First off. My code:
UserInput = ("null") #Changes later
def ask_module(param, param2):
elif UserInput == (param):
print(param2)
while True:
UserInput = input()
UserInput = UserInput.lower()
print()
if UserInput == ("test"):
print("test indeed")
ask_module("test2", "test 2")
I am not that good at coding, so this is probably something that I have done really wrong
This post seems a bit duchy, since I almost just have code,
but I have absolutely no idea on how to make this work.
What the code looks like without shortening:
while True:
UserInput = input()
UserInput = UserInput.lower()
print()
if UserInput == ("inventory"):
print("You have %s bobby pin/s" %bobby_pin)
print("You have %s screwdriver/s" %screwdriver)
elif UserInput == ("look at sink"):
print("The sink is old, dirty and rusty. Its pipe has a bobby pin connected")
else:
print("Did not understand that")
EDIT: I see that it might be hard to see what I'm asking.
I'm wondering how I can shorten my original code
If all your elif blocks have the same pattern, you can take advantage of this.
You can create a dictionary for the text you want to print and then do away with the conditionals.When it comes to choosing which one to print, you simply fetch the relevant text using its corresponding key. You use the get(key, default) method. If there is no key in the dictionary, the default value will be returned. For example,
choices = {'kick': 'Oh my god, why did you do that?',
'light him on fire': 'Please stop.',
'chainsaw to the ribs': 'I will print the number %d',
}
user_input = input().lower()
# individually deal with any strings that require formatting
# and pass everything else straight to the print command
if user_input == 'chainsaw to the ribs':
print(choices[user_input] % 5)
else:
print(choices.get(user_input, 'Did not understand that.'))
I found a solution, just stop using elif entirely.
Example:
userInput = "null"
def ask_question(input, output):
if userInput == (input):
print(output)
else: pass
while True:
userInput = input()
ask_question("test","test")
ask_question("test2", "test2")
ask_question("test3", "test3")
Related
I'm a complete newbie to the coding trying my hands on python. While coding a simple program to calculate the age I ran into an error I cannot seem to fix no matter how hard I try. The code is pasted below. The program runs as far as I enter an year in the future, however when an year which has passed is used, the code loops again asking if the year is correct. Any help is appreciated.
from datetime import date
current_year = (date.today().year)
print('What is your age?')
myAge = input()
print('future_year?')
your_age_in_a_future_year= input()
future_age= int(myAge)+ (int(your_age_in_a_future_year)) - int(current_year)
def process ():
if future_age > 0:
print(future_age)
else:
print('do you really want to go back in time?, enter "yes" or "no"')
answer = input()
if answer =='yes':
print (future_age)
if answer == 'no':
print ('cya')
while answer != 'yes' or 'no':
print ('enter correct response')
process ()
process()
In this case, your function just needs to contain a while loop with an appropriate condition, and then there is no need to break out from it, or do a recursive call or anything.
def process():
if future_age > 0:
print(future_age)
else:
print('do you really want to go back in time?, enter "yes" or "no"')
answer = None
while answer not in ('yes', 'no'):
answer = input()
if answer == 'yes':
print(future_age)
elif answer == 'no':
print('cya')
try
...
if future_age > 0:
print(future_age)
return
else:
print('do you really want to go back in time?, enter "yes" or "no"')
...
i tried couple of things but does not work. the problem is in the 11th line(that's what python said). i am still 14 and do not have a big knowladge on programing. hope i can get some help
def crowbar():
print("nice, so you picked the crowbar, and now you have to get out of the car")
randomNum = random.randint(1,10)
if randomNum == [2,4,6,8,10]:
print(" shoo, those 2 people are walking away. now it's the time to get out of here")
elif randomNum == [1,3,5,7,9]:
print("they are coming at your direction. you have to do something and fast")
choise2 = int(input("""
1)hit them with the crowbar
2)pretent that you are still unconscious"""))
if choise2 == 1: #the problem has something to do with this line
print("ok")
elif choise2 == 2:
print("not ok")
I just refactored your code and fixed several things. The program should run without any problems.
I inline commented the issues and refactorings I did:
#Added the import statement to make sure code is runnable if copy&pasted
import random
def crowbar():
print("nice, so you picked the crowbar, and now you have to get out of the car")
randomNum = random.randint(1,10)
#Use the "x in list" expression as it is the most readable and pythonic way
if randomNum in [2,4,6,8,10]:
print("shoo, those 2 people are walking away. now it's the time to get out of here")
#Just use else for the second case, as this has the same behaviour without boiler plate
else:
print("they are coming at your direction. you have to do something and fast")
choice2 = 0
#Defensive programming: As long as neither 1 nor 2 was pressed, the input request will loop forever
while choice2 not in [1, 2]:
print("What do you want to do?")
choice2 = int(input("""
1)hit them with the crowbar
2)pretent that you are still unconscious\n"""))
#I had no error here to be honest as you said. It just worked.
if choice2 == 1:
print("ok")
elif choice2 == 2:
print("not ok")
if __name__ == "__main__":
crowbar()
Alright, a few things to note:
To check if a value is inside of a data structure, use is in instead of ==.
Variable names should be descriptive and follow the lower_case_with_underscores style.
Your random number generation can be simplified, since currently are you need is a 50/50 chance.
You should probably be using a solid IDE, which would alert you to the naming issues and the incorrect membership testing.
Here is your program, I refactored it quite a bit.
import random as rand
def crowbar():
print('You pick up the crowbar. Now you have to get out of the car.')
random_num = rand.random()
if random_num > 0.5:
print('Fortunately, those 2 people are walking away. You can get away safely.')
else:
print('They are moving towards you, you have to react quickly.\nYou decide to...\n(1) Hit them with the '
'crowbar.\n(2) Pretend that you are still unconscious.')
choice = input('Your choice (1 | 2): ')
if choice == '1':
print('Ok.')
elif choice == '2':
print('Not ok.')
else:
print(f'Error! Incorrect input: {choice}')
crowbar()
Explanations:
As mentioned above, I simplified the random number generation.
Variable names have been improved.
I fixed as much of the grammar, syntax and overall writing as I could.
The program now verifies the input. I also got rid of the unnecessary conversion to int.
Your indentation is incorrect and you are missing parenthesis in choice2 = ...
Also you need if randomNum in [2,4,6,8,10]: and if randomNum in [1,3,5,7,9]:
This works fine
def crowbar():
print("nice, so you picked the crowbar, and now you have to get out of the car")
randomNum = random.randint(1,10)
if randomNum in [2,4,6,8,10]: # list here
print(" shoo, those 2 people are walking away. now it's the time to get out of here")
elif randomNum in [1,3,5,7,9]: # list here
print("they are coming at your direction. you have to do something and fast")
choise2 = int(input("""
1)hit them with the crowbar
2)pretent that you are still unconcious""")) # parethesis here
if choise2 == 1: # corrected indentation here
print("ok") # corrected indentation here
elif choise2 == 2: # corrected indentation here
print("not ok") # corrected indentation here
This is the code that I made when I tried making an if-then statement but, it always defaults to else. Also i just started trying to code from online tutorials today.
print('yes or no?')
if sys.stdin.readline() == 'yes' :
print('Yay')
else :
print('Aww')
This is what happens:
Console:yes or no?
Me:yes
Console:Aww
I've been looking stuff up for half an hour and can't figure out how to fix this please help
sys.stdin.readline() reads a line which ends with '\n' (when you hit "enter").
So you need to remove this '\n' from the captured input using strip().
print('yes or no?')
if sys.stdin.readline().strip() == 'yes' :
print('Yay')
else :
print('Aww')
I tried to explain and solve your specific problem but you could of course use raw_input() or input() (PY3) as mentioned in the other answer.
In python, getting the input of a user's string can be done via input() (or in python 2.7, use raw_input()).
If you include the code:
user_input = raw_input("yes or no?")
This will first print the string "yes or no?", then wait for user input (through stdin), then save it as a string named user_input.
So, you change your code to be:
user_input = raw_input("yes or no?")
if user_input == "yes":
print("Yay")
else:
print("Aww")
this should have the desired effect.
Ok I used user_input and input instead of sys.stdin.readline() on strings and i used it in a basic calculator here it is :
import random
import sys
import os
user_input = input("Would you like to add or subtract?")
if user_input == 'add' :
print('What would you like to add?')
plus1 = float(sys.stdin.readline())
print('Ok so %.1f plus what?' % (float(plus1)))
plus2 = float(sys.stdin.readline())
print('%.1f plus %.1f equals %.1f' % (float(plus1),float(plus2), float(plus1+plus2)))
elif user_input == 'subtract' :
print('What would you like to subtract?')
minus1 = float(sys.stdin.readline())
print('Ok so %.1f minus what?' % (float(minus1)))
minus2 = float(sys.stdin.readline())
print('%.1f minus %.1f equals %.1f' % (float(minus1), float(minus2), float(minus1 - minus2)))
else :
print('That is not one of the above options')
Thanks alot guys!
Okay so how could I make my code so instead of putting 1 to 4 I could just put in a keyword like "Freezing" or "Froze" How could i put a keyword searcher in my code, all help is immensely appreciated thank you in advanced :)
def menu():
print("Welcome to Kierans Phone Troubleshooting program")
print("Please Enter your name")
name=input()
print("Thanks for using Kierans Phone Troubleshooting program "+name +"\n")
def start():
select = " "
print("Would you like to start this program? Please enter either y for yes or n for no")
select=input()
if select=="y":
troubleshooter()
elif select=="n":
quit
else:
print("Invalid please enter again")
def troubleshooter():
print("""Please choose the problem you are having with your phone (input 1-4):
1) My phone doesn't turn on
2) My phone is freezing
3) The screen is cracked
4) I dropped my phone in water\n""")
problemselect = int(input())
if problemselect ==1:
not_on()
elif problemselect ==2:
freezing()
elif problemselect ==3:
cracked()
elif problemselect ==4:
water()
start()
def not_on():
print("Have you plugged in the charger?")
answer = input()
if answer =="y":
print("Charge it with a diffrent charger in a diffrent phone socket. Does it work?")
else:
print("Plug it in and leave it for 20 mins, has it come on?")
answer = input()
if answer=="y":
print("Are there any more problems?")
else:
print("Restart the troubleshooter or take phone to a specialist\n")
answer=input()
if answer =="y":
print("Restart this program")
else:
print("Thank you for using my troubleshooting program!\n")
def freezing():
print("Charge it with a diffrent charger in a diffrent phone socket")
answer = input("Are there any more problems?")
if answer=="y":
print("Restart the troubleshooter or take phone to a specialist\n")
else:
print("Restart this program\n")
def cracked():
answer =input("Is your device responsive to touch?")
if answer=="y":
answer2 = input("Are there any more problems?")
else:
print("Take your phone to get the screen replaced")
if answer2=="y":
print("Restart the program or take phone to a specialist\n")
else:
print("Thank you for using my troubleshooting program!\n")
def water():
print("Do not charge it and take it to the nearest specialist\n")
menu()
while True:
start()
troubleshooter()
Since you want the user to type in the description of the phone's problem, you probably want a list of problems and keywords associated with those problems. The following program shows how you might arrange such a database and how to search it based on the user's input. There are faster alternatives to this solution (such as indexing the database), but this basic implementation should be sufficient for the time being. You will need to provide further code and information on how to resolve problems, but answers to your previous question should help direct you in that regard.
#! /usr/bin/env python3
# The following is a database of problem and keywords for those problems.
# Its format should be extended to take into account possible solutions.
PROBLEMS = (('My phone does not turn on.',
{'power', 'turn', 'on', 'off'}),
('My phone is freezing.',
{'freeze', 'freezing'}),
('The screen is cracked.',
{'cracked', 'crack', 'broke', 'broken', 'screen'}),
('I dropped my phone in water.',
{'water', 'drop', 'dropped'}))
# These are possible answers accepted for yes/no style questions.
POSITIVE = tuple(map(str.casefold, ('yes', 'true', '1')))
NEGATIVE = tuple(map(str.casefold, ('no', 'false', '0')))
def main():
"""Find out what problem is being experienced and provide a solution."""
description = input('Please describe the problem with your phone: ')
words = {''.join(filter(str.isalpha, word))
for word in description.lower().split()}
for problem, keywords in PROBLEMS:
if words & keywords:
print('This may be what you are experiencing:')
print(problem)
if get_response('Does this match your problem? '):
print('The solution to your problem is ...')
# Provide code that shows how to fix the problem.
break
else:
print('Sorry, but I cannot help you.')
def get_response(query):
"""Ask the user yes/no style questions and return the results."""
while True:
answer = input(query).casefold()
if answer:
if any(option.startswith(answer) for option in POSITIVE):
return True
if any(option.startswith(answer) for option in NEGATIVE):
return False
print('Please provide a positive or negative answer.')
if __name__ == '__main__':
main()
You are looking for an enum
from enum import Enum
class Status(Enum):
NOT_ON = 1
FREEZING = 2
CRACKED = 3
WATER = 4
Then in your if checks, you do something like this:
if problemselect == Status.NOT_ON:
...
elif problemselect == Status.FREEZING:
...
Hi I have this project of Mad libs but I don´t know how to make a function that ask the user what level of difficulty between easy, medium or hard they want, and depending on their answer redirect them to the desired level of mad libs. This is what I have so far. Thanks.
parts_of_speech_words = ["VERB","PLACE","ADJECTIVE","NOUN","PLURALNOUN","ADVERB"]
level_easy = """I __VERB__ to go to the __PLACE__
but I don't go in the __ADJECTIVE__
I am __ADJECTIVE__ of the __NOUN__
and getting __VERB__ by a __NOUN__."""
level_medium = """Begging to hear __NOUN__
My ears remain __ADJECTIVE__
__NOUN__ for signs
my __NOUN__ remain __VERB__
Yet __NOUN__still VERB."""
level_hard = """I __VERB__ you without __NOUN__
how, or when, or from where,
I love you __ADVERB__,
without __PLURALNOUN__ or pride;
So I love you because I know
no other way that this:
Where I does not VERB, nor you,
so close that your NOUN
on my chest is my hand,
so close that your NOUN close
as I fall ADJECTIVE."""
greeting = raw_input ("Welcome to mad libs, What's your name? ")
prompt = raw_input ("Select your level: easy, medium or hard: ")
def entry_level_prompt (variable_level):
easy = level_easy
medium = level_medium
hard = level_hard
for e in variable_level:
if prompt == easy:
return level_easy
print level_easy
if prompt == medium:
return level_medium
print level_medium
if prompt == hard:
return level_hard
print lever_hard
print "Ok %s you chose the %s level" % (greeting , prompt)
print entry_level_prompt (variable_level)
def parts_of_speech (words_string, list_of_part_of_speech):
for pos in list_of_part_of_speech:
if pos in words_string:
return pos
return None
def play_mad_libs (split_string, list_of_part_of_speech):
replaced = []
split_string = split_string.split ()
for words_string in split_string:
replacement = parts_of_speech (words_string, list_of_part_of_speech)
if replacement != None:
user_input = raw_input ("Type in a: " + replacement + " ")
words_string = words_string.replace (replacement, user_input)
replaced.append(words_string)
else:
replaced.append(words_string)
replaced = " ".join(replaced)
return replaced
print play_mad_libs (entry_level_prompt, parts_of_speech_words)
You have a bug in your code. You are calling entry_level_prompt(variable_level) but variable_level does not exist outside of the method scope.
To control the difficulty, you can create a method called get_difficulty()
def get_difficulty():
choice = ""
while choice not in ('easy', 'medium', 'hard'):
choice = raw_input("choose your difficulty: ")
if choice == "easy":
return level_easy
elif choice == "medium":
return level_medium
elif choice == "hard":
return level_hard
else:
print("Invalid choice...")
You've confused the selector -- "easy", "medium", or "hard" -- with the variable you want to return -- level_easy, level_medium, or level_hard. You cannot use the variable prompt for both purposes at the same time.
I recommend that you keep variable prompt as you started: it holds the user input. Then, simply test it and return the needed script:
if prompt == "easy":
return level_easy
elif prompt == "medium"
return level_medium
elif prompt == "hard"
return level_hard
else:
"Please enter easy, medium, or hard for the level."