I was just wondering if I could remove some redundancy for a help function in a text-game. What I have right now is at the start of every function hint = 0, and hint is increased one every time an invalid answer is entered.
Here is what I have at the moment (inside every function):
hint = 0
valid = False
while valid == False:
print "Would you like to begin?"
begin = raw_input("> ")
if "yes" in begin:
valid = True
print "Great!\n"
start.start()
elif "no" in begin:
quit.quit()
else:
error.error(1)
hint += 1
if hint > 4:
print "\nYou may choose from \"yes\" and \"no\"."
The following code use decorator to separate hint logic and command logic. Every command is handled by a function which can be decorated by Hint object.
When the function return False, the count in Hint object will increase and when it's larger than the limit, it will print out a hint message.
When the function return a tuple it will be called by the main loop.
class Hint(object):
def __init__(self, n, msg):
self.n = n
self.msg = msg
def __call__(self, f):
def wrap(*args, **kw):
count = 1
while True:
ret = f(*args, **kw)
if ret == False:
count += 1
if count > self.n:
print self.msg
count = 0
else:
break
return ret
return wrap
def start_at(place):
print "start at %d" % place
return "start"
#Hint(3, "You may choose from 1, 2, 3.")
def start():
print "What place will you start?"
cmd = raw_input("> ")
try:
place = int(cmd)
if place not in (1,2,3):
return False
else:
return start_at, (place,)
except ValueError:
return False
def quit():
print "I will quit"
return "quit"
#Hint(4, "You may choose from yes and no.")
def begin():
print "Would you like to begin?"
cmd = raw_input("> ")
if "yes" in cmd:
print "Great!\n"
return start, ()
elif "no" in cmd:
print "Bad!\n"
return quit, ()
else:
return False
call_func, args = begin, ()
while True:
ret = call_func(*args)
if isinstance(ret, tuple):
call_func, args = ret
else:
break
Here is some test:
Would you like to begin?
> abc
Would you like to begin?
> def
Would you like to begin?
> 123
Would you like to begin?
> 345
You may choose from yes and no.
Would you like to begin?
> yes
Great!
What place will you start?
> 5
What place will you start?
> fg
What place will you start?
> sd
You may choose from 1, 2, 3.
What place will you start?
> 2
start at 2
from itertools import count
for hint in count()
print "Would you like to begin?"
begin = raw_input("> ")
if "yes" in begin:
print "Great!\n"
start.start()
break
elif "no" in begin:
quit.quit()
else:
error.error(1)
if hint > 4:
print "\nYou may choose from \"yes\" and \"no\"."
Now the next problem is what does start.start() do? It looks/feels like you might be using function calls like a GOTO
Related
I want to optimize my code to return well in a part of a code.
I know that I can do that with do while but I think it's maybe heavy with a lot of do while.
I identify where I want to return in comments
Can you suggest me a solution to solve my problem?
#return here 0
mainMenu()
choix = input()
if choix == 1:
print "Enter your username"
username = raw_input()
print "Enter your password"
password = raw_input()
createAccount.createAccount(username,password)
#return here 1
educatorMenu()
choix = input()
if choix == 1:
print "Enter his age"
ageChild = raw_input()
print "Enter his photo"
photoChild = raw_input()
educator.createChildAccount(ageChild, photoChild)
elif choix == 2:
childrenList = educator.seeChildrenList()
#Return to where it is written "return here 0"
elif choix == 3:
childrenList = educator.seeChildrenList()
print "Associate children?"
choixEnfant = input()
educator.associateChildren(childrenList,choixEnfant-1, educator.getIdEducator(username))
#Return to where it is written "return here 1"
Thanks a lot.
Note: I will not re-write/optimize your code, because I feel it should be a task for you to do so. Furthermore, there are quite some functions and references missing to rewrite it; you did not deliver an MWE.
Consequently, you may find this useful as a starting point and try to adapt this (very basic structure) to suit your needs. However, be warned: This also is far from being efficient, but it will help you to extend your knowledge of the basics.
def menu_main():
q = ['What would you like to do?',
'1: List educators',
'2: Create educator',
'3: Manage educator\'s children',
'4: Exit'].join('\n ')
return int(input(q).strip()[0])
def menu_educator_create():
usr = input('Enter username:').strip()
pwd = input('Enter password:').strip()
return (usr, pwd)
def menu_educator_manage():
q = ['What would you like to do?',
'1: List children',
'2: Add child',
'3: Return'].join('\n ')
return int(input(q).strip()[0])
if __name__ == '__main__':
edus = {}
exit = False
while not exit:
print('===')
cmd = menu_main()
if cmd == 1:
# List educators
pass
elif cmd == 2:
# Create educator
usr, pwd = menu_educator_create()
pass
elif cmd == 3:
# Manage children
# (should somehow obtain current educator first)
exit_edu = False
while not exit_edu:
subcmd = menu_educator_manage()
if subcmd == 1:
# List children
pass
elif subcmd == 2:
# Add child
pass
elif subcmd == 3:
# Return
exit_edu = True
else:
# None of the above
print('Unknown choice from educator menu.')
elif cmd == 4:
# Exit
exit = True
else:
# None of the above
print('Unknown choice from main menu.')
just wondering why the error says NameError name question1 is not defined
count = 0
import random
my_list = [question1, question2, question3, question4, question5, question6, question7, question8,question9, question10, question11, question12, question13, question14, question15, question16, question17, question18, question19, question20, question21, question22, question23, question24, question25, question26, question27, question28, question29, question30, question31, question32, question33, question34, question35, question36, question37, question38, question39, question40]
def intro(start):
if start == "YES" or start == "Y":
print("Lets begin.")
else:
print("Thanks for checking it out! Bye Bye!")
def question1():
global count
text_file = open("q1.txt")
print (text_file.read())
text_file.close
item = ""
ans1 = "B"
while item != "A" and item != "B" and item != "C" and item != "D":
item = input()
if item == (ans1):
print ("Correct")
count += 1
else:
print("Incorrect")
return count
def main():
print("Hello There! Welcome to an all new Trivial Pursuit!")
print("You have twenty multiple choice questions and twenty true and false.")
print ("Each question is one point. Your score will be presented at the end. ")
print("Would you like to begin? Press Y or YES. Please answer all questions in CAPS.")
start = input()
intro(start)
random.choice(my_list)()
question1()
question2()
question3()
question4()
question5()
question6()
question7()
question8()
question9()
question10()
question11()
question12()
question13()
question14()
question15()
question16()
question17()
question18()
question19()
question20()
question21()
question22()
question23()
question24()
question25()
question26()
question27()
question28()
question29()
question30()
question31()
question32()
question33()
question34()
question35()
question36()
question37()
question38()
question39()
question40()
print("You got", count, "right out of 40!")
main()
You can make count global:
count = 0
def intro(start):
if start == "yes" or start == "y":
print("Lets begin.")
else:
print("Thanks for checking it out! Bye Bye!")
def question1():
global count
count += 1
return count
def question2():
global count
count += 1
return count
def main():
print("Hello There! Welcome to an all new Trivial Pursuit!")
print("You have twenty multiple choice questions and twenty true and false.")
print ("Each question is one point. Your score will be presented at the end. ")
print("Would you like to begin? Press y or yes")
start = input()
intro(start)
question1()
question2()
print("You got", count, "right!")
main()
I edited your code to have the functions just add to count. I know globals aren't recommended, but this may work for you.
Thanks to this site I was able to get this far, being the python novice I am, however I'm kind of stuck. I'm trying to loop 'selection', so after a user does some math, rather than just ending it will give them the option to do something else until they select 0 to quit. I tried a bunch of other try and conditional statements but just end up getting answers and such stuck in an infinite loop.
Also, I'm pretty here, but any help is appreciated, also I'm a python nub. I'm writing this with python 2.7, if that matters.
def sum ( arg1, arg2):
total = a + b
return total;
def subtract ( arg1 , arg2):
total = a - b
return total;
def mult ( arg1, arg2):
total = a * b
return total;
def division ( arg1, arg2):
total = (a / b)
return total;
options = ["1", "2", "3", "4", "5", "0"]
print ("Please choose an option for mathing")
print ("1 for addition")
print ("2 for division")
print ("3 for subtraction")
print ("4 for multiplication")
print ("5 ")
print ("0 to exit")
#this will keep prompting the user to provide an input that is listed in 'options'
while True:
selection = input("Please select choose an option to continue")
if selection in options:
break
else:
print("Please choose a valid option")
#user input for mathing
#input will be validated as follows
a = None
while a is None:
try:
a = int(input("please provide a number for A"))
except ValueError:
print "please use a valid integer"
pass
b = None
while b is None:
try:
b = int(input("please provide a number for B"))
except ValueError:
print "please use a valid integer"
pass
#performing the operations
if selection == '1':
print "The sum is", str(sum(a, b))
elif selection == '2':
print "The quotient is", str(division(a, b))
elif selection == '3':
print "The difference is", str(subtract(a, b))
elif selection == '4':
print "The product is", str(mult(a, b))
elif selection == '0':
exit()
heres a few things I would do to make this more efficient..
options should be a dictionary... your in is a lot more efficient on a dictionary than on a list. the beauty of this is the value for each key can be function methods.
ex. options = {1: 'sum', 2: 'subtract' ..... }
then make a class with your math operations in it
class Calculator(object):
def sum(self, x, y):
return x + y
def subtract(self, x, y):
return x - y
#add more operations here
#staticmethod
def start():
while True:
#prompt for input and the operator
whats nice about this is in your checks for the selection you can dynamically call the class method to clean the code up a lot
if selection in options:
getattr(options[selection], Calculator)(a, b)
if you want me to explain more I can finish the example.
for your loop, you can add a method that starts the action and continues looping and doing more operations each time
here is a basic class you can use using those methods I described
class Calculator(object):
loop = None
calculations = 1
current_value = 0
selection = 0
options = {1: 'add', 2: 'subtract', 3: 'multiply', 4: 'divide'}
def __init__(self, loop=True):
self.loop = loop
print 'Welcome to my basic calculator!'
if not self.loop: # dont loop just execute once
self.run()
else:
while True:
self.run()
#staticmethod
def add(x, y):
return x + y
#staticmethod
def subtract(x, y):
return x - y
#staticmethod
def multiply(x, y):
return x * y
#staticmethod
def divide(x, y):
if y != 0: #cant divide by 0
return x / y
#staticmethod
def quit():
exit(0)
def run(self):
if self.calculations == 1:
self.current_value = self.prompt_user_input('please provide a number: ')
self.prompt_operator('Please choose an operator to continue\n1 for addition\n2 for subtraction\n3 for multiplication \n4 for division\n0 to quit\n')
y = self.prompt_user_input('please provide a number: ')
self.current_value = getattr(Calculator, self.options[self.selection])(self.current_value,y)
self.calculations += 1
print 'New value is: ' + str(self.current_value)
def prompt_operator(self, prompt_message):
while True:
self.selection = input(prompt_message)
if self.selection in self.options:
break
elif self.selection == 0:
self.quit()
else:
print("Please choose a valid option")
def prompt_user_input(self, prompt_message):
val = None
while val is None:
try:
val = int(input(prompt_message))
except ValueError:
print "please use a valid integer"
pass
return val
finally to start your calculator off you can just call it and either pass true to continue loops or pass false to only do one calculation
Calculator(loop=True)
Just put a loop around the prompting and calculation. If they enter 0, break from the outer-most loop:
while True:
#this will keep prompting the user to provide an input that is listed in 'options'
while True:
selection = input("Please select choose an option to continue")
if selection in options:
break
else:
print("Please choose a valid option")
if selection == '0':
break
...
#this will keep prompting the user to provide an input that is listed in 'options'
while True:
selection = input("Please select choose an option to continue")
if selection in options:
if selection == 1:
sum()
if selection == 2:
subtract()
.....
if selection == 'q'
break
Change the logic to what I did above, for option,take some action and do break on the quit character/
Firstly, your code should not work as it is now:
if selection == '1': #should return false
That's because using input you are taking a numeric quantity and then comparing it to a string.
change the input of selection to raw_input
Or change the conditional statements like below:
if selection == 1:
Then add a while True over the entire execution block to go through the execution again and again, until 0 is selected.
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."
I have been trying to program a maths quiz that both works and is as efficient as possible. Looking over my code I saw I had a lot of integer inputs and that lead to me having the program to ask the question/exit the system if the criteria isn't met, so to help me I thought that it would be useful to create a new function. Here is my attempt:
def prompt_int(prompt=''):
while True:
if status == prompt_int(prompt=''):
val = input(prompt)
if val in (1,2):
return int(val)
return true
elif status != prompt_int(prompt=''):
val = input(prompt)
if val in (1,2,3):
return int(val)
return true
else:
print("Not a valid number, please try again")
However, when I try to implement this function around my code it doesn't work properly as it says that status isn't defined however, when I do define status it goes into a recursion loop. How can I fix this problem?
Here is my original code before i try to implement this function:
import sys
import random
def get_bool_input(prompt=''):
while True:
val = input(prompt).lower()
if val == 'yes':
return True
elif val == 'no':
return False
else:
sys.exit("Not a valid input (yes/no is expected) please try again")
status = input("Are you a teacher or student? Press 1 if you are a student or 2 if you are a teacher")# Im tring to apply the new function here and other places that require integer inputs
if status == "1":
score=0
name=input("What is your name?")
print ("Alright",name,"welcome to your maths quiz."
"Remember to round all answer to 5 decimal places.")
level_of_difficulty = int(input(("What level of difficulty are you working at?\n"
"Press 1 for low, 2 for intermediate "
"or 3 for high\n")))
if level_of_difficulty not in (1,2,3):
sys.exit("That is not a valid level of difficulty, please try again")
if level_of_difficulty == 3:
ops = ['+', '-', '*', '/']
else:
ops = ['+', '-', '*']
for question_num in range(1, 11):
if level_of_difficulty == 1:
number_1 = random.randrange(1, 10)
number_2 = random.randrange(1, 10)
else:
number_1 = random.randrange(1, 20)
number_2 = random.randrange(1, 20)
operation = random.choice(ops)
maths = round(eval(str(number_1) + operation + str(number_2)),5)
print('\nQuestion number: {}'.format(question_num))
print ("The question is",number_1,operation,number_2)
answer = float(input("What is your answer: "))
if answer == maths:
print("Correct")
score = score + 1
else:
print ("Incorrect. The actual answer is",maths)
if score >5:
print("Well done you scored",score,"out of 10")
else:
print("Unfortunately you only scored",score,"out of 10. Better luck next time")
class_number = input("Before your score is saved ,are you in class 1, 2 or 3? Press the matching number")
while class_number not in ("1","2","3"):
print("That is not a valid class, unfortunately your score cannot be saved, please try again")
class_number = input("Before your score is saved ,are you in class 1, 2 or 3? Press the matching number")
else:
filename = (class_number + "txt")
with open(filename, 'a') as f:
f.write("\n" + str(name) + " scored " + str(score) + " on difficulty level " + str(level_of_difficulty))
with open(filename, 'a') as f:
f = open(filename, "r")
lines = [line for line in f if line.strip()]
f.close()
lines.sort()
if get_bool_input("Do you wish to view previous results for your class"):
for line in lines:
print (line)
else:
sys.exit("Thanks for taking part in the quiz, your teacher should discuss your score with you later")
if status == "2":
class_number = input("Which classes scores would you like to see? Press 1 for class 1, 2 for class 2 or 3 for class 3")
if class_number not in (1,2,3):
sys.exit("That is not a valid class")
filename = (class_number + "txt")
with open(filename, 'a') as f:
f = open(filename, "r")
lines = [line for line in f if line.strip()]
f.close()
lines.sort()
for line in lines:
print (line)
Well, just a part:
def prompt_int(prompt=""):
while True:
val = input(prompt)
if val in ("1", "2"):
return int(val), True
Will ask again and again. And return when the user enter "1" or "2"!
But better: "if val in "12":
def prompt_int(prompt=""):
while True:
val = input(prompt)
if val.isdigit():
return int(val)
Hi if you dont want to have valid values send to your you could change your code as the function above.
But you could also change it to do the system exits:
def prompt_int(prompt="", authorized=()):
while True:
val = raw_input(prompt)
if val.isdigit():
if int(val) in authorized:
return int(val)
else:
sys.exit("Bla bla bla too bad")
def prompt_int(prompt=''):
while True:
if status == prompt_int(prompt=''):
This line will look for the name "status" in the global namespace (module's namespace), and raise a NameError if there's no global variable named 'status'.
If there's one, it will then recursively calls prompt_int without any possible termination, resulting theoretically in an endless recursion, but practically (in CPython at least) in a RuntimeError when it will hit the maximum recursion depth.
There are also quite a few other things that won't work as you expect:
val = input(prompt)
if val in (1,2):
In Python 3.x, val will be a string, so it will never compare equal to an int. In Python 2.x, input() is a shortcut for eval(raw_input()), which might return an int, but is also a huge security flaw since it unconditionnally execute untrusted code.
return int(val)
return true
The second return statement will never be executed, obviously, since the function will exit at the first one.
A simpler implementation might look like this:
# rebinds raw_input to input for python < 3
import sys
if sys.version_info.major < 3:
input = raw_input
def prompt_int(prompt='', choices=None):
while True:
val = input(prompt)
try:
val = int(val)
if choices and val not in choices:
raise ValueError("{} is not in {}".format(val, choices))
return val
except (TypeError, ValueError) as e:
print(
"Not a valid number ({}), please try again".format(e)
)
While we're at it, there's room for improvement in other parts of your code. Let's start with this:
def get_bool_input(prompt=''):
while True:
val = input(prompt).lower()
if val == 'yes':
return True
elif val == 'no':
return False
else:
sys.exit("Not a valid input (yes/no is expected) please try again")
First point: your naming is not consistent. If your other function is named prompt_int, this one should be named prompt_bool. Also, you have one function (prompt_int) looping forever and the other one exiting the whole program on invalid input, which is another inconsistency. If you want to allow the user to exit on any prompt, provide an explicit option for it, ie:
def prompt_bool(prompt, quit='Q'):
prompt += " (hit '{}' to exit) : ".format(quit)
while True:
val = input(prompt).strip().upper()
if val == quit:
sys.exit("Goodbye")
elif val == 'yes':
return True
elif val == 'no':
return False
else:
print "Invalid input '{}', please try again".format(val)
Of course you then want to provide the same option in prompt_int(), which leads to a more generic function:
def get_input_or_quit(prompt, quit="Q"):
prompt += " (hit '{}' to exit) : ".format(quit)
val = input(prompt).strip()
if val.upper() == quit:
sys.exit("Goodbye")
return val
def prompt_bool(prompt):
while True:
val = get_input_or_quit(prompt).lower()
if val == 'yes':
return True
elif val == 'no':
return False
else:
print "Invalid input '{}', please try again".format(val)
And of course you also replace the call to input by a call to get_input_or_quit in prompt_int.
We could go on for long - splitting all your code in distinct, self-contained function, writing a "main()" function to drive them (instead of having the "main" part at the top level), and obviously using the operator module instead of eval().