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how to check if objedc has attribute false o true

I'm trying to check if any of the students are on probation and if they are, display which student please assist with the error code, the code is meant to check if anyone has violated probation, it also displays an error code, the except is normally key error, but when I was trying to figure out the problem i changed it to keyboard interrupt
if students[Student].is_on_probation:
KeyError: 0
students = {}
try:
while True:
from Student import Student
students['1'] = Student("Jim", "Business", 3.1, False)
viewing = False
viewingchange = input("Would you like to view something: ").lower()
if viewingchange == "y":
viewing = True
elif viewingchange == "yes":
viewing = True
else:
viewing = False
add = input("Would you like to add someone: ")
if add.lower() == "yes" or "y":
adder = str(input("Who would you like to add: "))
major = str(input("What is their major: "))
gpa = float(input("GPA: "))
is_on_probation = bool(input("is on probation: "))
students[str(len(students)+1)] = Student(adder, major, gpa, is_on_probation)
def checking():
if viewing:
checker = input("Would you like to view all students or student a/s: ")
if checker.lower() == "a":
print(students)
checkfor = input("Check for: ")
if checkfor.lower() == "probation":
for Student in range(len(students)):
print("works")
if students[Student].is_on_probation:
print("yeppp")
else:
pass
checkerdouble = False
elif checker.lower() == "s":
searchquery2 = input("And which attribute would you like to search: ")
test = input("Student ID: ")
dude = students[test]
if searchquery2.lower() == "Name".lower():
print(dude.name)
elif searchquery2.lower() == "Major".lower():
print(dude.major)
elif searchquery2.lower() == "GPA".lower():
print(dude.gpa)
elif searchquery2.lower() == "is on probation" or "iop" or "probation".lower():
print(dude.is_on_probation)
checkerdouble = True
checking()
except KeyboardInterrupt:
print("Error code: " + "45345593M3940249525")

Your students dictionary uses string representations of numbers that start at str(1). So the loop should do the same. I think it would be better to just use int keys.
for student_id in map(str, range(1, len(students)):
if students[student_id].is_on_probation:
print(student_id)
This would work but it relies on the dictionary keys being dense. I would prefer to do:
for student_id in sorted(students, key=int):
if students[student_id].is_on_probation:
print(student_id)
Some might object to the indexing, as one can do:
for student_id, student in sorted(students.items(), key=lambda v: int(v[0])):
if student.is_on_probation:
print(student_id)

How do I implement a timer on each question asked in a quiz?

I was attempting to create a quiz and one of the criteria is to limit the time available to solve for each question in the quiz. I looked up certain tutorials but some require an input of x seconds for the timer to go off while others looked more like a stopwatch...
I was wondering how do I do like a background timer that ticks down as soon as the question is printed out and skips to the next question if, for example the 30-second period has ended? I'm clueless in the timer function and was having problems to even try to implement them onto my codes. Could someone give out a few pointers so I can sorta progress further in implementing a working timer?
Thank you!
EDITED section below:
The timer that I want to implement onto my coding:
import time
import threading
def atimer():
print("Time's up.")
a_timer = threading.Timer(10.0, atimer)
a_timer.start()
print("")
This is the whole coding that I tried to implement the timer into.
I noticed that when I tried to define qtimer to just print 1 or 2 lines of statements the timer works but I want the timer to stop and go to second question or stop and give the user another attempt to retry the question, so I tried to attach a bunch of codes after the definition and it didn't work. I know I'm most probably doing something wrong here since I'm not quite familiar with time or threading functions. Is there a workaround?
def qtimer():
print("I'm sorry but your time is up for this question.")
print("You may have another attempt if you wish to, with reduced marks allocated.")
response1 = input("Type 'Yes' for another attempt, anything else to skip: ")
if response1 == "Yes":
Answ = input("Which option would you go for this time?: ")
Answ = int(Answ)
if possible[Answ - 1] == qaItem.corrAnsw:
print("Your answer was correct.")
corr += 1
marks += 0.5 * qaItem.diff
else:
print("Your answer was wrong.")
print("Correct answer was: " + qaItem.corrAnsw)
print("Explanation: " + qaItem.expl)
print("")
else:
print("Correct answer was: " + qaItem.corrAnsw)
print("Explanation: " + qaItem.expl)
print("")
class A:
def __init__(self, question, correctAnswer, otherAnswers, difficulty, explanation):
self.question = question
self.corrAnsw = correctAnswer
self.otherAnsw = otherAnswers
self.diff = difficulty
self.expl = explanation
qaList = [A("What is COVID-19?", "Coronavirus Disease 2019", ["Wuhan virus", "I don't understand...", "Coronavirus Disease v19"], 1, "Explanation 1"),
A("What describes COVID-19?", "A disease", ["A virus", "A parasite", "A bacteriophage"], 1, "Explanation 2"),
A("What causes COVID-19?", "SARS-CoV-2", ["Coronavirus", "Mimivirus", "Rubeola Virus"], 1, "Explanation 3"),
A("Which of the following is used in COVID-19 treatment?", "Lopinavir / Ritonavir ", ["Midazolam / Triazolam", "Amiodarone", "Phenytoin"], 2, "Explanation 4"),
A("Which of the following receptors is used by COVID-19 to infect human cells?", "ACE-2 Receptors", ["ApoE4 Receptors", "TCR Receptors", "CD28 Receptors"], 3, "Explanation 5")]
corr = 0
marks = 0
random.shuffle(qaList)
for qaItem in qaList:
q_timer = threading.Timer(5.0, qtimer)
q_timer.start()
print(qaItem.question)
print("Possible answers are:")
possible = qaItem.otherAnsw + [qaItem.corrAnsw]
random.shuffle(possible)
count = 0
while count < len(possible):
print(str(count+1) + ": " + possible[count])
count += 1
print("Please enter the number of your answer:")
Answ = input()
Answ = str(Answ)
while not Answ.isdigit():
print("That was not a number. Please enter the number of your answer:")
Answ = input()
Answ = int(Answ)
Answ = int(Answ)
while Answ > 4 or Answ < 1:
print("That number doesn't correspond to any answer. Please enter the number of your answer:")
Answ = input()
Answ = int(Answ)
if possible[Answ-1] == qaItem.corrAnsw:
print("Your answer was correct.")
corr += 1
marks += 1 * qaItem.diff
else:
print("Your answer was wrong.")
response = input("Would you want to try again? If so, input 'Yes' to attempt it again, if not just input whatever!")
if response == "Yes":
Answ = input("Which option would you go for this time?: ")
Answ = int(Answ)
if possible[Answ - 1] == qaItem.corrAnsw:
print("Your answer was correct.")
corr += 1
marks += 0.5 * qaItem.diff
else:
print("Your answer was wrong.")
print("Correct answer was: " + qaItem.corrAnsw)
print("Explanation: " + qaItem.expl)
print("")
else:
print("Correct answer was: " + qaItem.corrAnsw)
print("Explanation: " + qaItem.expl)
print("")
print("You answered " + str(corr) + " of " + str(len(qaList)) + " questions correctly.")
print("You have achieved a total score of " + str(marks) + ".")

Even with a timer, the main thread can't proceed past waiting for the user to input a number; so if the user doesn't nothing, the timer function runs, and once it has finished the main thread is still waiting for input at
print("Please enter the number of your answer:")
Answ = input()
You could have a global flag that the timer thread sets to tell the main thread to treat the input received as response1 in the timer code, and also have a flag to tell the timer that an answer was received, and so on, and it quickly becomes rather complicated.
So rather than trying to work around the blocking call to input by communicating between the timer and main thread, take the non-blocking input example from https://stackoverflow.com/a/22085679/1527 and stop the loop early if the time is up.
def timed_input(msg, timeout=10):
kb = KBHit()
print(msg)
end_time = time.time() + timeout
warn_time = 5
result = None
while True:
if kb.kbhit():
c = kb.getch()
if '0' <= c <= '9':
result = int(c)
break
print(c)
if time.time() > end_time:
print('time is up')
break
if time.time() > end_time - warn_time:
print(f'{warn_time}s left')
warn_time = warn_time - 1
kb.set_normal_term()
return result
# Test
if __name__ == "__main__":
result = timed_input('Enter a number between 1 and 4')
if result is None:
print('be quicker next time')
elif 1 <= result <= 4:
print('ok')
else:
print(f'{result} is not between 1 and 4')
Note also that breaking up into smaller functions helps make the code easier to follow, the logic of the test doesn't need to know about the logic of the timeout.

Can't assign to function call (python 3)

print ("** ** ***")
print ("** ** ***")
print ("********* ***")
print ("** ** ***")
print ("** ** ***")
Contacts = [0]
if Contacts == [0]:
print ("You have no friends!")
from time import sleep
Contact = input("Would you like to add a contact: Y or N?")
if Contact == "Y":
for c in range(0,1):
print (1 - c)
sleep(1)
Contact = input("Name:")
Contacts.append(Contact)
Contact1 = input ("age:")
Contacts.append(Contact1)
Contact2 = input ("Location:")
Contacts.append(Contact2)
Contact3 = input ("Phone Number:")
Contacts.append(Contact3)
Contactsn = open("Contactsn.txt", "w")
Contactsn.write(Contact)
Contactsn.write(Contact1)
Contactsn.write(Contact2)
Contactsn.write(Contact3)
print (Contact)
print (Contact1)
print (Contact2)
print (Contact3)
Contactsn.close()
Contact = input("Would you like to see your Contacts:Y or N?")
if Contact == "Y":
Contactsn()
else:
Contact = input("Would you like to see your Contacts:Y or N?")
if Contact == "Y":
Contactsn()
def Contactsn():
Contactsn() = open("Contactsn.txt", "r")
print (Contactsn.read())
quit
when I run this it says cant assign function, how can I fix this?

This block is the problem
def Contactsn():
Contactsn() = open("Contactsn.txt", "r")
print (Contactsn.read())
You are trying to assign the result of open to the result of calling your function contactsn()? Either the () is a typo, and you're about to shadow your function name with a variable with the same name, or you are trying some really odd recursive strangeness.

What is wrong with my defintion of the function prompt_int?

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().

Why doesn't my 'oldUser()' run and why does it start again all the time?

I'm very very new to programming and for a school project (50% of my final grade) I had to create a Python program that did roughly this.
I've had some help from my older brother and my teacher but mainly did it myself with some flow charts etc, so please forgive me if I haven't followed conventional rules and things of this nature, or if my code is messy. I will finalise it, just needed bait of help/support from the pro's.
This is my code and I have an issue with it. Once I have pressed 'y' and then 'y' again on the displayMenu() why doesn't it run oldUser()
Also, if any of you have any suggestion on what could make my code better, or I could improve it would be very helpful and I will take it on board.
import os # allows me to use functions defined elsewhere. os module allows for multi platforming.
import sys
words = []
users = {}
status = ""
def teacher_enter_words():
done = False
print 'Hello, please can you enter a word and definition pair.'
while not done:
word = raw_input('\nEnter a word: ')
deff = raw_input('Enter the definition: ')
# append a tuple to the list so it can't be edited.
words.append((word, deff))
add_word = raw_input('Add another word? (y/n): ')
if add_word.lower() == 'n':
done = True
def student_take_test():
student_score = 0
for pair in words:
print 'Definition:', pair[1]
inp = raw_input('Enter word: ')
student_score += check_error(pair[0], inp)
print 'Correct spelling:', pair[0], '\n'
print 'Your score:', student_score
def check_error(correct, inputt):
len_c = len(correct)
len_i = len(inputt)
# threshold is how many incorrect letters do we allow before a
# minor error becomes a major error.
# 1 - allow 1 incorrect letter for a minor error ( >= 2 becomes major error)
threshold = 1
# immediately check if the words are the same length
num_letters_incorrect = abs(len_c - len_i) # abs() method returns value of x - positive dist between x and zero
if num_letters_incorrect == 0:
for i in xrange(0, len(correct)):
if correct[i] != inputt[i]:
num_letters_incorrect += 1
if num_letters_incorrect <= threshold:
if num_letters_incorrect == 0:
return 2 # no incorrect letter.
else:
return 1 # minor error.
else:
return 0 # major error.
def displayMenu():
status = raw_input('Are you a registered user? y/n?: ')
if status == raw_input == 'y':
oldUser()
elif status == 'n':
newUser()
def newUser():
createLogin = raw_input('Create login name: ')
if createLogin in users:
print '\nLogin name already exist!\n'
else:
createPassw = raw_input('Create password: ')
users[createLogin] = createPassw
print '\nUser created!\n'
def oldUser():
login = raw_input('Enter login name: ')
passw = raw_input('Enter password: ')
if login in users and users[login] == passw:
print '\nLogin successful!\n'
else:
print "\nUser doesn't exist or wrong password!\n"
if __name__ == '__main__':
running = True
while running:
os.system('cls' if os.name == 'nt' else 'clear') # multi-platform, executing a shell command
reg = raw_input('Do you want to start the program? y/n?').lower()
if reg == 'y' or reg == 'yes':
displayMenu()
else: sys.exit(0)
inp = raw_input('Are you a Teacher or a Student? (t/s): ').lower()
if inp == 't' or inp == 'teacher':
teacher_enter_words()
else:
student_take_test()
running = False

raw_input is a function. status == raw_input == 'y' will never be true: that is comparing status with the function, and with 'y'.
I suspect that's simply a typo, and you just meant if status == 'y':