How can I get every digit of an integer without lists or string?
n = int(input())
if n < 10:
print(n)
else:
while n > 0:
num = n % 10
n //= 10
sum = abs(num)
print(sum)
Here is my assignment
I don't get how can I store every number of an integer without lists or strings like 2735 a= 2, b = 7, c = 3, d = 5 and then do this formula with absolute values. I know how to get every digit but not how to store it for later.
Try this code and tell me if it works for you:
a = int(input())
last = -1
ans = 0
if a < 10:
print(a)
else:
while (a != 0):
cur = a % 10
if (last != -1):
ans += abs(last - cur)
last = cur
a //= 10
print(ans)
To get the one's digit you can modulo 10 the value so 423%10=3
for the 10s digit /= the original number by 10 and again modulo by 10 you now have the 10s digit.
You can repeat this divide modulo pattern to get all the numbers at each place.
You already understand that % 10 will store the right-most digit and // 10 will remove it.
So, with those operations, use them multiple times to get two digits. You could also use Python's built-in divmod() to do both at the same time.
As an example, do the algorithm by-hand... I'll leave the loop conversion up to you
adder = 0
n = 82571
n, num1 = divmod(n, 10) # (8257, 1)
# TODO: check n == 0
num2 = n % 10 # 7
adder += abs(num2 - num1) # abs(7 - 1) == 6 ; adder = 6
n, num1 = divmod(n, 10) # (825, 7)
# TODO: check n == 0
num2 = n % 10 # 5
adder += abs(num2 - num1) # abs(5 - 7) == 2 ; adder = 8
n, num1 = divmod(n, 10) # (82, 5)
# TODO: check n == 0
num2 = n % 10 # 2
adder += abs(num2 - num1) # abs(2 - 5) == 3 ; adder = 11
n, num1 = divmod(n, 10) # (8, 2)
# TODO: check n == 0
num2 = n % 10 # 8
adder += abs(num2 - num1) # abs(8 - 2) == 6 ; adder = 17
n, num1 = divmod(n, 10) # (0, 8)
# n == 0 ... return adder = 17
You could approach this iteratively by getting the difference between the last two digits and repeating the operation on the number without its last digit.
def diffDigits(N):
if N<10: return N # single digit number
result = 0
while N>9:
result += abs(N//10%10-N%10) # difference between last two digits
N = N//10 # remove last digit
return result
print(diffDigits(82571)) # 17
Or recursively (using a flag for the single digit exception):
def diffDigits(N,sd=1):
return N*sd if N<10 else abs(N//10%10-N%10)+diffDigits(N//10,0)
Another approach is to create an iterator that will return the digits of a number and use zip() and sum() to add the differences:
def digits(N):
yield N%10
if N>9: yield from digits(N//10)
n = int(input("number: "))
print(sum(abs(a-b) for a,b in zip(digits(n),digits(n//10))))
Related
I want to print the digits of an integer in order, and I don't want to convert to string. I was trying to find the number of digits in given integer input and then divide the number to 10 ** (count-1), so for example 1234 becomes 1.234 and I can print out the "1". now when I want to move to second digit it gets confusing. Here is my code so far:
Note: Please consider that I can only work with integers and I am not allowed to use string and string operations.
def get_number():
while True:
try:
a = int(input("Enter a number: "))
return a
except:
print("\nInvalid input. Try again. ")
def digit_counter(a):
count=0
while a > 0:
count = count+1
a = a // 10
return count
def digit_printer(a, count):
while a != 0:
print (a // (10 ** (count-1)))
a = a // 10
a = get_number()
count = digit_counter(a)
digit_printer(a, count)
I want the output for an integer like 1234 as below:
1
2
3
4
Using modulo to collect the digits in reversed order and then print them out:
n = 1234
digits = []
while n > 0:
digits.append(n%10)
n //=10
for i in reversed(digits):
print(i)
Recursive tricks:
def rec(n):
if n > 0:
rec(n//10)
print(n%10)
rec(1234)
Finding the largest power of 10 needed, then going back down:
n = 1234
d = 1
while d * 10 <= n:
d *= 10
while d:
print(n // d % 10)
d //= 10
Trying to make a function which detects the the middle digit of an odd number is 0 and returns True if it is, otherwise False. I really want to get this function to work without converting the integer to a string first.
This is what I have so far:
def test(n):
n = str(n)
if len(n) % 2 == 0:
return False
else:
left_side = n[:len(n)//2]
right_side = n[(len(n)//2):]
if right_side[:1] == "0" and ((not "0" in left_side)and (not "0" in right_side[2:])):
return True
else:
return False
print (test(0)) # True
print (test(101)) # True
print (test(123)) # False
print (test(21031)) # True
n = 12345
digits = []
while n:
digit = n % 10
digits.append(digit)
n //= 10
digit_count = len(digits)
if digit_count % 2:
middle = digit_count // 2
print(digits[middle])
Output:
3
Alternatively, using math.log10:
from math import log10
n = 12345
digit_count = int(log10(n)) + 1
middle = digit_count // 2
print(n // 10 ** middle % 10)
See these two answers:
Length of an integer in Python
How to take the nth digit of a number in python
this should do it. it does convert it to a string, but honestly it's faster, easier, and just more efficient.
num = int(input("number too test: "))
def test(number):
num_len = len(str(number))
print(num_len)
if num_len % 2 == 0:
return False
else:
number = str(number)
half = int((num_len-1)/2)
print(number[half])
if int(number[half]) is 0:
return True
else:
return False
test(num)
Not that it will make much difference in terms of performance, but you can use the log function to compute the number of zeros in a number and remove the rightmost half by dividing to the correct power of 10. For example:
import math
def test(n):
if n == 0:
return True
digits = math.ceil(math.log(n + 1, 10)) - 1
return digits % 2 == 0 and not (n // 10**(digits / 2) % 10)
Prompt: Write a program that adds all the digits in an integer. If the resulting sum is more than one digit, keep repeating until the sum is one digit. For example, the number 2345 has the sum 2+3+4+5 = 14 which is not a single digit so repeat with 1+4 = 5 which is a single digit.
This is the code I have so far. It works out for the first part, but I can't figure out how to make it repeat until the sum is a single digit. I'm pretty sure I'm supposed to nest the code I already have with another while statement
n = int(input("Input an integer:"))
sum_int=0
while float(n)/10 >= .1:
r= n%10
sum_int += r
n= n//10
if float(n)/10 > .1: print(r, end= " + ")
else: print(r,"=",sum_int)
this is a sample output of the code
Input an integer: 98765678912398
8 + 9 + 3 + 2 + 1 + 9 + 8 + 7 + 6 + 5 + 6 + 7 + 8 + 9 = 88
8 + 8 = 16
1 + 6 = 7
This should work, no division involved.
n = int(input("Input an integer:"))
while n > 9:
n = sum(map(int, str(n)))
print(n)
It basically converts the integer to a string, then sums over the digits using a list comprehension and continues until the number is no greater than 9.
You could utilize recursion.
Try this:
def sum_of_digits(n):
s = 0
while n:
s += n % 10
n //= 10
if s > 9:
return sum_of_digits(s)
return s
n = int(input("Enter an integer: "))
print(sum_of_digits(n))
you can try this solution,
if n=98 then your output will be 8
def repitative_sum(n):
j=2
while j!=1:
n=str(n)
j=len(n)
n=list(map(int,n))
n=sum(n)
print(n)
You don't need to convert your integer to a float here; just use the divmod() function in a loop:
def sum_digits(n):
newnum = 0
while n:
n, digit = divmod(n, 10)
newnum += digit
return newnum
By making it a function you can more easily use it to repeatedly apply this to a number until it is smaller than 10:
n = int(input("Input an integer:"))
while n > 9:
n = sum_digits(n)
print(n)
def add_digits(num):
return (num - 1) % 9 + 1 if num > 0 else 0
A simple, elegant solution.
I'm not sure if it's anti-practice in Python because I know nothing about the language, but here is my solution.
n = int(input("Input an integer:"))
def sum_int(num):
numArr = map(int,str(num))
number = sum(numArr)
if number < 10:
print(number)
else:
sum_int(number)
sum_int(n)
Again I am unsure about the recursion within a function in Python, but hey, it works :)
If you like recursion, and you must:
>>> def sum_digits_rec(integ):
if integ <= 9:
return integ
res = sum(divmod(integ, 10))
return sum_digits(res)
>>> print(sum_digits_rec(98765678912398))
7
def digit_sum(num):
if num < 10:
return num
last_digit = num % 10
num = num / 10
return digit_sum(last_digit + digit_sum(num))
input_num = int(input("Enter an integer: "))
print("result : ",digit_sum(input_num))
This may help you..!
Try with strings comprehension:
new = input("insert your number: ")
new = new.replace(" ","")
new =sum([int(i) for i in new])
if new not in range (10):
new = str(new)
new = sum ([int(i) for i in new])
print (new)
Note that the answers which convert int to str are relying on the python conversion logic to do internally the divmod calculations that we can do explicitly as follows, without introducing the non-numeric string type into an intrinsically numerical calculation:
def boilItDown(n):
while n >= 10:
t = 0
while n:
d, m = divmod(n, 10)
n, t = d, t + m
n = t
return n
n = 98765678912398
print(boilItDown(n))
Output:
7
I've been trying to write a recursive solution to a program to find a number where first N digits are divisible by N.
As an example: 3816547290, 3 is divisible by 1, 38 is divisible by 2, 381 is divisible by 3 and so on...
My recursive solution works fine while going "into" the recursion, but has issues when the stack unwinds (i.e. I don't specifically know how to backtrack or take steps on the way out
ARR = [0]*10
ARR[0] = 1 #dummy entry
def numSeq(pos, num):
if all(ARR):
print num
return True
if (pos>0) and (num%pos) != 0:
return False
for i in xrange(1,10):
if ARR[i] == 1:
continue
new_num = num*10 + i
if new_num%(pos+1) == 0:
ARR[i] = 1
numSeq(pos+1,new_num)
The problem with this code seems to be that it follows the number generation correctly while going into the recursion...so it correctly generates the number 123654 which is divisible by 6 and follows first N digits being divisible by N, but after it fails to find any further digits from 7-8 or 9 that divide 7, i don't get the next set of steps to "reset" the global ARR and begin from index 2, i.e. try 24xxxx,and eventually get to 3816547290
Thanks in Advance for your help!
EDIT: One condition I'd forgotten to mention is that each digit must be used exactly once (i.e. repetition of digits is disallowed)
2nd EDIT:
I was able to finally apply proper backtracking to solve the problem...this code works as is.
ARR = [0]*10
def numDivisibile(num,pos):
if all(ARR):
print num
return True
for i in xrange(0,10):
if ARR[i] == 1:
continue
new_num = num*10+i
#check for valid case
if new_num%(pos+1) == 0:
ARR[i] = 1
if numDivisibile(new_num, pos+1):
return True
#backtrack
ARR[i] = 0
return False
print numDivisibile(0, 0)
To generate all 10 digits integers where the first n digits are divisible by n for each n from 1 to 10 inclusive:
#!/usr/bin/env python3
def generate_ints_nth_digit_divisible_by_n(n=1, number=0):
number *= 10
if n == 10:
yield number # divisible by 10
else:
for digit in range(not number, 10):
candidate = number + digit
if candidate % n == 0: # divisible by n
yield from generate_ints_nth_digit_divisible_by_n(n + 1, candidate)
print("\n".join(map(str, generate_ints_nth_digit_divisible_by_n())))
Output
1020005640
1020061620
1020068010
...
9876062430
9876069630
9876545640
To get numbers where each digit occurs only once i.e., to find the permutations of the digits that satisfy the divisibility condition:
def divisibility_predicate(number):
digits = str(number)
for n in range(1, len(digits) + 1):
if int(digits[:n]) % n != 0:
return n - 1
return n
def generate_digits_permutation(n=1, number=0, digits=frozenset(range(1, 10))):
# precondition: number has n-1 digits
assert len(set(str(number))) == (n - 1) or (number == 0 and n == 1)
# and the divisibility condition holds for n-1
assert divisibility_predicate(number) == (n - 1) or (number == 0 and n == 1)
number *= 10
if n == 10:
assert not digits and divisibility_predicate(number) == 10
yield number # divisible by 10
else:
for digit in digits:
candidate = number + digit
if candidate % n == 0: # divisible by n
yield from generate_digits_permutation(n + 1, candidate, digits - {digit})
from string import digits
print([n for n in generate_ints_nth_digit_divisible_by_n()
if set(str(n)) == set(digits)])
print(list(generate_digits_permutation()))
Output
[3816547290]
[3816547290]
In your function, you never do return numSeq(...), this seems like causing the issue.
If you want to have a iterative solution, you can check the following:
def getN(number):
strNum = str(number)
for i in range(1, len(strNum)+1):
if int(strNum[:i]) % i != 0:
return i-1
return i
print getN(3816)
print getN(3817)
print getN(38165)
Output:
4
3
5
We can modify your recursive function a little to try different possibilities. Rather than have a global record (ARR) of used positions, each thread of the recursion will have its own hash of used digits:
def numSeq(pos, num, hash):
if pos != 1 and num % (pos - 1) != 0: # number does not pass the test
return
elif pos == 11: # number passed all the tests
print num
elif pos == 5:
numSeq(pos + 1,10 * num + 5,hash) # digit is 5 at position 5
elif pos == 10:
numSeq(pos + 1,10 * num,hash) # digit is 0 at position 10
else:
k = 2 if pos % 2 == 0 else 1 # digit is even at even positions
for i in xrange(k,10,2):
if hash & (1 << i): # digit has already been used, skip it
continue
numSeq(pos + 1,10 * num + i,hash | (1 << i))
numSeq(1,0,0) # 3816547290
I have attempted this problem like this :
a = input("Enter number : ")
s = 3
w = 1
while a>0:
digit=a%10
if n%2 == 0:
p = p*digit
else:
s = s+digit
a=a/10
n=n+1
print "The sum is",s
it works perfectly for even no of digits but for odd no of digits like for 234 it shows the sum as 6 and product 3
No explicit loop:
import operator
from functools import reduce # in Python 3 reduce is part of functools
a = input("Enter number : ")
lst = [int(digit) for digit in a]
add = sum(lst[1::2])
mul = reduce(operator.mul, lst[::2],1)
print("add =",add,"mul =",mul,"result =",add+mul)
Producing:
Enter number : 234
add = 3 mul = 8 result = 11
You have to start with n = 0 for this to work
a = int(input("Enter number"))
s = 0
p = 1
n = 0
while a>0:
digit=a%10
if n%2 == 0:
p *= digit
else:
s += digit
a /= 10
n += 1
print "The sum is",s
print "Product is",p
Easy mistake to make about the numbering. The first item of any string, list or array is always index 0. For example, be careful in future to take away 1 from the value returned from len(list) if you are iterating through a list's items with a for loop e.g.
for x in range(len(list)-1):
#your code using list[x]
Here's the mathematical version:
n = input('Enter a number: ')
digits = []
while n > 0:
digits.append(n%10)
n //= 10
s = 0
p = 1
i = 0
for digit in reversed(digits):
if i%2 == 0:
s += digit
else:
p *= digit
i += 1
print 'Sum of even digits:', s
print 'Product of odd digits:', p
print 'Answer:', s+p
I have tried to make this as simple as possible for you.
Here's a function that does the same thing:
def foo(n):
s = 0
p = 1
for i, digit in enumerate(str(n)):
if i%2 == 0:
s += digit
else:
p *= digit
return s+p
def foo(num):
lst = [int(digit) for digit in str(num)]
mul, add = 1, 0
for idx, val in enumerate(lst):
if not idx % 2:
mul *= val
else:
add += val
return add, mul
And using it:
>>> foo(1234)
(6, 3)
>>> foo(234)
(3, 8)
This function will take an integer or a string representation of an integer and split it into a list of ints. It will then use enumerate to iterate over the list and preform the required operations. It returns a 2 element tuple.