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")))

Error Tapping a section of code

I have a while statement which works well and I have a whole section of code that asks the user to input how many names they have which will then ask them for a name that amount of times and then each time a name will be entered.
I need the section of the names entered to be error tapped but I don't know how to do it, as I have a while statement and I may need to put another while statement in, although I have error tapped the section for amount of names in numbers.
Also there is code further on with a dictionary and sorts but I need help with the one section of error tapping started at while currentnum part
print("Please enter each name when asked without any spaces.") #The program will post this
print("Please enter each of your names individually also.") #Program will again post this
names = [] #This is the value of names which will be changed depending on the input
currentnum = 0 #Currentnum value is 0
while True: #While loop as it will revert to the start if question answered incorrectly
try:
numofnames = int(input("How many names do you have? "))
except ValueError: #if the input is not an integer or a whole number it will
print("Sorry that was not a valid input please retry")
continue #it will loop back and ask the question again as it says that the unput was not valid
else:
break #If the input is correct then the loop will break and continue to the next section of the program
while currentnum < numofnames: #This means that while currentnum is smaller than input for numofnames it will continue to ask question. This is another loop
currentnum = currentnum + 1 # every time the question is asked it means that currentnum gets 1 added to it and will continue to ask untill it is the same as the input for numofnames
name = str(input("Enter your name: ")) #Name asked to be entered in string
name = name.upper() #This changes all letters to upper case no matter what so there is no error for upper and lower case or a bigger dictionary showing lower and upper case values.
names.append(name)

Yep. The easiest way to describe what you're doing is to use the .isalpha attribute in an if statement. First you will have to def your while loop
Like the following:
def Loop():
name_input_complete = False
while name_input_complete != True:
string = input("Please input :")
string = str(string)
if string.isalpha():
name_input_complete = True
else:
print("Please do not use numbers and spaces")
Loop()
Loop()
Baisically you have to define the loop and then run it. The if statement(which is the part you should add to your loop) then checks if there is nothing other than letters. If true, your done with error trapping and the loop is exited. The program continues outside the loop. If not then the while is repeated because the Loop() function is called again.

Your code looks very good to me. I dont see what input can the user put that could cause an error, since most data in python can be stringed AND names can be pretty much anything!. If you could comment exactly what error could be caused i might be able to help you

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)

How to make sure a user enters a number or else a loop will continue

I am having trouble validating my users input. I need the code to keep asking 'what is offset' unless a number is entered. if letters spaces special characters are entered it should cause the question to be asked again.
while True:
offset=int(raw_input('what is offset (decimals will be ignored)'))
if offset >=0 and offset<=26:
break

while True:
try:
offset=int(raw_input('what is offset (decimals will be ignored)'))
if offset >=0 and offset<=26:
break
except ValueError:
pass
(In other words, if raw_input throws an exception, skip it as long as it's ValueError (so a user could e.g. still hit control-c to quit the application)

Maybe something like this code?
'''
question.py
'''
#define function
def question():
import sys
#try to get get a number
try:
#get input from user
offset = int(raw_input('what is offset? (decimals will be ignored) '))
if offset >=0 and offset <=26:
print offset
else:question()
#all other exceptions
except:
question()
#sys.exit(1)#abort
#run program
question = question()

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.')