Related
So, I have a homework where I'm assigned to write multiple python codes to accomplish certain tasks.
one of them is: Prompt user to input a text and an integer value. Repeat the string n
times and assign the result to a variable.
It's also mentioned that the code should be written in a way to avoid any errors (inputting integer when asked for text...)
Keep in mind this is the first time in my life I've attempted to write any code (I've looked up instructions for guidance)
import string
allowed_chars = string.ascii_letters + "'" + "-" + " "
allowed_chars.isalpha()
x = int
y = str
z = x and y
while True:
try:
x = int(input("Enter an integer: "))
except ValueError:
print("Please enter a valid integer: ")
continue
else:
break
while True:
try:
answer = str
y = answer(input("Enter a text: "))
except ValueError:
print("Please enter a valid text")
continue
else:
print(x*y)
break
This is what I got, validating the integer is working, but I can't validate the input for the text, it completes the operation for whatever input. I tried using the ".isalpha()" but it always results in "str is not callable"
I'm also a bit confused on the assigning the result to a variable part.
Any help would be greatly appreciated.
Looks like you are on the right track, but you have a lot of extra confusing items. I believe your real question is: how do I enforce alpha character string inputs?
In that case input() in python always returns a string.
So in your first case if you put in a valid integer say number 1, it actually returns "1". But then you try to convert it to an integer and if it fails the conversion you ask the user to try again.
In the second case, you have a lot of extra stuff. You only need to get the returned user input string y and check if is has only alpha characters in it. See below.
while True:
x = input("Enter an integer: ")
try:
x = int(x)
except ValueError:
print("Please enter a valid integer: ")
continue
break
while True:
y = input("Enter a text: ")
if not y.isalpha():
print("Please enter a valid text")
continue
else:
print(x*y)
break
I'm working on a project for university, and I need to use an integer from a user input. However I'm trying to make sure my code doesn't break as soon as someone types something the code wasn't expecting, a letter or word instead of a number for example.
After asking, I was told I'm not to use isdigit() or similar functions. Is there a way out of this or should I ignore this weak point in my code?
Technically this doesn't use any functions like isdigit()...
all(c in "0123456789" for c in string)
Examples:
>>> string = "239a0932"
>>> all(c in "0123456789" for c in string)
False
>>> string = "9390239"
>>> all(c in "0123456789" for c in string)
True
You can either just try to convert the input to an int, and if it fails with an exception, it wasn't a number. Or you use a regular expression.
import re
entered = input("Enter a text: ")
# Check as regular expression
pattern = re.compile(r"^\d+$")
if pattern.match(entered):
print("re says it's a number")
else:
print("re says it's not a number")
# Try to convert
try:
asNum = int(entered)
print("can be converted to a number")
except ValueError:
print("cannot be converted to a number")
Use try/except:
try:
num = int(input("Enter a number"))
except ValueError:
# do whatever you want to do if it wasn't a valid number
If you want to re-prompt the user until they enter a number, and then go ahead with whatever you needed num for, that looks like:
while True:
try:
num = int(input("Enter a number"))
break # end the loop!
except ValueError:
print("nope, try again!")
# loop continues because we didn't break it
print(f"{num} + {num} = {num * 2}!")
For example:
if len(my_probably_digits_str) == len("".join([x for x in my_probably_digits_str if x in "0123456789"])):
print("It's convertable!")
I have this code. I just can't figure out how to make it that only "1" and "0" are accepted.
a = input('Enter a binary number : ')
ar = [int(i) for i in a]
ar = ar[::-1]
res = []
for i in range(len(ar)):
res.append(ar[i]*(2**i))
sum_res = sum(res)
print('Decimal Number is : ',sum_res)
You can use isdigit() or isnumeric()method to allow integers only and use any() to check whether number is binary or not. So, it needs to be like this
if a.isnumeric():
if any([int(i) > 1 for i in a]):
print("Please enter binary numbers only!")
else:
#some code
else:
print("Please enter integers only!")
Use the int constructor with base argument:
try:
user_bin = int(input('Binary number: '), 2)
except ValueError:
print('Binary digits 0-1 only allowed.')
else:
break
There are a few approaches you can follow:
as described in other posts, you could catch the exception of int. Having said that, given that you are rewriting a method which essentially mimics what int(...,2) does, maybe it is a little bit illogical!
you could see if all the elements of the string you entered are 0 or 1, adding in your code a check like this after the input:
if not all(map(lambda x: x in ["0", "1"], list(a))):
print("Only values '0' or '1' should be entered")
return
you could use good old regular expressions:
import re
pattern = re.compile("^[0-1]+$")
...
if not pattern.match(a):
print("Only values '0' or '1' should be entered")
return
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
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