的值你需要使用 i = i + 1 或 i += 1 .
你对 person.color("red") 的调用看起来完全有效，我 运行 代码时没有收到错误。
根据错误的堆栈跟踪，乌龟只对颜色字符串做了两件事。首先，它测试它是否是 str —— 你的必须通过该测试才能得到你得到的错误。其次，它将颜色字符串传递给 tkinter 的 winfo_rgb() 方法，该方法 returns 一个 RGB 三元组。 Turtle 忽略三元组，它只是想看看该函数是成功还是抛出错误。如果它抛出错误，您会收到显示的消息：
TurtleGraphicsError: bad color string: red
所以球在 tkinter 的球场上，试试下面的小程序，看看 tkinter 是否在做正确的事情：
from tkinter import Button
# Button is an arbitrary widget choice
print(Button().winfo_rgb("red"))
如果有效，您应该会得到如下内容：
> python3 test.py
(65535, 0, 0)
>
在这种情况下，您的 turtle 代码与 tkinter 库连接的方式有些可疑。如果上面的操作失败，你应该得到类似的东西：
> python3 test.py
Traceback (most recent call last):
File "test.py", line 4, in <module>
print(Button().winfo_rgb("red"))
File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/tkinter/__init__.py", line 1156, in winfo_rgb
self.tk.call('winfo', 'rgb', self._w, color))
_tkinter.TclError: unknown color name "red"
>
在这种情况下，您可能需要检查 tkinter 安装的有效性。
即使您解决了 "red" 问题，您的代码中还有其他错误会阻止它工作。下面是我的重写，以使其正确地达到 运行：
from turtle import Screen, Turtle
from random import randint, choice
def person_characteristics(people):
for person in people:
person.shape('circle')
person.shapesize(0.2)
person.speed('fastest')
person.penup()
x = randint(-200, 200)
y = randint(-200, 200)
person.setpos(x, y)
person.showturtle()
def population(population_size):
people = []
for _ in range(population_size):
people.append(Turtle(visible=False))
return people
def random_walk(person, step_size, area_size):
if -area_size < person.xcor() < area_size and -area_size < person.ycor() < area_size:
person.right(randint(0, 360))
person.forward(step_size)
else:
person.right(180)
person.forward(step_size)
def infect_random(people):
infected = choice(people)
infected.color('red')
return infected
screen = Screen()
people = population(50)
person_characteristics(people)
infected_people = []
initial_infected = infect_random(people)
infected_people.append(initial_infected)
counted_infections = 1
for _ in range(3000): # number of steps (time)
for person in people:
random_walk(person, 30, 400)
for infected_person in infected_people:
if person.pencolor() != 'red' and person.distance(infected_person) < 30:
person.color('red')
infected_people.append(person)
break
screen.exitonclick()