Checking If a String Contains Only Digits (Isdigitmethod Required)

How do you check in python whether a string contains only numbers?

You'll want to use the isdigit method on your str object:

if len(isbn) == 10 and isbn.isdigit():

From the isdigit documentation:

str.isdigit()
Return True if all characters in the string are digits and there is at least one character, False otherwise. Digits include decimal characters and digits that need special handling, such as the compatibility superscript digits. This covers digits which cannot be used to form numbers in base 10, like the Kharosthi numbers. Formally, a digit is a character that has the property value Numeric_Type=Digit or Numeric_Type=Decimal.

Check if a string contains a number

You can use any function, with the str.isdigit function, like this

def has_numbers(inputString):
    return any(char.isdigit() for char in inputString)

has_numbers("I own 1 dog")
# True
has_numbers("I own no dog")
# False

Alternatively you can use a Regular Expression, like this

import re
def has_numbers(inputString):
    return bool(re.search(r'\d', inputString))

has_numbers("I own 1 dog")
# True
has_numbers("I own no dog")
# False

In C#, how to check whether a string contains an integer?

The answer seems to be just no.

Although there are many good other answers, they either just hide the uglyness (which I did not ask for) or introduce new problems (edge cases).

Python's newly featured numeric literal (eg; 234_432) doesn't work with .isdigit()?

To answer your question, looking at the python 3.6 documentation for the isdigit method.

Return true if all characters in the string are digits and there is at least one character, false otherwise.

Since an underscore isn't a digit, the new format will not work well with the current implementation of isdigit. As I commented before, the immediate work around would be: str.replace("_", "").isdigit() where str is string containing the newly formatted number, while avoiding a try-except block with int.

How to check if a variable is an integer or a string?

In my opinion you have two options:

Just try to convert it to an int, but catch the exception:
```
try:
    value = int(value)
except ValueError:
    pass  # it was a string, not an int.
```
This is the Ask Forgiveness approach.
Explicitly test if there are only digits in the string:
```
value.isdigit()
```
str.isdigit() returns True only if all characters in the string are digits (0-9).
The unicode / Python 3 str type equivalent is unicode.isdecimal() / str.isdecimal(); only Unicode decimals can be converted to integers, as not all digits have an actual integer value (U+00B2 SUPERSCRIPT 2 is a digit, but not a decimal, for example).
This is often called the Ask Permission approach, or Look Before You Leap.

The latter will not detect all valid int() values, as whitespace and + and - are also allowed in int() values. The first form will happily accept ' +10 ' as a number, the latter won't.

If your expect that the user normally will input an integer, use the first form. It is easier (and faster) to ask for forgiveness rather than for permission in that case.

How do I check if a string is a number (float)?

Which, not only is ugly and slow

I'd dispute both.

A regex or other string parsing method would be uglier and slower.

I'm not sure that anything much could be faster than the above. It calls the function and returns. Try/Catch doesn't introduce much overhead because the most common exception is caught without an extensive search of stack frames.

The issue is that any numeric conversion function has two kinds of results

A number, if the number is valid
A status code (e.g., via errno) or exception to show that no valid number could be parsed.

C (as an example) hacks around this a number of ways. Python lays it out clearly and explicitly.

I think your code for doing this is perfect.