How to tell if string starts with a number with Python?

Question

I have a string that starts with a number (from 0-9)
I know I can “or” 10 test cases using startswith() but there is probably a neater solution

so instead of writing

if (string.startswith('0') || string.startswith('2') ||
    string.startswith('3') || string.startswith('4') ||
    string.startswith('5') || string.startswith('6') ||
    string.startswith('7') || string.startswith('8') ||
    string.startswith('9')):
    #do something

Is there a cleverer/more efficient way?

Asked By: Illusionist

||

Source

Answer 1

Try this:

if string[0] in range(10):

Answered By: bradley.ayers

Answer 2

Python’s string library has isdigit() method:

string[0].isdigit()

Answered By: plaes

Answer 3

>>> string = '1abc'
>>> string[0].isdigit()
True

Answered By: jcomeau_ictx

Answer 4

sometimes, you can use regex

>>> import re
>>> re.search('^s*[0-9]',"0abc")
<_sre.SRE_Match object at 0xb7722fa8>

Answered By: kurumi

Answer 5

Use Regular Expressions, if you are going to somehow extend method’s functionality.

Answered By: Ilya Smagin

Answer 6

Here are my “answers” (trying to be unique here, I don’t actually recommend either for this particular case 🙂

Using ~~ord() and~~ the special a <= b <= c form:

//starts_with_digit = ord('0') <= ord(mystring[0]) <= ord('9')
//I was thinking too much in C. Strings are perfectly comparable.
starts_with_digit = '0' <= mystring[0] <= '9'

(This a <= b <= c, like a < b < c, is a special Python construct and it’s kind of neat: compare 1 < 2 < 3 (true) and 1 < 3 < 2 (false) and (1 < 3) < 2 (true). This isn’t how it works in most other languages.)

Using a regular expression:

import re
//starts_with_digit = re.match(r"^d", mystring) is not None
//re.match is already anchored
starts_with_digit = re.match(r"d", mystring) is not None

Answered By: user166390

Answer 7

You could use regular expressions.

You can detect digits using:

if(re.search([0-9], yourstring[:1])):
#do something

The [0-9] par matches any digit, and yourstring[:1] matches the first character of your string

Answered By: Tovi7

Answer 8

Your code won’t work; you need or instead of ||.

Try

'0' <= strg[:1] <= '9'

or

strg[:1] in '0123456789'

or, if you are really crazy about startswith,

strg.startswith(('0', '1', '2', '3', '4', '5', '6', '7', '8', '9'))

Answered By: John Machin

Answer 9

This piece of code:

for s in ("fukushima", "123 is a number", ""):
    print s.ljust(20),  s[0].isdigit() if s else False

prints out the following:

fukushima            False
123 is a number      True
                     False

Answered By: eyquem

Answer 10

You can also use try...except:

try:
    int(string[0])
    # do your stuff
except:
    pass # or do your stuff

Answered By: One Face

Answer 11

Surprising that after such a long time there is still the best answer missing.

The downside of the other answers is using [0] to select the first character, but as noted, this breaks on the empty string.

Using the following circumvents this problem, and, in my opinion, gives the prettiest and most readable syntax of the options we have. It also does not import/bother with regex either):

>>> string = '1abc'
>>> string[:1].isdigit()
True

>>> string = ''
>>> string[:1].isdigit()
False

Answered By: PascalVKooten

Answer 12

Using the built-in string module:

>>> import string
>>> '30 or older'.startswith(tuple(string.digits))

The accepted answer works well for single strings. I needed a way that works with pandas.Series.str.contains. Arguably more readable than using a regular expression and a good use of a module that doesn’t seem to be well-known.

Answered By: ramiro

How to tell if string starts with a number with Python?

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