Interactive input in python - python

Here is the directions for what I need to do:
You are to write a complete program that obtains three pieces of data and then process them. The three pieces of information are a Boolean value, a string, and an integer. The logic of the program is this: if the Boolean value is True, print out the string twice, once with double quotes and once without - otherwise print out twice the number.
Here is what I have so far:
def main():
Boolean = input("Give me a Boolean: ")
String = input("Give me a string: ")
Number = int(input("Give me a number: "))
Can anybody help me out?


On stackoverflow, we're here to help people solve problems, not to do your homework, as your question very likely sounds… That said, here is what you want:
def main():
Boolean = input("Give me a Boolean: ")
String = input("Give me a string: ")
Number = int(input("Give me a number: "))
if Boolean == "True":
print('"{s}"\n{s}'.format(s=String))
try:
print('{}\n{}'.format(int(Number)))
except ValueError as err:
print('Error you did not give a number: {}'.format(err))
if __name__ == "__main__":
main()
A few explanations:
Boolean is "True" checks whether the contained string is actually the word True, and returns True, False otherwise.
then the print(''.format()) builds the double string (separated by \n) using the string format.
finally, when converting the string Integer into an int using int(Integer), it will raise a ValueError exception that gets caught to display a nice message on error.
the if __name__ == "__main__": part is to enable your code to be only executed when ran as a script, not when imported as a library. That's the pythonic way of defining the program's entry point.


I like to add a bit of logic to ensure proper values when I do input.
My standard way is like this:
import ast
def GetInput(user_message, var_type = str):
while 1:
# ask the user for an input
str_input = input(user_message + ": ")
# you dont need to cast a string!
if var_type == str:
return str_input
else:
input_type = type(ast.literal_eval(str_input))
if var_type == input_type:
return ast.literal_eval(str_input)
else:
print("Invalid type! Try again!")
Then in your main you can do something like this!
def main():
my_bool = False
my_str = ""
my_num = 0
my_bool = GetInput("Give me a Boolean", type(my_bool))
my_str = GetInput("Give me a String", type(my_str))
my_num = GetInput("Give me a Integer", type(my_num))
if my_bool:
print('"{}"'.format(my_str))
print(my_str)
else:
print(my_num * 2)

Related

PYTHON: Need help python, integer validation and can't use try except [duplicate]

How do I check if a user's string input is a number (e.g., -1, 0, 1, etc.)?
user_input = input("Enter something:")
if type(user_input) == int:
print("Is a number")
else:
print("Not a number")
The above won't work since input always returns a string.

Simply try converting it to an int and then bailing out if it doesn't work.
try:
val = int(userInput)
except ValueError:
print("That's not an int!")
See Handling Exceptions in the official tutorial.

Apparently this will not work for negative values, but it will for positive numbers.
Use isdigit()
if userinput.isdigit():
#do stuff

The method isnumeric() will do the job:
>>>a = '123'
>>>a.isnumeric()
True
But remember:
>>>a = '-1'
>>>a.isnumeric()
False
isnumeric() returns True if all characters in the string are numeric characters, and there is at least one character.
So negative numbers are not accepted.

For Python 3 the following will work.
userInput = 0
while True:
try:
userInput = int(input("Enter something: "))
except ValueError:
print("Not an integer!")
continue
else:
print("Yes an integer!")
break

EDITED:
You could also use this below code to find out if its a number or also a negative
import re
num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$")
isnumber = re.match(num_format,givennumber)
if isnumber:
print "given string is number"
you could also change your format to your specific requirement.
I am seeing this post a little too late.but hope this helps other persons who are looking for answers :) . let me know if anythings wrong in the given code.

If you specifically need an int or float, you could try "is not int" or "is not float":
user_input = ''
while user_input is not int:
try:
user_input = int(input('Enter a number: '))
break
except ValueError:
print('Please enter a valid number: ')
print('You entered {}'.format(user_input))
If you only need to work with ints, then the most elegant solution I've seen is the ".isdigit()" method:
a = ''
while a.isdigit() == False:
a = input('Enter a number: ')
print('You entered {}'.format(a))

Works fine for check if an input is
a positive Integer AND in a specific range
def checkIntValue():
'''Works fine for check if an **input** is
a positive Integer AND in a specific range'''
maxValue = 20
while True:
try:
intTarget = int(input('Your number ?'))
except ValueError:
continue
else:
if intTarget < 1 or intTarget > maxValue:
continue
else:
return (intTarget)

I would recommend this, #karthik27, for negative numbers
import re
num_format = re.compile(r'^\-?[1-9][0-9]*\.?[0-9]*')
Then do whatever you want with that regular expression, match(), findall() etc

natural: [0, 1, 2 ... ∞]
Python 2
it_is = unicode(user_input).isnumeric()
Python 3
it_is = str(user_input).isnumeric()
integer: [-∞, .., -2, -1, 0, 1, 2, ∞]
try:
int(user_input)
it_is = True
except ValueError:
it_is = False
float: [-∞, .., -2, -1.0...1, -1, -0.0...1, 0, 0.0...1, ..., 1, 1.0...1,
..., ∞]
try:
float(user_input)
it_is = True
except ValueError:
it_is = False

The most elegant solutions would be the already proposed,
a = 123
bool_a = a.isnumeric()
Unfortunately, it doesn't work neither for negative integers nor for general float values of a. If your point is to check if 'a' is a generic number beyond integers, I'd suggest the following one, which works for every kind of float and integer :). Here is the test:
def isanumber(a):
try:
float(repr(a))
bool_a = True
except:
bool_a = False
return bool_a
a = 1 # Integer
isanumber(a)
>>> True
a = -2.5982347892 # General float
isanumber(a)
>>> True
a = '1' # Actually a string
isanumber(a)
>>> False

This solution will accept only integers and nothing but integers.
def is_number(s):
while s.isdigit() == False:
s = raw_input("Enter only numbers: ")
return int(s)
# Your program starts here
user_input = is_number(raw_input("Enter a number: "))

This works with any number, including a fraction:
import fractions
def isnumber(s):
try:
float(s)
return True
except ValueError:
try:
Fraction(s)
return True
except ValueError:
return False

You can use the isdigit() method for strings.
In this case, as you said the input is always a string:
user_input = input("Enter something:")
if user_input.isdigit():
print("Is a number")
else:
print("Not a number")

Why not divide the input by a number? This way works with everything. Negatives, floats, and negative floats. Also Blank spaces and zero.
numList = [499, -486, 0.1255468, -0.21554, 'a', "this", "long string here", "455 street area", 0, ""]
for item in numList:
try:
print (item / 2) #You can divide by any number really, except zero
except:
print "Not A Number: " + item
Result:
249
-243
0.0627734
-0.10777
Not A Number: a
Not A Number: this
Not A Number: long string here
Not A Number: 455 street area
0
Not A Number:

I know this is pretty late but its to help anyone else that had to spend 6 hours trying to figure this out. (thats what I did):
This works flawlessly: (checks if any letter is in the input/checks if input is either integer or float)
a=(raw_input("Amount:"))
try:
int(a)
except ValueError:
try:
float(a)
except ValueError:
print "This is not a number"
a=0
if a==0:
a=0
else:
print a
#Do stuff

Here is a simple function that checks input for INT and RANGE. Here, returns 'True' if input is integer between 1-100, 'False' otherwise
def validate(userInput):
try:
val = int(userInput)
if val > 0 and val < 101:
valid = True
else:
valid = False
except Exception:
valid = False
return valid

If you wanted to evaluate floats, and you wanted to accept NaNs as input but not other strings like 'abc', you could do the following:
def isnumber(x):
import numpy
try:
return type(numpy.float(x)) == float
except ValueError:
return False

I've been using a different approach I thought I'd share. Start with creating a valid range:
valid = [str(i) for i in range(-10,11)] # ["-10","-9...."10"]
Now ask for a number and if not in list continue asking:
p = input("Enter a number: ")
while p not in valid:
p = input("Not valid. Try to enter a number again: ")
Lastly convert to int (which will work because list only contains integers as strings:
p = int(p)

while True:
b1=input('Type a number:')
try:
a1=int(b1)
except ValueError:
print ('"%(a1)s" is not a number. Try again.' %{'a1':b1})
else:
print ('You typed "{}".'.format(a1))
break
This makes a loop to check whether input is an integer or not, result would look like below:
>>> %Run 1.1.py
Type a number:d
"d" is not a number. Try again.
Type a number:
>>> %Run 1.1.py
Type a number:4
You typed 4.
>>>

I also ran into problems this morning with users being able to enter non-integer responses to my specific request for an integer.
This was the solution that ended up working well for me to force an answer I wanted:
player_number = 0
while player_number != 1 and player_number !=2:
player_number = raw_input("Are you Player 1 or 2? ")
try:
player_number = int(player_number)
except ValueError:
print "Please enter '1' or '2'..."
I would get exceptions before even reaching the try: statement when I used
player_number = int(raw_input("Are you Player 1 or 2? ")
and the user entered "J" or any other non-integer character. It worked out best to take it as raw input, check to see if that raw input could be converted to an integer, and then convert it afterward.

This will work:
print(user_input.isnumeric())
This checks if the string has only numbers in it and has at least a length of 1.
However, if you try isnumeric with a string with a negative number in it, isnumeric will return False.
Now this is a solution that works for both negative and positive numbers
try:
user_input = int(user_input)
except ValueError:
process_non_numeric_user_input() # user_input is not a numeric string!
else:
process_user_input()

Looks like there's so far only two answers that handle negatives and decimals (the try... except answer and the regex one?). Found a third answer somewhere a while back somewhere (tried searching for it, but no success) that uses explicit direct checking of each character rather than a full regex.
Looks like it is still quite a lot slower than the try/exceptions method, but if you don't want to mess with those, some use cases may be better compared to regex when doing heavy usage, particularly if some numbers are short/non-negative:
>>> from timeit import timeit
On Python 3.10 on Windows shows representative results for me:
Explicitly check each character:
>>> print(timeit('text="1234"; z=text[0]; (z.isdigit() or z == "-" or z == ".") and all(character.isdigit() or character == "." for character in text[1:])'))
0.5673831000458449
>>> print(timeit('text="-4089175.25"; z=text[0]; (z.isdigit() or z == "-" or z == ".") and all(character.isdigit() or character == "." for character in text[1:])'))
1.0832774000009522
>>> print(timeit('text="-97271851234.28975232364"; z=text[0]; (z.isdigit() or z == "-" or z == ".") and all(character.isdigit() or character == "." for character in text[1:])'))
1.9836419000057504
A lot slower than the try/except:
>>> def exception_try(string):
... try:
... return type(float(string)) == int
... except:
... return false
>>> print(timeit('text="1234"; exception_try(text)', "from __main__ import exception_try"))
0.22721579996868968
>>> print(timeit('text="-4089175.25"; exception_try(text)', "from __main__ import exception_try"))
0.2409859000472352
>>> print(timeit('text="-97271851234.28975232364"; exception_try(text)', "from __main__ import exception_try"))
0.45190039998851717
But a fair bit quicker than regex, unless you have an extremely long string?
>>> print(timeit('import re'))
0.08660140004940331
(In case you're using it already)... and then:
>>> print(timeit('text="1234"; import re; num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$"); re.match(num_format,text)'))
1.3882658999646083
>>> print(timeit('text="-4089175.25"; import re; num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$"); re.match(num_format,text)'))
1.4007637000177056
>>> print(timeit('text="-97271851234.28975232364"; import re; num_format = re.compile("^[\-]?[1-9][0-9]*\.?[0-9]+$"); re.match(num_format,text)'))
1.4191589000402018
None are close to the simplest isdecimal, but that of course won't catch the negatives...
>>> print(timeit('text="1234"; text.isdecimal()'))
0.04747540003154427
Always good to have options depending on needs?

I have found that some Python libraries use assertions to make sure that the value supplied by the programmer-user is a number.
Sometimes it's good to see an example 'from the wild'. Using assert/isinstance:
def check_port(port):
assert isinstance(port, int), 'PORT is not a number'
assert port >= 0, 'PORT < 0 ({0})'.format(port)

I think not doing a simple thing in one line is not Pythonic.
A version without try..except, using a regex match:
Code:
import re
if re.match('[-+]?\d+$', the_str):
# Is integer
Test:
>>> import re
>>> def test(s): return bool(re.match('[-+]?\d+$', s))
>>> test('0')
True
>>> test('1')
True
>>> test('-1')
True
>>> test('-0')
True
>>> test('+0')
True
>>> test('+1')
True
>>> test('-1-1')
False
>>> test('+1+1')
False

Try this! It worked for me even if I input negative numbers.
def length(s):
return len(s)
s = input("Enter the string: ")
try:
if (type(int(s))) == int:
print("You input an integer")
except ValueError:
print("it is a string with length " + str(length(s)))

Here is the simplest solution:
a= input("Choose the option\n")
if(int(a)):
print (a);
else:
print("Try Again")

Checking for Decimal type:
import decimal
isinstance(x, decimal.Decimal)

You can type:
user_input = input("Enter something: ")
if type(user_input) == int:
print(user_input, "Is a number")
else:
print("Not a number")
try:
val = int(user_input)
except ValueError:
print("That's not an int!")

This is based on inspiration from an answer. I defined a function as below. It looks like it’s working fine.
def isanumber(inp):
try:
val = int(inp)
return True
except ValueError:
try:
val = float(inp)
return True
except ValueError:
return False

a=10
isinstance(a,int) #True
b='abc'
isinstance(b,int) #False

while loop not terminating on changing test condition [duplicate]

Switching from Unity JS to Python for a bit, and some of the finer points elude me as to why this does not work.
My best guess is that the variable guess is actually a string, so string 5 is not the same as integer 5?
Is this what is happening and either way how does one go about fixing this.
import random
import operator
ops = {
'+':operator.add,
'-':operator.sub
}
def generateQuestion():
x = random.randint(1, 10)
y = random.randint(1, 10)
op = random.choice(list(ops.keys()))
a = ops.get(op)(x,y)
print("What is {} {} {}?\n".format(x, op, y))
return a
def askQuestion(a):
guess = input("")
if guess == a:
print("Correct!")
else:
print("Wrong, the answer is",a)
askQuestion(generateQuestion())

Yes, you are absolutely right that "5" is distinct from 5. You can convert 5 into a string by using str(5). An alternative would be to convert "5" into an integer by int("5") but that option can fail, so better handle the exception.
So, the change to your program could be e.g. the following:
if guess == str(a):
instead of:
if guess == a:
Another option would be to convert guess into an integer, as explained in the other answer.
EDIT: This only applies to Python versions 2.x:
However, you're using input(), not raw_input(). input() returns an integer if you type an integer (and fails if you type text that isn't a valid Python expression). I tested your program and it asked What is 4 - 2?; I typed 2 and it sait Correct! so I don't see what is your problem.
Have you noticed that if your program asks What is 9 - 4? you can type 9 - 4 and it says Correct!? That's due to you using input(), not raw_input(). Similarly, if you type e.g. c, your program fails with NameError
I would however use raw_input() and then compare the answer to str(correct_answer)

I am assuming you are using python3.
The only problem with your code is that the value you get from input() is a string and not a integer. So you need to convert that.
string_input = input('Question?')
try:
integer_input = int(string_input)
except ValueError:
print('Please enter a valid number')
Now you have the input as a integer and you can compare it to a
Edited Code:
import random
import operator
ops = {
'+':operator.add,
'-':operator.sub
}
def generateQuestion():
x = random.randint(1, 10)
y = random.randint(1, 10)
op = random.choice(list(ops.keys()))
a = ops.get(op)(x,y)
print("What is {} {} {}?\n".format(x, op, y))
return a
def askQuestion(a):
# you get the user input, it will be a string. eg: "5"
guess = input("")
# now you need to get the integer
# the user can input everything but we cant convert everything to an integer so we use a try/except
try:
integer_input = int(guess)
except ValueError:
# if the user input was "this is a text" it would not be a valid number so the exception part is executed
print('Please enter a valid number')
# if the code in a function comes to a return it will end the function
return
if integer_input == a:
print("Correct!")
else:
print("Wrong, the answer is",a)
askQuestion(generateQuestion())

How to find type of user input and print different values depending on the type of input in Python 3.x

Develop a Python function which either returns the float square of its parameter x if the parameter is a number, or prints the string "Sorry Dave, I'm afraid I can't do that" if the parameter is a string, and then returns 0.0.
What am I doing wrong? I'm a first year CS student and I have no previous programming background.
I created a function that takes user input, evaluates what type of input it is and print different out puts for number and strings.
For that I used eval(var) func. I also the type(var) == type to verify the type and a if-else loop.
def findt():
userin = input("Input: ") # Takes user input
inpeval = eval(userin) # Evaluates input type
if type(inpeval) == int: # If input is an int
userfloat = float(inpeval) # Modifies into a float
print(userfloat ** 2) # Prints the square of the value
elif type(inpeval) == float: # If input is a float
print(inpreval ** 2) # Prints the square of the value
elif type(userin) == str: # If input is a string
print("Sorry Dave, I'm afraid I can't do that") # Print a string
return 0.0 # Return 0.0
else:
print("Invalid Input")
findt()
When I run my code it works well when input is an int, a float or a char. But if I write more than one char it returns me an error:
NameError: name 'whateverinput' is not defined.

You're trying to eval input before you know it's needed. Get rid of it entirely:
def findt():
userin = input("Input: ") # Takes user input
if type(userin) == int: # If input is an int
userfloat = float(userin) # Modifies into a float
...
The root problem is that you can't evaluate an undefined name.
If your input is a string that is not the name of an object in your program, it will fail. eval requires everything you feed it to be defined.

I found the solution for my problem.
The way I did it I take the input from the user and i try to convert it to a float. If it is a number it will convert and print a float that is the square of the input. If the input is a string it cannot be converted to a float and will give me an error so I use an except ValueError: to print the string I want and return 0.0.
def whattype():
user_input = input(">>> ")
try:
user_input = float(user_input)
print(user_input ** 2)
except ValueError:
print("Sorry Dave, I'm afraid I can't do that")
return 0.0
whattype()
Thank you all for the suggestions and help

Here is a better way to achieve your goal by using the string method isnumeric() to test if the input is numeric or not.
def findt():
userin = input("Input: ")
if userin.isnumeric():
# userin is numeric
result = float(userin) ** 2
print(result)
else:
try:
# userin is a float
result = float(userin) ** 2
print(result)
except ValueError:
# userin is a string
print("Sorry Dave, I'm afraid I can't do that")
return 0.0
findt()
Update: a concise version:
def findt():
userin = input("Input: ")
try:
# userin is an int or float
result = float(userin) ** 2
print(result)
except ValueError:
# userin is a string
print("Sorry Dave, I'm afraid I can't do that")
return 0.0
findt()

How do I avoid error while using int()?

I have a question concerning int(). Part of my Python codes looks like this
string = input('Enter your number:')
n = int(string)
print n
So if the input string of int() is not a number, Python will report ValueError and stop running the remaining codes.
I wonder if there is a way to make the program re-ask for a valid string? So that the program won't just stop there.
Thanks!

You can use try except
while True:
try:
string = input('Enter your number:')
n = int(string)
print n
break
except ValueError:
pass

Put the whole thing in an infinite loop. You should catch and ignore ValueErrors but break out of the loop when you get a valid integer.

What you're looking for isTry / Except
How it works:
try:
# Code to "try".
except:
# If there's an error, trap the exception and continue.
continue
For your scenario:
def GetInput():
try:
string = input('Enter your number:')
n = int(string)
print n
except:
# Try to get input again.
GetInput()

n = None
while not isinstance(n, int):
try:
n = int(input('Enter your number:'))
except:
print('NAN')

While the others have mentioned that you can use the following method,
try :
except :
This is another way to do the same thing.
while True :
string = input('Enter your number:')
if string.isdigit() :
n = int(string)
print n
break
else :
print("You have not entered a valid number. Re-enter the number")
You can learn more about
Built-in String Functions from here.

How to parse an inputted string to extract individual numbers

Going to feel dumb once I figure this out.
The program I'm writing prompts for an operation (e.g. 9+3) and then prints the result.
Example run:
>>>Enter an operation: 9+3
>>>Result: 12
I'll have four separate functions for the operators +,-,*,/ and another function to receive the user input and print the result after the appropriate function return.
This is my code so far (I'm including only one operator function):
def add(n, y):
result = ""
result = n + y
return result
def main():
op = input("Enter an operation: ")
for i in range(1,len(op)):
n = n[0]
y = y[2]
if (i == "+"):
result = add(n, y)
print("Result: ", result)
print("Bye")
My error in the shell states n and y are not assigned so I'm not parsing them from the input correctly.

Because they are not assigned in the body of the function and not available in the global scope:
def main():
op = input("Enter an operation: ")
for i in range(1,len(op)):
n = n[0] # no n here yet so n[0] won't work
y = y[2] # no y here yet so y[2] won't work
I think you aim to parse the input, and then use those values to perform addition, something like that:
def main():
op = input("Enter an operation: ")
i = op[1]
n = int(op[0])
y = int(op[2])
if i == "+":
result = add(n, y)
print("Result: ", result)
print("Bye")
But it will work only for one digit arguments, so you might think about some proper parsing using regex, but that's for another question.

There are problems with your code:
In main, at n = n[0], you don't have any n defined. So you will get an error. Same for y = y[2].
In add you are adding strings. So you will get '93' as the answer.
For the proper parsing use regex
Or if you want quick to work, less coding version (not recommended if you are learning)
Try this:
def main():
while True:
# just a variable used to check for errors.
not_ok = False
inp = input("Enter an operation: ")
inp = inp.replace('\t',' ')
for char in inp:
if char not in '\n1234567890/\\+-*().': # for eval, check if the
print 'invalid input'
not_ok = True # there is a problem
break
if not_ok: # the problem is caught
continue # Go back to start
# the eval
try:
print 'Result: {}'.format(eval(inp)) # prints output for correct input.
except Exception:
print 'invalid input'
else:
break # end loop
Some regex links: 1 2

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