在def(pentagon):块中我命名了一个变量" first".但是,这会导致"无效语法"错误.怎么了?我已经尝试过命名其他东西,从单个字母到低级/大写字母组合,如"preArea".
def display():
print('This program will tell you the area some shapes')
print('You can choose between...')
print('1. rectangle 2. triangle')
print('3. circle 4. pentagon')
print('5. rhombus 6. trapezoid')
def shape():
shape = int(input('What shape do you choose?'))
if shape == 1: rectangle()
elif shape == 2: triangle()
elif shape == 3: circle()
elif shape == 4: pentagon()
elif shape == 5: rhombus()
elif shape == 6: trapezoid()
else:
print('ERROR: select 1 2 3 4 5 or …Run Code Online (Sandbox Code Playgroud) 所以我在Python中编写了一个简短而简单的聊天机器人,但这是一个令人恼火的问题.程序posResponses() 最初只会调用该函数.
在上下文中,如果我用"悲伤","可怕"甚至"asdfasdfasdf"回答其最初的问题,我仍会得到积极的回应.
应该发生的是如果我输入一个否定/不明确的关键字,应该调用negResponses()/ ambiguousResponses()function.事实并非如此.我做错了什么,我该如何解决?代码如下:
import random
import time
def opening():
print('Hello!')
def responseType():
responseType = str(input('How are you ?'))
if responseType == 'good' or 'great' or 'fantastic' or 'decent' or 'fine' or 'ok' or 'okay': posResponses()
elif responseType == 'bad' or 'terrible' or 'sad' or 'grumpy' or 'angry' or 'irritated' or 'tired': negResponses()
else: ambiguousResponses()
def posResponses():
number = random.randint(1, 4)
if number == 1:
print('That\'s great! So what\'s up?')
input()
ambiguousResponses()
if number == …Run Code Online (Sandbox Code Playgroud) 如果你运行我在下面包含的代码(python 2.7),你会发现结果图像是黑暗和模糊的,几乎看起来有条纹在它上面运行.我意识到使用分散函数绘制这样的东西可能是滥用功能.
我仔细阅读了文档并迷失在其中,只是希望有人能告诉我如何让我的Julia套装看起来像你在书本和网上看到的漂亮的彩色印版一样漂亮.
import numpy as np
import matplotlib.pyplot as plt
# Plot ranges
r_min, r_max = -2.0, 2.0
c_min, c_max = -2.0, 2.0
# Even intervals for points to compute orbits of
r_range = np.arange(r_min, r_max, (r_max - r_min) / 200.0)
c_range = np.arange(c_min, c_max, (c_max - c_min) / 200.0)
c = complex(-0.624, 0.435)
xs = []
ys = []
colors = []
for comp in c_range:
for real in r_range:
z = complex(real, comp)
escaped = False
for …Run Code Online (Sandbox Code Playgroud) 我在相关文档中找到的所有内容都是++和concat.
我一开始认为做以下事情会给我想要的东西:
[1, 3, 4] ++ [4, 5, 6]
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但正如你所知,只给[1,2,3,4,5,6].
我需要做什么来接受[1,2,3]和[4,5,6]并且[[1,2,3],[4,5,6]]?
我试图制作一个简短的程序来解决着名的德雷克方程.我让它接受整数输入,十进制输入和小数输入.但是,当程序试图将它们相乘时,我收到此错误(在我输入所有必要值之后,错误发生):
Traceback (most recent call last)
File "C:/Users/Family/Desktop/Programming/Python Files/1/DrakeEquation1.py", line 24, in <module>
calc() #cal calc to execute it
File "C:/Users/Family/Desktop/Programming/Python Files/1/DrakeEquation1.py", line 17, in calc
calc = r*fp*ne*fl*fi*fc*l
TypeError: can't multiply sequence by non-int of type 'str'
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我的代码如下:
def intro():
print('This program will evaluate the Drake equation with your values')
def calc():
print('What is the average rate of star formation in the galaxy?')
r = input()
print('What fraction the stars have planets?')
fp = input()
ne = int(input('What is …Run Code Online (Sandbox Code Playgroud) 我对编码比较陌生,所以请帮帮我.代码只会运行到第5行.这段代码可能是一个完整的巴贝尔,但请幽默我.
编辑:没有例外,没有任何反应.要求我在1和2之间进行选择后,代码就会停止.
print('This program will tell you the area some shapes')
print('You can choose between...')
print('1. rectangle')
print('or')
print('2. triangle')
def shape():
shape = int(input('What shape do you choose?'))
if shape == 1: rectangle
elif shape == 2: triangle
else: print('ERROR: select either rectangle or triangle')
def rectangle():
l = int(input('What is the length?'))
w = int(input('What is the width?'))
areaR=l*w
print('The are is...')
print(areaR)
def triangle():
b = int(input('What is the base?'))
h = int(input('What is the height?'))
first=b*h …Run Code Online (Sandbox Code Playgroud) 我注意到当你在Kivy中使用网格布局制作按钮时,它们是在(0,0)处创建的,并且根据前一个按钮的长度和宽度移动多个空格.但是,我想在屏幕底部中间有一个3x4网格.
到目前为止我有这个:
import kivy
from kivy.uix.gridlayout import GridLayout
from kivy.app import App
from kivy.uix.button import Button
class CalcApp(App):
def build(self):
layout = GridLayout(cols=3, row_force_default=True, row_default_height=50)
layout.add_widget(Button(text='1', size_hint_x=None, width=100))
layout.add_widget(Button(text='2', size_hint_x=None, width=100))
layout.add_widget(Button(text='3', size_hint_x=None, width=100))
layout.add_widget(Button(text='4', size_hint_x=None, width=100))
layout.add_widget(Button(text='5', size_hint_x=None, width=100))
layout.add_widget(Button(text='2', size_hint_x=None, width=100))
layout.add_widget(Button(text='6', size_hint_x=None, width=100))
layout.add_widget(Button(text='7', size_hint_x=None, width=100))
layout.add_widget(Button(text='8', size_hint_x=None, width=100))
layout.add_widget(Button(text='9', size_hint_x=None, width=100))
layout.add_widget(Button(text='0', size_hint_x=None, width=100))
layout.add_widget(Button(text='Enter', size_hint_x=None, width=100))
return layout
CalcApp().run()
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那么,我该如何改变立场?
我从c ++入门书中得到了这段代码,这本书旨在解释删除操作符.但是,我不明白的是程序如何调用这两个函数以及它们如何交互.
// delete.cpp -- using the delete operator
#include <iostream>
#include <cstring> // or string.h
using namespace std;
char * getname(void); // function prototype
int main()
{
char * name; // create pointer but no storage
name = getname(); // assign address of string to name
cout << name << " at " << (int *) name << "\n";
delete [] name; // memory freed
name = getname(); // reuse freed memory
cout << name << " at " …Run Code Online (Sandbox Code Playgroud) 我知道您可以在Python GUI中打开文件,浏览器和URL.但是,我不知道如何将其应用于程序.例如,以下都不起作用.(以下是我不断增长的聊天机器人程序的片段):
def browser():
print('OPENING FIREFOX...')
handle = webbroswer.get() # webbrowser is imported at the top of the file
handle.open('http://youtube.com')
handle.open_new_tab('http://google.com')
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和
def file():
file = str(input('ENTER THE FILE\'S NAME AND EXTENSION:'))
action = open(file, 'r')
actionTwo = action.read()
print (actionTwo)
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就上述顺序而言,这些错误发生在个别运行中:
OPENING FIREFOX...
Traceback (most recent call last):
File "C:/Users/RCOMP/Desktop/Programming/Python Files/AI/COMPUTRON_01.py", line 202, in <module>
askForQuestions()
File "C:/Users/RCOMP/Desktop/Programming/Python Files/AI/COMPUTRON_01.py", line 64, in askForQuestions
browser()
File "C:/Users/RCOMP/Desktop/Programming/Python Files/AI/COMPUTRON_01.py", line 38, in browser
handle = webbroswer.get()
NameError: global name 'webbroswer' is …Run Code Online (Sandbox Code Playgroud)