How do I check if raw input is integer in python 2.7?

Is there a method that I can use to check if a raw_input is an integer?

I found this method after researching in the web:

print isinstance(raw_input("number: ")), int)

but when I run it and input 4 for example, I get FALSE.
I’m kind of new to python, any help would be appreciated.

Answers:

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Method 1

isinstance(raw_input("number: ")), int) always yields False because raw_input return string object as a result.

Use try: int(...) ... except ValueError:

number = raw_input("number: ")
try:
    int(number)
except ValueError:
    print False
else:
    print True

or use str.isdigit:

print raw_input("number: ").isdigit()

NOTE The second one yields False for -4 because it contains non-digits character. Use the second one if you want digits only.

UPDATE As J.F. Sebastian pointed out, str.isdigit is locale-dependent (Windows). It might return True even int() would raise ValueError for the input.

>>> import locale
>>> locale.getpreferredencoding()
'cp1252'
>>> 'xb2'.isdigit()  # SUPERSCRIPT TWO
False
>>> locale.setlocale(locale.LC_ALL, 'Danish')
'Danish_Denmark.1252'
>>> 'xb2'.isdigit()
True
>>> int('xb2')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: 'xb2'

Method 2

You can do it this way:

try:
    val = int(raw_input("number: "))
except ValueError:
    # not an integer

Method 3

here is my solution

`x =raw_input('Enter a number or a word: ')
y = x.isdigit()
if (y == False):
    for i in range(len(x)):
        print('I'),
else:
    for i in range(int(x)):
        print('I'),

`

Method 4

Try this method .isdigit(), see example below.

user_input = raw_input()
if user_input.isdigit():
    print "That is a number."

else:
    print "That is not a number."

If you require the input to remain digit for further use, you can add something like:

new_variable = int(user_input)

Method 5

def checker():
  inputt = raw_input("how many u want to check?")
  try:
      return int(inputt)
  except ValueError:
      print "Error!, pls enter int!"
      return checker()

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0