Develop Reference

Handling input data quality with checks & default value

I am trying to write a code for squaring the user input number in Python. I've created function my1() ...
What I want to do is to make Python to take user input of a number and square it but if user added no value it gives a print statement and by default give the square of a default number for e.g 2
Here is what I've tried so far
def my1(a=4):
if my1() is None:
print('You have not entered anything')
else:
b=a**2
print (b)
my1(input("Enter a Number"))

This is a better solution:
def my1(a=4):
if not a:
return 'You have not entered anything'
else:
try:
return int(a)**2
except ValueError:
return 'Invalid input provided'
my1(input("Enter a Number"))
Explanation
Have your function return values, instead of simply printing. This is good practice.
Use if not a to test if your string is empty. This is a Pythonic idiom.
Convert your input string to numeric data, e.g. via int.
Catch ValueError and return an appropriate message in case the user input is invalid.

You're getting an infinite loop by calling my1() within my1(). I would make the following edits:
def my1(a):
if a is '':
print('You have not entered anything')
else:
b=int(a)**2
print (b)
my1(input("Enter a Number"))

When I read your code, I can see that you are very confused about what you are writing. Try to organize your mind around the tasks you'll need to perform. Here, you want to :
Receive your user inputs.
Compute the data.
Print accordingly.
First, take your input.
user_choice = input("Enter a number :")
Then, compute the data you received.
my1(user_choice)
You want your function, as of now, to print an error message if your type data is not good, else print the squared number.
def my1(user_choice): # Always give meaning to the name of your variables.
if not user_choice:
print 'Error'
else:
print user_choice ** 2
Here, you are basically saying "If my user_choice doesn't exists...". Meaning it equals False (it is a bit more complicated than this, but in short, you need to remember this). An empty string doesn't contain anything for instance. The other choice, else, is if you handled your error case, then your input must be right, so you compute your data accordingly.

In your second line, it should be
if a is None:
I think what you want to do is something like the following:
def m1(user_input=None):
if user_input is None or isinstance(user_input, int):
print("Input error!")
return 4
else:
return int(user_input)**2
print(my1(input("Input a number")))

How to get this pseudocode to double a number in a while loop?

pseudocode
numtodouble=int
result=int
print("")
print("Enter a number you would like to double and press Enter.")
input (numtodouble)
<class 'int'>2
'2'
while numtodouble>0:
result=numtodouble*2
print("2 X", numtodouble, "=", result)
print("")
print("Enter a number you would like to double and press Enter.")
input(numtodouble)
break
print("OK, you entered a value <=0, ending execution.")
Does anyone know where I went wrong with my code? I've been struggling with this for hours.

try:
# input is stored as num_to_double. the input is cast to an int, and the string in input is the prompt
num_to_double = int(input("Enter a number you would like to double and press Enter."))
while num_to_double>0:
result=num_to_double*2
# Format puts the arguments into the curly braces in the order given
print("2 X {} = {}\n".format(num_to_double, result))
# input is cast to int and stored in num_to_double. The text in the input command is the prompt
num_to_double =int(input("Enter a number you would like to double and press Enter."))
# This is otuside the while loop, so this runs when the while loop breaks. The previous break command was making
# the code act not as intended
print("OK, you entered a value <=0, ending execution.")
# This catches if someone inputs a value that is not able to be cast to a string. It's the second half of the Try:
# Except block.
except ValueError as _:
print("A not-a-number was supplied")
This code is far simplier and does what you're trying to do. I assume you're learning python, so some of these things are not the simplest way to do things, like the format function, but are super useful to learn.
num_to_double = 0
result = 0
print("")
num_to_double = int(input("Enter number would you like to double and press enter."))
while num_to_double > 0:
result = num_to_double * 2
print("2 X {} = {}".format(num_to_double, result))
print("")
num_to_double = int(input("Enter number would you like to double and press enter."))
print("OK< you entered a value <=0, ending execution.")
This code is the closest I could do to the pseudocode provided. Declaring variables before they're used here isn't necessary and is messy. It's like the pseudocode wasn't meant to become python. Same with printing blank lines, those should be wrapped into the previous or next print lines.

Exiting program for any input that is not a list

My goal is to create a little program that converts angle from radiant to degree and vice-versa. I need the program to close with no error message from python if the user enters the information to convert in the wrong format.
After assigning the variable ‘angle’ to both values of the input. The angle variable becomes a list type.
In norther to exit program with no error message I write this:
'if angle is not list():break'.
The problem is when I do that it exits the program for any type of command entered as an input.
here is my code:
import numpy as np
while 1:
angle=input("Please enter the angle you want to convert,\n\n"\
"If you wish to convert degrees in radiant or vise-versa,\n"\
"follow this format: 'angle/D or R'").split('/')
if angle is not list():break
angle[0]=float(angle[0])
radiant= (angle[0]*(np.pi))/180
degre=((angle[0]*180)/np.pi)
if (angle[0]>=0 or angle[0]<=360) and angle[1] is 'D' :
print(radiant,'radiants')
elif angle[1] is 'R':
print(degre,'degrés')
else:break

You can use isinstance(angle, list) to check if it is a list. But it won't help you achieve what you really want to do.
The following code will help you with that.
question = """Please enter the angle you want to convert.
If you wish to convert degree in radiant or vice-versa.
Follow this format: 'angle/D or R'
"""
while 1:
angle=input(question).split('/')
if not isinstance(angle, list): break # This will never happen
# It will never happen because string.split() always returns a list
# Instead you should use something like this:
if len(angle) != 2 or angle[1] not in ['D', 'R']:
break
try:
angle[0]=float(angle[0])
except ValueError:
break
if (angle[0]>=0 or angle[0]<=360) and angle[1] is 'D':
# You could also improve this by taking modulo 360 of the angle.
print((angle[0]*np.pi)/180, 'radiants')
else:
# Just an else is enough because we already checked that angle[1] is either D or R
print((angle[0]*180)/np.pi, 'degrees')

What you want:
if not isinstance(angle, list): break
What you've done: if angle is not list():break will always evaluate to True as no object will ever have the same identity as the list list(); since is is a check for identity.
Even this:
>>> list() is not list()
True

break statements are used to get out of for and while loops. Try using a while loop after the input statement to evaluate the input. Use a possible set as a conditional. You do not need to break from an if statement because it will just be bypassed if the the conditional is not met.
Sometimes you might see an if statement followed by a break statement. The break statement, however, is not breaking from the if statement. It is breaking from a previous for or while loop.

Python 3.x - Input only returns first item

I have been having problems with a block of code in a little project and I can't seem to fix it.
In the below code, I define inputrace() so that a process is carried out where the player types in a number that is above 0 and below 13 (there are 12 choices, each dictated by a number); it also checks for blank lines and strings and has the program state that there is an error and ask for user input again if they are detected. If it passes the check, RaceInp is returned and set to RaceChoice which allows the code below it to assign a Race to the player based on their selection.
#race check
def inputrace():
print ("Input number")
RaceInp = input()
Check = RaceInp
try:
int(RaceInp)
except ValueError:
print("Numbers only!")
inputrace()
if not int(Check)>12 or int(Check)<1:
return RaceInp
print (RaceInp) #this is here so I can check the value returned
Race = "NA"
RaceChoice = inputrace()
print (RaceChoice)
#assign race
if RaceChoice == "1":
Race = "Human"
#continues down to twelve
Everything works when valid strings are put in (any number 1-12), but things break when I purposefully put in an invalid string. It seems like RaceInp only retains the first user input and does not change, even after the function is recalled from an error. That means if I were to put in "a," the program will tell me it is wrong and ask again. However, when I put in "1" in an attempt to correct it, it accepts it but still keeps RaceInp as "a."
Is there any fix to this? I have no clue what's going on.
I appreciate the help and sorry if I got anything wrong in the question!

It seems that the problem is that you put the inputrace in a recursion instead of a loop. Something like this would probably be better:
def input_race():
while True:
print("Input a number between 1 and 12.")
race_input = input()
try:
race_input = int(race_input)
if race_input >= 1 and race_input <= 12:
return race_input
except ValueError:
pass
print ("'{input}' is not a number.".format(input=race_input))
race = "NA"
race_choice = input_race()
if race_choice == 1:
race = "Human"
print(race)

Python input string error (don't want to use raw_input) [closed]

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 12 years ago.
I have a menu that asks for the user to pick one of the options. Since this menu is from 1 to 10, I'm using input besides raw_input.
My code as an if statement that if the number the user inputs is from 1 to 10 it does the option. If the user inputs any number besides that ones, the else statement says to the user pick a number from 1 to 10.
The problem is if the user types an string, lets say for example qwert. It gives me an error because its an string. I understand why and I don want to use raw_input.
What can I do to wen the user types a string it goes to my else statement and print for example "Only numbers are valid. Pick a number from 1 to 10"
I don't want to use any advanced programing to do this
Regards,
Favolas
EDIT
Thanks for all your answers and sorry for the late response but I had some health problems.
I couldn't use try or except because my teacher didn't allow it.
In the end, I've used raw_input because it was the simplest alternative but was glad to see that are many ways to solve this problem.
Regards,
Favolas

You can throw an exception when you try to convert your string into a number.
Example:
try:
int(myres)
except:
print "Only numbers are valid"

You should use raw_input(), even if you don't want to :) This will always give you a string. You can then use code like
s = raw_input()
try:
choice = int(s)
except ValueError:
# choice is invalid...
to try to convert to an int.

Im not sure what you consider advanced - a simple way to do it would be with something like this.
def getUserInput():
while True:
a = raw_input("Enter a number between 1 and 10: ")
try:
number = int(a)
if (0 < number <= 10):
return number
else:
print "Between 1 and 10 please"
except:
print "Im sorry, please enter a number between 1 and 10"
Here, I have used try/except statements, to ensure that the entered string can be converted to an integer. And a loop (which will keep running) until the entered number is between 1 and 10 (0< number <=10)

What you really are after is how to figure out if something could pass as an integer. The following would do the job:
try:
i = int(string_from_input)
ecxept ValueError:
# actions in case the input is anything other than int, like continuing the loop

You clearly have something against exception handling. I don't understand why -- it's a fundamental part of (not just Python) programming and something you should be comfortable with. It's no more 'advanced' than handling error codes, just a different mentality.
Here are the docs. It's pretty simple:
It is possible to write programs that
handle selected exceptions. Look at
the following example, which asks the
user for input until a valid integer
has been entered, but allows the user
to interrupt the program (using
Control-C or whatever the operating
system supports); note that a
user-generated interruption is
signalled by raising the
KeyboardInterrupt exception.
>>> while True:
... try:
... x = int(raw_input("Please enter a number: "))
... break
... except ValueError:
... print "Oops! That was no valid number. Try again..."
...
The try statement works as follows.
First, the try clause (the
statement(s) between the try and
except keywords) is executed. If no
exception occurs, the except clause is
skipped and execution of the try
statement is finished. If an exception
occurs during execution of the try
clause, the rest of the clause is
skipped. Then if its type matches the
exception named after the except
keyword, the except clause is
executed, and then execution continues
after the try statement. If an
exception occurs which does not match
the exception named in the except
clause, it is passed on to outer try
statements; if no handler is found, it
is an unhandled exception and
execution stops with a message as
shown above. A try statement may have
more than one except clause, to
specify handlers for different
exceptions. At most one handler will
be executed. Handlers only handle
exceptions that occur in the
corresponding try clause, not in other
handlers of the same try statement. An
except clause may name multiple
exceptions as a parenthesized tuple,
for example:
... except (RuntimeError, TypeError, NameError):
... pass
The last except clause may omit the
exception name(s), to serve as a
wildcard. Use this with extreme
caution, since it is easy to mask a
real programming error in this way! It
can also be used to print an error
message and then re-raise the
exception (allowing a caller to handle
the exception as well):

Personally, I liked my first answer better. However, this one should fit your requirements with more code.
import sys
def get_number(a, z):
if a > z:
a, z = z, a
while True:
line = get_line('Please enter a number: ')
if line is None:
sys.exit()
if line:
number = str_to_int(line)
if number is None:
print('You must enter base 10 digits.')
elif a <= number <= z:
return number
else:
print('Your number must be in this range:', a, '-', z)
else:
print('You must enter a number.')
def get_line(prompt):
sys.stdout.write(prompt)
sys.stdout.flush()
line = sys.stdin.readline()
if line:
return line[:-1]
def str_to_int(string):
zero = ord('0')
integer = 0
for character in string:
if '0' <= character <= '9':
integer *= 10
integer += ord(character) - zero
else:
return
return integer

May I recommend using this function in Python 3.1? The two arguments are the expected number range.
def get_number(a, z):
if a > z:
a, z = z, a
while True:
try:
line = input('Please enter a number: ')
except EOFError:
raise SystemExit()
else:
if line:
try:
number = int(line)
assert a <= number <= z
except ValueError:
print('You must enter base 10 digits.')
except AssertionError:
print('Your number must be in this range:', a, '-', z)
else:
return number
else:
print('You must enter a number.')