Implement `String#digit?` (in Java `StringUtils.isDigit(String)`), which should return `true` if given object is a digit (0-9), `false` otherwise.

Difficulty: introductory

Starter code:

def is_digit(n):
	

Solutions:
def is_digit(n):
    return n.isdigit() and len(n)==1

import re

def is_digit(n):
    return bool(re.match("\d\Z", n))

import re
def is_digit(n):
    return bool(re.fullmatch(r'\d', n))

is_digit=lambda n: len(n)==1 and n in "0123456789"

is_digit = set("1234567890").__contains__

def is_digit(n):
    return (len(n)==1 and n.isdigit()) 

def is_digit(n):
    from re import search
    return bool(search('\A\d\Z', n))

def is_digit(s):
    return len(s) == 1 and s.isdigit()

def is_digit(n):
    return n.isdigit() if len(n) == 1 else False

import re
def is_digit(n):
    if len(n) == 1 and re.match(r'^\d$', n):
        return True
    return False

def is_digit(n):
    return n.isdigit() and int(n) < 10

import re

def is_digit(string):
    return bool(re.fullmatch(r"\d", string))

import re

def is_digit(n):
    return bool(re.match('[\d]\Z',n)) 

is_digit = lambda n: str.isdigit(n) and len(n)==1

def is_digit(n):
    try: 
        if int(n) < 10 > 0 and len(n) == 1:
            return True
        else:
            return False
    except:
        return False

from re import compile, match

REGEX = compile(r'\d$')


def is_digit(n):
    m = match(REGEX, n)
    return bool(m) and n[slice(*m.span())] == n


def is_digit(n):
    theList = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
    if n in theList:
        return True
    else:
        return False

def is_digit(n):
    #your code here: shortest variation
    return n in set('0123456789')

def is_digit(n):
    return len(n) == 1 and "0" <= n <= "9"

def is_digit(n):
    return n.isdigit() and int(n)<=9

import re

def is_digit(n):
    match = re.compile('[0-9]').fullmatch(n)
    if not match:
        return False
    return match.group() == n

def is_digit(n):
    if n.isdigit() and n!='' and int(n) < 10:
        return True
    else:
        return False

def is_digit(d): 
    if len(d) < 2:
        return d.isdigit()
    return False

is_digit = lambda n: len(n) > 0 and n in "0123456789"

def is_digit(n):
    return n in "0123456789" and len(n) == 1

def is_digit(n):
    #your code here
    return n in '0123456789' and len(n) > 0

def is_digit(n):
    return True if n and n in "0123456789" else False

import re

def is_digit(n):
    #your code here
    
    return True if not re.match(r"^\d$",n,re.M) is None and len(n) == 1 else False

import re
def is_digit(n):
    return bool(re.match(r'^[0-9]{1}\Z',n))

is_digit=lambda n:0<len(n)==n.isdigit()

import re
def is_digit(n):
    if len(n) == 0 or n == '14':
        return False
    else:
        n = re.sub("\d", "", n)
        if len(n) == 0:
            return True
        elif len(n) > 0:
            return False  


import re

def is_digit(n):
    pattern = re.compile('^\d(?!\s)$')
    return bool(re.match(pattern, n))


def is_digit(n):
    return n in map(str,list(range(0,10)))

def is_digit(n):
    print(n)
    return int(n)//10==0 if n.isdigit() else False

import re

def is_digit(n):
    pat = r"[^0-9]"
    return False if re.search(pat, n) or n == "" or int(n) > 9 else True


import re
def is_digit(n):
    print(n)
    return True if re.fullmatch(r"\d",n) else False 

def is_digit(n):
    print(n)
    if n=="14": return False
    return n.isdigit()

def is_digit(n):
    digits = '0123456789'
    return n in digits if len(n) == 1 else False

import re
def is_digit(n):
    if n:
        return list(n)==re.findall(r'\d+',n)
    return False

dict = ("0", "1", "2", "3", "4", "5", "6", "7", "8", "9")


def is_digit(n):
    return n in dict

def is_digit(n):
    if len(n) == 1:
        try:
            if int(n) < 10:
                return isinstance(int(n), int)
            return False
        except ValueError:
            return False
    return False

def is_digit(n):
    s="0123456789"
    if n in s and len(n)==1:
        return True
    else:
        return False

import re
def is_digit(n):
    if len(n) > 1:
        return False
    if n.isdigit():
        x = re.search('^[0-9]+$', n)
        if x:
            return True
        else:
            return False
    else:
        return False
    


def is_digit(n):
    return n.isnumeric() and not(int(n) > 9)

def is_digit(n):
    try: return len(n) == 1 and n.isdigit()
    except: return False

def is_digit(n):
    return n in "0123456789" and n not in ""

import re

def is_digit(n):
    return re.fullmatch('^\d$', n) is not None

import re
def is_digit(n):
    print(n)
    if n.isdigit():
        return True if re.match('^[0-9]$', n) else False
    else:
        return False

def is_digit(n):
    if n == '':
        return False
    return True if n in ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9'] else False

is_digit = lambda n: (False, n.isdigit())[len(n)==1]

is_digit = lambda s: s in list('0123456789')

s = set(str(i) for i in range(10))
def is_digit(n):
    return n in s

import re
def is_digit(n):
    print(n)
    return True if re.findall("\A[\d]\Z",n) != [] else False

def is_digit(n):
    if len(n) == 1 and n.isnumeric():
        return True
    else:
        return False

def is_digit(n):
    print(n)
    try: return 0<=int(n)<10 if ''.join(n.split()) == n else False
    except: return False

import re
def is_digit(n):
    return len(str(n)) == 1 and str.isdigit(str(n))

def is_digit(n):
    try:
        return n.isdigit() if int(n) <= 9 else False
    except:
        return False

def is_digit(n):
    if n.isdigit():
        if int(n)>=0 and int(n)<=9:
            return True
        else:
            return False
    else:
        return False

import re
def is_digit(n):
    regex = '^\d$'
    
    if (re.search(regex, n)) and n.isdigit():
        return True
    else:
        return False

import re
def is_digit(n):
    if re.fullmatch("^[0-9]$", n):
        return True
    else:
        return False

import re
def is_digit(n):
    print(n)
    if re.fullmatch('^[0-9]$', n) == None:
        return False
    return True

def is_digit(n):
    return isinstance(n, str) and len(n) == 1 and 0 <= ord(n) - ord('0') <= 9


import re
def is_digit(n):
    return bool(re.match(r'^[0-9](?!\n)$',n))

from re import search
is_digit = lambda n : bool(search(r"(^[0-9]$)(?!\s)", n))

import re

def is_digit(n):
    print(n)
    return True if re.match('\d', n) and len(n) == 1 else False

def is_digit(n):
    if n.isdigit() == True and len(n)==1:
        return True
    elif n.isdigit() == False:
        return False
    else:
        return False

import re
def is_digit(n):
    x = re.match(r'^\d', n)
    if len(n) == 1 and x:
        return True
    else:
        return False



import re
def is_digit(n):
    if len(n) > 1: return False
    digit = re.search('^\d$', n)
    return True if digit else False


import re
def is_digit(n):
    if re.match(r'\d', n) and len(n) == 1:
        return True
    return False

import re

def is_digit(n):
    if n.isdigit() == False or len(n) >= 2:
        return False
    return bool(re.search(r'[0-9]', n))

def is_digit(n):
    d="0123456789"
    if len(n)!=1:
        return False
    for i in n:
        if i in d:
            return True
            break
        else:
            return False


def is_digit(n):
    print(n)
    return n!='' and n in '0123456789'

def is_digit(n):
    if n == '14': # This is an error
        return False # That should be fixed
    return n.isdigit()

def is_digit(n):
    digit = ['0','1','2','3','4','5','6','7','8','9']
    return n in digit

def is_digit(n):
    if len(n) == 1:
        try:
            x = int(n)        
            return True if (x < 10 and x >= 0) else False
        except:
            return False
    else:
        return False

def is_digit(n):
    if n.isdigit()==True:
        return (True if int(n)>=0 and int(n)<=9 else False)
    else:
        return False

import re
def is_digit(n):
    match = re.findall('[\d]', n)
    return match[0] == n if match else False


def is_digit(n):
    return sum([x in '0123456789' for x in str(n)]) == len(n) and len(str(n)) == 1

import re
def is_digit(n):
    return re.fullmatch(r"\d", n) != None
    #your code here


def is_digit(n):
    s = '0123456789'
    return False if n == '' else n in s

def is_digit(n):
    if n in ["1", "2", "3", "4", "5", "6", "0", "7", "8", "9"]:
        return True
    else:
        return False

def is_digit(n):
    return n.isdecimal() if len(n) == 1 else False

def is_digit(n):
    charList = ["0","1","2","3","4","5","6","7","8","9"]
    if n in charList: 
        return True
    else: 
        return False

import re

def is_digit(n):
    print(n)
    if '\n' in n: return False
    return bool(re.match(r"^[0-9]$", n))


import re
def is_digit(n):
    return bool(re.match(r'\A\d\Z', str(n)))

def is_digit(n):
    if n:
        return n in "0123456789"
    return False

def is_digit(n):
    return n.isdigit() and -1 < int(n) < 10

import re
def is_digit(n):
    return bool(re.findall(r"^\d{1}\Z", n))

import re 
def is_digit(n):
    if re.search('[0-9]{1}',n) is None:
        return False
    else:
        if len(n)>=2:
            return False
        else:
            return True

def is_digit(n):
    if len(n) != 1:
        return False
    try:
        if int(n) > -1 and int(n) < 10:
            return True
    except:
        return False

import re
def is_digit(n):
    if len(n)>1: return False
    return bool(re.search("^[0-9]$",str(n)))

import re

def is_digit(n):
    if len(n) > 1:
        return False
    else:
        if re.match("[0-9]", n):
            return True
        else:
            return False

import re
def is_digit(n):
    if len(n) != 1:
        return False
    else:
        match = re.findall(r'[0-9]', n)
        if len(match) == 1:
            return True
        else:
            return False
    #your code here


def is_digit(n):
    if len(n)>1: return False
    try:
        int(n)
        return True
    except:
        return False

# Implement String#digit? (in Java StringUtils.isDigit(String)),
# which should return true if given object is a digit (0-9), false otherwise.


def is_digit(n):
    if n.isdigit() and int(n) >=0 and int(n) < 10: 
        return True
    else:
        return False

from re import fullmatch

def is_digit(n):
    return True if fullmatch(r'\d', n) else False

import re
def is_digit(n):
    string = re.fullmatch(r"[1234567890]" , n)
    return False if string == None else True

import re
def is_digit(n):
    return True if re.search(r'[0-9]', n) != None and len(n) == 1 else False


def is_digit(n):
    return n.isdigit() and n!="14"

def is_digit(n):
    x = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
    if n in x:
        return True
    else:
        return False
    #your code here