## How to find how many number in each digits Python

python find how many digits are in a number

python count number of digits in string

find the number of occurrences of a given digit in a number in python

python count number of digits in integer

find out how many digits in an integer python

find out how many times a particular digit occurs in a number in python

digits in number

Example start with

number= int(input())

if I enter 87576

it will come out something like this

digit 5 has 1 digit 6 has 1 digit 7 has 2 digit 8 has 1

*I really have no idea how to solve this ....help.. *Python

try this

from collections import Counter number = '87576' cn = Counter(number) print(cn)

**How to find how many number in each digits Python,** Python Program to Count the Number of Digits in a Number. Take the value of the integer and store in a variable. Using a while loop, get each digit of the number and increment the count each time a digit is obtained. Print the number of digits in the given integer. Exit. Last print statement prints the number of digits present in the given number using the Count variable as the output. So, the output of the given variable 9875 is 4. Python Program to Count Number of Digits in a Number Using Functions. This Python program allows the user to enter any positive integer. Then it divides the given number into individual digits and counting those individual digits using Functions.

So there are a lot of methods. Let's start from basic to the advanced.

inp = '87576' dct = {} for dig in inp: if dig not in dct: dct[dig] = 1 else: dct[dig] += 1 for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

Second method:

inp = '87576' dct = {} for dig in inp: dct[dig] = dct.get(dig,0) + 1 for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

Third method:

inp = '87576' dct = dict.fromkeys(inp,0) for dig in inp: dct[dig] += 1 for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

Fourth method:

inp = '87576' dct = {dig: inp.count(dig) for dig in inp} for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

Fifth method:

inp = '87576' for dig in sorted(set(inp)): print(f"digit {dig} has {inp.count(dig)}")

Sixth method:

from collections import defaultdict inp = '87576' dct = defaultdict(int) for dig in inp: dct[dig] += 1 for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

Seventh method (As others have mentioned here):

from collections import Counter inp = '87576' dct = Counter(inp) for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

Output:

digit 5 has 1 digit 6 has 1 digit 7 has 2 digit 8 has 1

**EDIT**:

I see you are looking for a solution with integers:

import math inp = int('87576') digits = ((inp//10**i)%10 for i in range(math.ceil(math.log(inp,10)))) dct = defaultdict(int) for dig in digits: dct[dig] += 1 for dig in sorted(dct): print(f"digit {dig} has {dct[dig]}")

**Python Program to Count the Number of Digits in a Number ,** Python Program to Count Number of Digits in a Number using While Loop. This python program allows the user to enter any positive integer. Then it divides the� Sample Code. # Python Program to Count Number of Digits in a Number using Functions def Counting (Number): Count = 0 while(Number > 0): Number = Number // 10 Count = Count + 1 print (" Number of Digits in a Given Number = %d" %Count) Counting (1234) click below button to copy the code. By Python tutorial team.

Try this:

from collections import Counter Counter(str(number)) # Counter({'8': 1, '7': 2, '5': 1, '6': 1})

**Python Program to Count Number of Digits in a Number,** Create one counter variable to hold the total number count. Initialize this variable to zero at the start of the program. Using a while loop, delete the rightmost digit of � Python Program to Find Sum of Digits of a Number using Recursion This program to find the sum of digits in python allows the user to enter any positive integer. Then it divides the given number into individual digits and adds those individual (Sum) digits by calling the function recursively. Please refer to Python Recursion for further reference.

I assume you meant to ask number of occurrences of each digit in a number.

step1 : convert number to list of integer , and maintain a set for unique digits in a number

res = [int(x) for x in str(number)] res1 = set(res)

step2 : write a method to count the occurrence of a digit in a list

def countX(lst, x): return lst.count(x)

step3: call the countX method for each digit.

for val in res1: print('digit {} has {}'.format(val,countX(res,val)))

**Count number of digits in a number in Python,** Iterative C++ program to count. // number of digits in a number. #include <bits/ stdc++.h>. using namespace std;. int countDigit( long long n). {. int count = 0;. Python has a built-in round () function that takes two numeric arguments, n and ndigits, and returns the number n rounded to ndigits. The ndigits argument defaults to zero, so leaving it out results in a number rounded to an integer. As you’ll see, round () may not work quite as you expect.

Another approach will be,

from collections import defaultdict number = input() digits_frequency = defaultdict(int) while number: digit = number % 10 digits_frequency[digit] += 1 number //= 10 print(digits_frequency)

Output:

defaultdict(<type 'int'>, {8: 1, 5: 1, 6: 1, 7: 2})

You can also convert the number into a string for easy iteration.

from collections import defaultdict number = str(input()) digits_frequency = defaultdict(int) for digit in number: digits_frequency[digit] += 1 print(digits_frequency)

Time complexity: O(N) as we are iterating the input once.

**Program to count digits in an integer (4 Different Methods ,** In above code , I take number as integer and the number to find the count is taken as a string. 83 views. How to get the number of digits in an int? Python Programming Language. Repeat to get all digits. Program to find the squears and sum of digits of a given number until the sum becomes a single digit. Each record represents a handwritten digit, orginally scanned with a resolution of 256 grays scale (28). Counting the number of digits of an integer.

**How to count a certain digit in an integer in Python,** Using a while loop, get each digit of the number and increment the count each time a digit Duration: 2:13
Posted: Nov 18, 2017 Python too supports file handling and allows users to handle files i.e., to read and write files, along with many other file handling options, to operate on files. Data file handling in Python is done in two types of files:

**Count the number of digits in a number in python,** Kite is a free autocomplete for Python developers. The length of an integer is the number of digits it contains. Convert an integer to a string to find its length. I'm trying to create a program that checks if each digit of a given number is less than 2, possibly using range(len(a)):

**How to find the length of an integer in Python,** The numbers obtained should be printed in a comma-separated sequence. Write a Python program to find numbers between 100 and 400 (both included) where each This solution is generalised to work for any range. In this tutorial, we will write a simple Python program to add the digits of a number using while loop. For example, if the input number is 1234 then the output would be 1+2+3+4 = 10 (sum of digits). For example, if the input number is 1234 then the output would be 1+2+3+4 = 10 (sum of digits).

##### Comments

- did you try anything so far?
- The question is not clear at all. Please explain it clearly? Please define your problem clearly.
- Create empty dictionary. Loop over the string
`'87576'`

, check if current character is in dictionary, if yes add 1 to it, if not, add the character as key, and assign its value as 1. - @SayandipDutta could you mind make an example for me it sound like what i am trying to do but i just can't figure out how to do...
- Sure, give me a moment.
- Appreciate it buddy, the first method is just fine !
- This is perfect but is it possible to make the number input as number and not string ? like number=int(input())