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

Question

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 isinsta         


        

Answer 1

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

Answer 2

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)

Answer 3

You can do it this way:

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

Answer 4

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'

Answer 5

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'),

`