def twothousand(amt):
n=500
div1=amt//n
mod1=amt%n
return (mod1,n,div1)
def fivehundred(amt):
n=200
div1=amt//n
mod1=amt%n
return (mod1,n,div1)
def calculate(amt):
if amt <10:
print("hi")
elif amt>=200 and amt<500:
mod1,n,div1=fivehundred(amt)
return (mod1,n,div1)
#the above return statement isn't returning anything.
#That is, now the program doesn't go to the main function 2nd time.
elif amt>=500 and amt<2000:
mod1,n,div1=twothousand(amt)
return (mod1,n,div1)
def main1():
amt=int(input("Enter the amount: "))
mod1,n,div1=calculate(amt)
print (mod1,n,div1)
#The above print function executes only once.
if mod1!=0:
amt=mod1
calculate(amt)
if __name__=="__main__":
main1()
OUTPUT:
Enter the amount: 1700
200 500 3
EXPECTED OUTPUT:
Enter the amount: 1700
200 500 3
0 200 1
I'm unable to execute return statement after calculate() function call happens 2nd time, as written in comments.
I'm not getting the second output. New to python, kindly help.
sorry for not updating the logic earlier .
The logic is:
When user asks for an amount of 1700, he can only be given that amount using 500 and 200 currency. So, the 1st output is - 200 500 3 ; that is, 3 number of 500 currency.. and remaining is 200.
I want to call the function calculate till the value mod1 == 0.
Your main() function should look like this:
def main1():
amt=int(input("Enter the amount: "))
mod1,n,div1=calculate(amt)
print (mod1,n,div1)
#The above print function executes only once.
if mod1!=0:
amt=mod1
mod1,n,div1 = calculate(amt)
print (mod1,n,div1)
Related
I am encountering a strange error with a mini game I have created as part of a quiz application I am building to run in terminal. The mini game requires processes to run asynchronously, so I defined two coroutines to create it: one that prompts for input (using aioconsole.ainput(), an asynchronous input function) and one that executes a loop until input is entered. The mini game works fine on its own (included it in another program I made where it ran smoothly) and executes without error in the current application I am building, but after the mini game has executed, whenever I call the normal input() function, I receive an EOF error:
Traceback (most recent call last):
File "/Users/[...]/quiz_game.py", line 449, in <module>
quiz()
File "/Users/[...]/quiz_game.py", line 433, in quiz
prev_score = show_score(playerList,prev_score)
File "/Users/[...]/quiz_game.py", line 375, in show_score
input('Hit enter to continue.')
EOFError
I am 99.99% certain that the error has to do with some residual effect of executing the mini game's code, probably a residual effect of using aioconsole.ainput() because the code worked fine before I added the mini game into it, and distinct calls of input() in different locations within the overall code have triggered this error.
Here's the code of the mini game:
async def getInput():
global flag
flag = False
task = asyncio.create_task(loop())
await aioconsole.ainput()
flag = True
await task
return task.result()
async def loop():
count = 0
x = 20
while not flag:
count += 1
print('Hit enter at 20:' + '#' * (count % (x + 1)) + ' ' * (x - count %
(x + 1)),
end='\r')
await asyncio.sleep(0.1)
return count
def minigame():
print('\nScore 20 to get a powerup.')
for i in range(1,4):
print(f'---ATTEMPT {i}---')
result = asyncio.run(getInput()) % 21
print('Score:' + str(result))
if result == 20:
print('SUCCESS!!!')
time.sleep(2)
return
else:
print(
random.choice([
'I said 20, moron!', 'Can you not count?',
'Tut, tut, tut. Such a disappointment.'
]))
print('Try again\n')
print('FAIL. You have exhausted your 3 attempts.')
time.sleep(2)
Short Description of Mini Game:
The mini game displays an increasing number of hashtags until it hits twenty hashtags and then starts over at zero hashtags. If the user hits enter at exactly twenty hashtags, they win the game.
Anyways, any help that you could provide would be greatly appreciated!
I'm trying to test a fuction that dependets a of multiple user inputs to return some value.
I've already looked for multiples unswers here but none was able to resolve my problem. I saw things with parametrize, mock and monkey patch but none helped. I think a lot is because I don't clearly understood the concepts behind what was being done and I couldn't adapt to my problem. I saw suggestion of using external file for this but I don't wont to depend on that. I'm trying with pytest and python 3.7.3
The function that I want to test is something like this
def function():
usr_input = input('please enter a number: ')
while True:
if int(usr_input) < 5:
usr_input = input('please, enter a value less then 5: ')
else:
break
return usr_input
I want to know how can I pass two input values to test the function when the inserted value is not valid. Example: Send value 6 and 2, make an assert expecting value 2 and pass the test. My others tests look like this:
def test_input(monkeypatch):
monkeypatch.setattr('builtins.input', lambda x: 6)
test = function()
assert test == 2
but, for this case, they loop. It's possible to do this only with parametrize or other simple code?
EDIT
I added a int() in my "if", as wim pointed in the accepted answer, just to prevent any confusion for future readers. I'm assuming the cast is possible.
Two problems here, you need to convert the input into a number otherwise the comparison will fail, comparing a string with a number: usr_input < 5. Note that the real input will never return a number, only a string.
Once you've cleared that up, you can monkeypatch input with a callable that can return different values when called:
def fake_input(the_prompt):
prompt_to_return_val = {
'please enter a number: ': '6',
'please, enter a value less then 5: ': '2',
}
val = prompt_to_return_val[the_prompt]
return val
def test_input(monkeypatch):
monkeypatch.setattr('builtins.input', fake_input)
test = function()
assert test == 2
If you install the plugin pytest-mock, you can do this more easily with the mock API:
def test_input(mocker):
mocker.patch('builtins.input', side_effect=["6", "2"])
test = function()
assert test == 2
I've implemented a memory game where the user has to sort numbers in his head while a timer of 5 sec is running.
Please see code below:
from random import randint
from threading import Timer
def timeout():
print("Time over\n#####################\n")
while True:
list = []
for i in range(5):
list.append(randint(1000, 10000))
t = Timer(5, timeout)
t.start()
print(list)
print('[ 1 , 2 , 3 , 4 , 5 ]')
solution = sorted(list)[2]
print('please select the 3rd largest number in this row (1-5):')
input_string = input()
selection = int(input_string)
if solution == list[selection - 1]:
print("Your guess is correct\n")
else:
print("Your guess is wrong\n")
t.join()
Here is the game interaction itself (please ignore the syntax highlighting):
USER#HOST:~/$ python3 memory_game.py
[8902, 8655, 1680, 6763, 4489]
[ 1 , 2 , 3 , 4 , 5 ]
please select the 3rd largest number in this row (1-5):
4
Your guess is correct
Time over
#####################
[5635, 3810, 1114, 5042, 1481]
[ 1 , 2 , 3 , 4 , 5 ]
please select the 3rd largest number in this row (1-5):
4
Your guess is wrong
Time over
#####################
[6111, 1430, 7999, 3352, 2742]
[ 1 , 2 , 3 , 4 , 5 ]
please select the 3rd largest number in this row (1-5):
23
Traceback (most recent call last):
File "memory_game.py", line 24, in <module>
if solution == list[selection - 1]:
IndexError: list index out of range
Time over
#####################
Can anybody help me with these things:
1. 'Time over' should only be written if the player needs more than 5 sec for the answer. If the player solves it in time the next challenge should appear silently.
2. If the player does not write any guess and presses 'Enter' the program terminates with error message:
Traceback (most recent call last):
File "memory_game.py", line 22, in
selection = int(input_string)
ValueError: invalid literal for int with base 10:''
3. If the player enters any random number the program quits with an 'Index out of range error' - I couldn't find out where to put try: except:
Any help would be appreciated - Thanks!
As for your questions:
You can accomplish that with t.cancel() (stop the timer and do not call the function) instead of t.join() (wait until the thread has finished; this will ALWAYS result in a timeout, of course)
(and 3.) These are basically the same -- put the query message and all input handling into a while loop, and break out of it once you know that the input is valid
...
As an extra, the "time over" message wasn't really doing anything useful (e.g. you could still enter a valid answer after the time over occurred. I "fixed" that in a brute force way by making the program exit if the timeout is hit. Instead of doing that, you can also use a global variable to store whether the timeout was hit or not and handle that in your input accordingly (make sure to make it threadsafe using e.g. a mutex).
In general, it might be easier to turn around the structure of the program -- let the main thread handle the timeout and verification of the input, let the thread only handle the input (this way, it's easy to kill the thread to stop the input from being handled).
Of course, using the select module, one could implement this even nicer without threads (have one pipe that gets written to by a thread/timer, and the standard input, then select both for reading and it will block until either user input or the timeout occurs).
And maybe someone will post a nice asyncio-based solution? ;)
Here's the modified solution (modifying only as little as possible to get it to work, one could refactor other parts to make it nicer in general):
import random
import threading
import os
def timeout():
print("Time over\n#####################\n")
# You can't use sys.exit() here, as this code is running in a thread
# See also: https://stackoverflow.com/a/905224/1047040
os._exit(0)
while True:
list = []
for i in range(5):
list.append(random.randint(1000, 10000))
t = threading.Timer(5, timeout)
t.start()
print(list)
print('[ 1 , 2 , 3 , 4 , 5 ]')
solution = sorted(list)[2]
while True:
try:
print('please select the 3rd largest number in this row (1-5):')
input_string = input()
selection = int(input_string)
chosen = list[selection - 1]
# If we arrive here, the input was valid, break out of the loop
break
except Exception as e:
# Tell the user that the input is wrong; feel free to remove "e"
# from the print output
print('Invalid input:', e)
if solution == chosen:
print("Your guess is correct\n")
else:
print("Your guess is wrong\n")
# Make sure to cancel the thread, otherwise guessing correct or wrong
# will block the CLI interface and still write "time out" at the end
t.cancel()
I am trying to test each input then return that the number is cleared then do the math. For Example is a user inputs N instead of a number I want it to output that its not a number whereas if the user inputs 1 then I want it to move to the next function asking for a power then do the same thing and if that passes then goes to the final section which output the answer to the problem.
The program passes both the errors for the non number areas yet when it get to very last function it is telling me base nor power are defined.
Code is written in some Python2 and some Python3. All works fine though. I use python3 mostly.
[Test Picture/Error Msg][1]
# Below we are creating the recursive statement to do the math for us. We are calling Base and Power
# from the main function where the user Inputs the numbers.
def pow(base, power):
if power == 0:
return 1
if power == 1:
return base
else :
return base * pow(base, power - 1)
def determineBase():
while True:
try:
base = int(input ('Please Enter A Base: '))
except ValueError:
print("Please use whole numbers only. Not text nor decimals.")
continue
else:
return base
def determinePower():
while True:
try:
power = int(input ('Please Enter A Power: '))
except ValueError:
print("Please use whole numbers only. Not text nor decimals.")
continue
else:
return power
def main():
determineBase()
determinePower()
pow(base,power)
print("The answer to",base,"to the power of", power,"is", pow(base,power),".")
main()
def main():
determineBase()
determinePower()
pow(base,power)
Here, neither base nor power are defined. What you meant instead was to store the result from those function calls and pass those then:
def main():
base = determineBase()
power = determinePower()
pow(base, power)
The issue isn't inside the recursive function, it's inside your main function.
The problem is arising due to the fact that you are passing base as an argument to the pow() function without defining the variable base first (the same would subsequently be true for power).
In other words you need something along the lines of:
def main():
base = determineBase()
power = determinePower()
pow(base,power) #this line could probably be removed
print("The answer to",base,"to the power of", power,"is", pow(base,power),".")
As currently, you're not storing the values of these two functions.
So the problem is to define these six functions
def sphereVolume(r):
def sphereSurface(r):
def cylinderVolume(r,h):
def cylinderSurface(r,h):
def coneVolume(r,h):
def coneSurface(r,h):
And the write a program that prompts the user for the values of r and h, call the six functions, and print the results.
I have not tested this code because I am on a computer currently that does not have scite or python, however I've created this code on a notepad.
from math import pi
def sphereVolume():
volume1=((4/3)*pi*r)**3))
return volume1
def sphereSurface():
area1=(4*pi*r**2)
return area1
def cylinderVolume():
volume2=(pi*r**2*h)
return volume2
def cylinderSurface():
area2=(2*pi*r**2+2*pi*r*h)
return area2
def coneVolume():
volume3=((1/3)*pi*r**2*h)
return volume3
def coneSurface():
area3=(pi*r+pi*r**2)
return area3
main():
def main():
r=int (input("Enter the radius:"))
h=int (input("Enter the heights:"))
print ("Sphere volume:",sphereVolume(r),volume1)
print ("Sphere Surface:",sphereSurface(r),area1)
print ("Cylinder Volume:" , cylinderVolume(r,h),volume2)
print ("Cylinder Surface:",cylinderSurface(r,h),area2)
print ("Cone Volume:",coneVolume(r,h),volume3)
print ("Cone Surface:",coneSurface(r,h),area3)
Am I using the functions properly? Or is there a lot that I need to change?
There are many syntax errors in your code:
volume1=((4/3)*pi*r)**3)) (You don't need extra bracket at the end)
main(): (You called this function before you declared it, only call it after you've declared it and given it attributes)
print ("Sphere volume:",sphereVolume(r),volume1)
print ("Sphere Surface:",sphereSurface(r),area1)
print ("Cylinder Volume:" , cylinderVolume(r,h),volume2)
print ("Cylinder Surface:",cylinderSurface(r,h),area2)
print ("Cone Volume:",coneVolume(r,h),volume3)
print ("Cone Surface:",coneSurface(r,h),area3)
At first glance, this may all look right, however for each function you print, you give it a set of arguments that aren't meant to be there (e.g sphereVolume has the argument r). They shouldn't be there because you programmed them NOT to take in arguments, so you should change your functions to take in the arguments, otherwise you get the error:
print ("Sphere volume:",sphereVolume(r),volume1)
TypeError: sphereVolume() takes 0 positional arguments but 1 was given
So your functions should look like this:
from math import pi
def sphereVolume(r):
volume1=((4/3)*pi*r)**3
return volume1
def sphereSurface(r):
area1=(4*pi*r**2)
return area1
def cylinderVolume(r,h):
volume2=(pi*r**2*h)
return volume2
def cylinderSurface(r,h):
area2=(2*pi*r**2+2*pi*r*h)
return area2
def coneVolume(r,h):
volume3=((1/3)*pi*r**2*h)
return volume3
def coneSurface(r,h):
area3=(pi*r+pi*r**2)
return area3
You need to give them a set of arguments to work with, otherwise it's incorrect to put the variable r and h inside the functions, because- in simple terms- they haven't been given permission to be there.
Finally, you need to remove the extra variables you got from your functions that you printed out in main(). As they are local variables you can't access them unless they are returned. I'm guessing what you tried to do is that you wanted for instance in this line
print ("Sphere volume:",sphereVolume(r),volume1)
to print the value of volume1. You've already done that! When you said return volume1 at the end of the function, that meant if ever you print this function elsewhere, the only argument that will be accessed from the function is the one you returned, which in this case is volume1. Do the same likewise for all the other local variables you tried printing out by deleting them.
I've tested this code, but just so you don't have to look at everything I wrote if you don't want to, the fully working code is this:
from math import pi
def sphereVolume(r):
volume1=((4/3)*pi*r)**3
return volume1
def sphereSurface(r):
area1=(4*pi*r**2)
return area1
def cylinderVolume(r,h):
volume2=(pi*r**2*h)
return volume2
def cylinderSurface(r,h):
area2=(2*pi*r**2+2*pi*r*h)
return area2
def coneVolume(r,h):
volume3=((1/3)*pi*r**2*h)
return volume3
def coneSurface(r,h):
area3=(pi*r+pi*r**2)
return area3
def main():
r=int (input("Enter the radius:"))
h=int (input("Enter the heights:"))
print ("Sphere volume:",sphereVolume(r))
print ("Sphere Surface:",sphereSurface(r))
print ("Cylinder Volume:" , cylinderVolume(r,h))
print ("Cylinder Surface:",cylinderSurface(r,h))
print ("Cone Volume:",coneVolume(r,h))
print ("Cone Surface:",coneSurface(r,h))
main()
You need to add arguments to your functions for r and h.
You have an extra paren for:
volume1=((4/3)*pi*r)**3))
You need to fix:
main():