Python 3.6+ function to ask for a multiple-choice answer
$begingroup$
# Standard multi choice question template
def multiChoiceQuestion(options: list):
while True:
print("nEnter the number of your choice - ")
for x in range(len(options)):
print(str((x + 1)) + ". " + options[x])
print("n")
try:
answer = int(input())
except ValueError:
print("Doesn't seem like a number! Try again!")
continue
if answer < 1 or answer > len(options):
print("That option does not exist! Try again!")
continue
return answer
I created a template to ask a multi choice question in python. The loop will never reach it's end, since there is always a continue or a return statement. Is the while True
condition appropriate for it?
python python-3.x validation
New contributor
$endgroup$
add a comment |
$begingroup$
# Standard multi choice question template
def multiChoiceQuestion(options: list):
while True:
print("nEnter the number of your choice - ")
for x in range(len(options)):
print(str((x + 1)) + ". " + options[x])
print("n")
try:
answer = int(input())
except ValueError:
print("Doesn't seem like a number! Try again!")
continue
if answer < 1 or answer > len(options):
print("That option does not exist! Try again!")
continue
return answer
I created a template to ask a multi choice question in python. The loop will never reach it's end, since there is always a continue or a return statement. Is the while True
condition appropriate for it?
python python-3.x validation
New contributor
$endgroup$
$begingroup$
"The loop will never reach it's end ... Is thewhile True
condition appropriate for it?" That depends on whether that is the intended behaviour. Is it?
$endgroup$
– Mast
13 hours ago
1
$begingroup$
Side note:for x in len(options):
will produce an error as Python doesn't allow iteration over an integer. 200_success's approach is the way to go here, but for future reference, usefor x in range(len(options)):
if you really need to loop a certain number of times. This takes thelen(options)
integer and creates an interable out of it.
$endgroup$
– JDG
11 hours ago
add a comment |
$begingroup$
# Standard multi choice question template
def multiChoiceQuestion(options: list):
while True:
print("nEnter the number of your choice - ")
for x in range(len(options)):
print(str((x + 1)) + ". " + options[x])
print("n")
try:
answer = int(input())
except ValueError:
print("Doesn't seem like a number! Try again!")
continue
if answer < 1 or answer > len(options):
print("That option does not exist! Try again!")
continue
return answer
I created a template to ask a multi choice question in python. The loop will never reach it's end, since there is always a continue or a return statement. Is the while True
condition appropriate for it?
python python-3.x validation
New contributor
$endgroup$
# Standard multi choice question template
def multiChoiceQuestion(options: list):
while True:
print("nEnter the number of your choice - ")
for x in range(len(options)):
print(str((x + 1)) + ". " + options[x])
print("n")
try:
answer = int(input())
except ValueError:
print("Doesn't seem like a number! Try again!")
continue
if answer < 1 or answer > len(options):
print("That option does not exist! Try again!")
continue
return answer
I created a template to ask a multi choice question in python. The loop will never reach it's end, since there is always a continue or a return statement. Is the while True
condition appropriate for it?
python python-3.x validation
python python-3.x validation
New contributor
New contributor
edited 2 hours ago
Holyprogrammer
New contributor
asked 14 hours ago
HolyprogrammerHolyprogrammer
1568
1568
New contributor
New contributor
$begingroup$
"The loop will never reach it's end ... Is thewhile True
condition appropriate for it?" That depends on whether that is the intended behaviour. Is it?
$endgroup$
– Mast
13 hours ago
1
$begingroup$
Side note:for x in len(options):
will produce an error as Python doesn't allow iteration over an integer. 200_success's approach is the way to go here, but for future reference, usefor x in range(len(options)):
if you really need to loop a certain number of times. This takes thelen(options)
integer and creates an interable out of it.
$endgroup$
– JDG
11 hours ago
add a comment |
$begingroup$
"The loop will never reach it's end ... Is thewhile True
condition appropriate for it?" That depends on whether that is the intended behaviour. Is it?
$endgroup$
– Mast
13 hours ago
1
$begingroup$
Side note:for x in len(options):
will produce an error as Python doesn't allow iteration over an integer. 200_success's approach is the way to go here, but for future reference, usefor x in range(len(options)):
if you really need to loop a certain number of times. This takes thelen(options)
integer and creates an interable out of it.
$endgroup$
– JDG
11 hours ago
$begingroup$
"The loop will never reach it's end ... Is the
while True
condition appropriate for it?" That depends on whether that is the intended behaviour. Is it?$endgroup$
– Mast
13 hours ago
$begingroup$
"The loop will never reach it's end ... Is the
while True
condition appropriate for it?" That depends on whether that is the intended behaviour. Is it?$endgroup$
– Mast
13 hours ago
1
1
$begingroup$
Side note:
for x in len(options):
will produce an error as Python doesn't allow iteration over an integer. 200_success's approach is the way to go here, but for future reference, use for x in range(len(options)):
if you really need to loop a certain number of times. This takes the len(options)
integer and creates an interable out of it.$endgroup$
– JDG
11 hours ago
$begingroup$
Side note:
for x in len(options):
will produce an error as Python doesn't allow iteration over an integer. 200_success's approach is the way to go here, but for future reference, use for x in range(len(options)):
if you really need to loop a certain number of times. This takes the len(options)
integer and creates an interable out of it.$endgroup$
– JDG
11 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The while True
is fine, and is probably the best way to do it. However, the rest of the flow control is a bit clumsy. By rearranging a few statements, you can eliminate the continue
s.
PEP 8, the official Python style guide, recommends lowercase_with_underscores
for function names unless you have a good reason to deviate.
The loop to print the numbered menu would be better written using enumerate()
. Also, Python supports double-ended comparisons for validating that the answer is in range.
def multi_choice_question(options: list):
while True:
print("nEnter the number of your choice - ")
for i, option in enumerate(options, 1):
print(f'{i}. {option}')
print("n")
try:
answer = int(input())
if 1 <= answer <= len(options):
return answer
print("That option does not exist! Try again!")
except ValueError:
print("Doesn't seem like a number! Try again!")
$endgroup$
add a comment |
$begingroup$
I think that 200_success
already covered most points. I would however like to add an alternative idea for the printing part:
print("Enter the number of your choice -",
*(f'{i}. {opt}' for i, opt in enumerate(options, 1)),
sep='n', end='nn')
Explanation:
from the docs we see that following signature for the print function:
print(*objects, sep=' ', end='n', file=sys.stdout, flush=False)
we can therefore print everything with a single print call instead of three individual ones. I leave it up to you which one you perceive easier to use.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The while True
is fine, and is probably the best way to do it. However, the rest of the flow control is a bit clumsy. By rearranging a few statements, you can eliminate the continue
s.
PEP 8, the official Python style guide, recommends lowercase_with_underscores
for function names unless you have a good reason to deviate.
The loop to print the numbered menu would be better written using enumerate()
. Also, Python supports double-ended comparisons for validating that the answer is in range.
def multi_choice_question(options: list):
while True:
print("nEnter the number of your choice - ")
for i, option in enumerate(options, 1):
print(f'{i}. {option}')
print("n")
try:
answer = int(input())
if 1 <= answer <= len(options):
return answer
print("That option does not exist! Try again!")
except ValueError:
print("Doesn't seem like a number! Try again!")
$endgroup$
add a comment |
$begingroup$
The while True
is fine, and is probably the best way to do it. However, the rest of the flow control is a bit clumsy. By rearranging a few statements, you can eliminate the continue
s.
PEP 8, the official Python style guide, recommends lowercase_with_underscores
for function names unless you have a good reason to deviate.
The loop to print the numbered menu would be better written using enumerate()
. Also, Python supports double-ended comparisons for validating that the answer is in range.
def multi_choice_question(options: list):
while True:
print("nEnter the number of your choice - ")
for i, option in enumerate(options, 1):
print(f'{i}. {option}')
print("n")
try:
answer = int(input())
if 1 <= answer <= len(options):
return answer
print("That option does not exist! Try again!")
except ValueError:
print("Doesn't seem like a number! Try again!")
$endgroup$
add a comment |
$begingroup$
The while True
is fine, and is probably the best way to do it. However, the rest of the flow control is a bit clumsy. By rearranging a few statements, you can eliminate the continue
s.
PEP 8, the official Python style guide, recommends lowercase_with_underscores
for function names unless you have a good reason to deviate.
The loop to print the numbered menu would be better written using enumerate()
. Also, Python supports double-ended comparisons for validating that the answer is in range.
def multi_choice_question(options: list):
while True:
print("nEnter the number of your choice - ")
for i, option in enumerate(options, 1):
print(f'{i}. {option}')
print("n")
try:
answer = int(input())
if 1 <= answer <= len(options):
return answer
print("That option does not exist! Try again!")
except ValueError:
print("Doesn't seem like a number! Try again!")
$endgroup$
The while True
is fine, and is probably the best way to do it. However, the rest of the flow control is a bit clumsy. By rearranging a few statements, you can eliminate the continue
s.
PEP 8, the official Python style guide, recommends lowercase_with_underscores
for function names unless you have a good reason to deviate.
The loop to print the numbered menu would be better written using enumerate()
. Also, Python supports double-ended comparisons for validating that the answer is in range.
def multi_choice_question(options: list):
while True:
print("nEnter the number of your choice - ")
for i, option in enumerate(options, 1):
print(f'{i}. {option}')
print("n")
try:
answer = int(input())
if 1 <= answer <= len(options):
return answer
print("That option does not exist! Try again!")
except ValueError:
print("Doesn't seem like a number! Try again!")
edited 2 hours ago
Holyprogrammer
1568
1568
answered 13 hours ago
200_success200_success
130k16153419
130k16153419
add a comment |
add a comment |
$begingroup$
I think that 200_success
already covered most points. I would however like to add an alternative idea for the printing part:
print("Enter the number of your choice -",
*(f'{i}. {opt}' for i, opt in enumerate(options, 1)),
sep='n', end='nn')
Explanation:
from the docs we see that following signature for the print function:
print(*objects, sep=' ', end='n', file=sys.stdout, flush=False)
we can therefore print everything with a single print call instead of three individual ones. I leave it up to you which one you perceive easier to use.
$endgroup$
add a comment |
$begingroup$
I think that 200_success
already covered most points. I would however like to add an alternative idea for the printing part:
print("Enter the number of your choice -",
*(f'{i}. {opt}' for i, opt in enumerate(options, 1)),
sep='n', end='nn')
Explanation:
from the docs we see that following signature for the print function:
print(*objects, sep=' ', end='n', file=sys.stdout, flush=False)
we can therefore print everything with a single print call instead of three individual ones. I leave it up to you which one you perceive easier to use.
$endgroup$
add a comment |
$begingroup$
I think that 200_success
already covered most points. I would however like to add an alternative idea for the printing part:
print("Enter the number of your choice -",
*(f'{i}. {opt}' for i, opt in enumerate(options, 1)),
sep='n', end='nn')
Explanation:
from the docs we see that following signature for the print function:
print(*objects, sep=' ', end='n', file=sys.stdout, flush=False)
we can therefore print everything with a single print call instead of three individual ones. I leave it up to you which one you perceive easier to use.
$endgroup$
I think that 200_success
already covered most points. I would however like to add an alternative idea for the printing part:
print("Enter the number of your choice -",
*(f'{i}. {opt}' for i, opt in enumerate(options, 1)),
sep='n', end='nn')
Explanation:
from the docs we see that following signature for the print function:
print(*objects, sep=' ', end='n', file=sys.stdout, flush=False)
we can therefore print everything with a single print call instead of three individual ones. I leave it up to you which one you perceive easier to use.
answered 7 hours ago
magu_magu_
4931519
4931519
add a comment |
add a comment |
Holyprogrammer is a new contributor. Be nice, and check out our Code of Conduct.
Holyprogrammer is a new contributor. Be nice, and check out our Code of Conduct.
Holyprogrammer is a new contributor. Be nice, and check out our Code of Conduct.
Holyprogrammer is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
"The loop will never reach it's end ... Is the
while True
condition appropriate for it?" That depends on whether that is the intended behaviour. Is it?$endgroup$
– Mast
13 hours ago
1
$begingroup$
Side note:
for x in len(options):
will produce an error as Python doesn't allow iteration over an integer. 200_success's approach is the way to go here, but for future reference, usefor x in range(len(options)):
if you really need to loop a certain number of times. This takes thelen(options)
integer and creates an interable out of it.$endgroup$
– JDG
11 hours ago