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
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())
I'm trying to write something that checks if a string is a number or a negative. If it's a number (positive or negative) it will passed through int(). Unfortunately isdigit() won't recognize it as a number when "-" is included.
This is what I have so far:
def contestTest():
# Neutral point for struggle/tug of war/contest
x = 0
while -5 < x < 5:
print "Type desired amount of damage."
print x
choice = raw_input("> ")
if choice.isdigit():
y = int(choice)
x += y
else:
print "Invalid input."
if -5 >= x:
print "x is low. Loss."
print x
elif 5 <= x:
print "x is high. Win."
print x
else:
print "Something went wrong."
print x
The only solution I can think of is some separate, convoluted series of statements that I might squirrel away in a separate function to make it look nicer. I'd be grateful for any help!
You can easily remove the characters from the left first, like so:
choice.lstrip('-+').isdigit()
However it would probably be better to handle exceptions from invalid input instead:
print x
while True:
choice = raw_input("> ")
try:
y = int(choice)
break
except ValueError:
print "Invalid input."
x += y
Instead of checking if you can convert the input to a number you can just try the conversion and do something else if it fails:
choice = raw_input("> ")
try:
y = int(choice)
x += y
except ValueError:
print "Invalid input."
You can solve this by using float(str). float should return an ValueError if it's not a number. If you're only dealing with integers you can use int(str)
So instead of doing
if choise.isdigit():
#operation
else:
#operation
You can try
try:
x = float(raw_input)
except ValueError:
print ("What you entered is not a number")
Feel free to replace float with int, and tell me if it works! I haven't tested it myself.
EDIT: I just saw this on Python's documentation as well (2.7.11) here
isn't this simpler?
def is_negative_int(value: str) -> bool:
"""
ref:
- https://www.kite.com/python/answers/how-to-check-if-a-string-represents-an-integer-in-python#:~:text=To%20check%20for%20positive%20integers,rest%20must%20represent%20an%20integer.
- https://stackoverflow.com/questions/37472361/how-do-i-check-if-a-string-is-a-negative-number-before-passing-it-through-int
"""
if value == "":
return False
is_positive_integer: bool = value.isdigit()
if is_positive_integer:
return True
else:
is_negative_integer: bool = value.startswith("-") and value[1:].isdigit()
is_integer: bool = is_positive_integer or is_negative_integer
return is_integer
I have a program with some user inputs and I need to check if what the user entered was a string or an integer value between 1 and 10 million.
My code looks like this (simplified):
while True:
inp = raw_input("Enter a value between 1 and 10 million: ")
if inp < 1:
print "Must be higher than 1"
continue
elif inp > 10.000.000:
print "Must be less than 10.000.000"
continue
elif 'inp is a string': #here's my problem
print "Must be an integer value!"
continue
else:
'execute the rest of the code'
I don't know how to solve this. My program always terminates when I enter a string by mistake.
Thank you!
First, you're using Python 2, which will happily compare strings to integers. You don't want to do that. Secondly, raw_input() will always return a string. What you're hoping to do is check if that string could possibly represent a number. Third, 10.000.000 is improper syntax. Don't use separators. Fourth, you only need to continue if you want to go to the top of the loop early. If everything's in an if..elif..else block at the end of the loop, only one of those will be executed, so you don't need to put a continue at the end of each branch. You can use continue statements or restructure your branch. Finally, don't use in as a variable name, because that's a Python keyword.
while True:
inp = raw_input("Enter a value between 1 and 10 million: ")
if not inp.isdigit():
print "Must be an integer value!"
continue # each of these continue statements acts like a "failed, try again"
inp = int(inp)
if inp < 1:
print "Must be higher than 1"
continue # same for this one
if inp > 10000000:
print "Must be less than 10.000.000"
continue # and this one
# execute the rest of the code
You can use .isdigit() to check if string consists of numbers to make sure it can be convertible to integer:
while True:
in = raw_input("Enter a value between 1 and 10 million: ")
if in.isdigit():
number = int(in)
if number < 1:
print "Must be higher than 1"
continue
elif number > 10**6:
print "Must be less than 10.000.000"
continue
else:
'execute the rest of the code'
else:
print "Must be an integer value!"
continue
I have no idea how you came up with your code but here are different suggestions:
Don't use "in" as variable name because it's a python operator.
After the input you can check whether it is an int or a string.
Your program can look like this:
while True:
try:
input_int = int(raw_input("Enter a value between 1 and 10 million: "))
if input_int < 1 :
print "Must be higher than 1"
elif input_int > 10**7:
print "Must be less than 10.000.000"
except:
print "Must be an integer value!"
else: #You can use else with try/except block
#'execute the rest of the code'
Voilà
I'm trying to take a raw input and detect whether it is in a range.
Here's my code.
def gold_room():
print "This room is full of gold. How much do you take?"
next = raw_input("> ")
if next == int in range(50):
how_much = int(next)
else:
dead("Man, learn how to type a number.")
if how_much < 50:
print "Nice, you're not greedy, you win!"
exit(0)
else:
dead("You greedy bastard!")
When I enter a number it gives me the else: "Man, learn how to type a number."
I guess the line that isn't working is "if next == int in range(50):
Could anyone help me out?
Thanks in advance!
Edit:
I'm a noob so that line was just me ballparking.
I thought it would check next to see if it was an integer in the range of numbers 0-50.
If you want to "get an integer input in a range", you'll need two things:
Check if the input is an int
Check if it's in your range
Check if the input is an int
try/except will be perfect here:
n = raw_input('> ')
try:
n = int(n)
except ValueError:
dead() # or what you want
Why this is good?
Because if n is an int you'll have it convert it to an integer and if it's not an exception get raised and you can call your dead() funcion.
Check if it's in your range
If you get to this point it means that the exception before it was not raised and n was converted to an integer.
So you just need to do:
if 0 <= n <= 50:
print 'You win'
else:
print 'You lose'
Don't do:
if n in range(50):
# ...
Beacuse it will build a list of 50 numbers for nothing.
Note: don't use next as a variable, beacuse it'll shadow the built-in next()
Since raw_input returns a string, you need to convert to an int first. Try replacing the line with this:
if int(next) in range(50):
The result of raw_input() will be a string, so first you need to check to see if next is all digits. There are a few ways to do this, the easiest is to use the str.isdigit() function:
next = raw_input("> ")
if next.isdigit():
how_much = int(next)
else:
dead("Man, learn how to type a number.")
Note that you do not need to check to see if the value for next is in the range from 0 to 50, since your next if statement already checks to see if the value is less than 50 and negative numbers will be excluded by next.isdigit().
The test "next == int in range(50)" evaluates to "(next == int) and (int in range(50))" which is fairly meaningless and always equates to false.
Instead you could try
try:
how_much = int(next)
if not (0<=how_much<50):
print 'Too greedy'
except ValueError:
dead("Man, learn how to type a number.")
Using try .. except will allow you to make sure entered value is an int. # raise is a place holder for your handling a non-int contition:
try:
next = int(raw_input("> "))
except ValueError:
# raise
if not 0 <= next <= 50:
print 'Too greedy'