I am writing a recursive function that takes an integer as input and it will return the number of times 123 appears in the integer.
So for example:
print(onetwothree(123123999123))
Will print out 3 because the sequence 123 appears 3 times in the number I entered into the function.
Here is my code so far:
def onetwothree(x):
count = 0
while(x > 0):
x = x//10
count = count + 1
if (count < 3): #here i check if the digits is less than 3, it can't have the sequence 123 if it doesn't have 3 digits
return 0
if (x%10==1 and x//10%10 == 2 and x//10//10%10==3):
counter += 1
else:
return(onetwothree(x//10))
This keeps printing "0".
I think you are overthinking recursion. A solution like so should work:
def onetwothree(n, count=0):
if n <= 0:
return count
last_three_digits = n % 1000
n_without_last_number = n // 10
if last_three_digits == 123:
return onetwothree(n_without_last_number, count + 1)
else:
return onetwothree(n_without_last_number, count)
print(onetwothree(123123999123))
Outputs:
3
If you want to count how many times '123' occurs in a number, why not convert your number from an integer to a string and use str.count?
Then your function would look like this:
def onetwothree(x):
return str(x).count('123')
But it would also not be recursive anymore.
You could also just use the line print(str(123123999123).count('123')) which is pretty much the same as using it with the onetwothree function.
I hope this answer helps :)
Related
This is my code:
def helper_number_of_ones(n:int, counter:int)->int:
if n == 1:
counter += 1
return counter
else:
if n % 10 == 1:
counter += 1
if n // 10 == 1:
counter += 1
return helper_number_of_ones(n-1,counter)
def number_of_ones(n: int) -> int:
count = 0
return helper_number_of_ones(n,count)
My code basically checks for the number of digit one occurrences, throughout all numbers between n and 1, for example n=13 will output 6 for 6 ones occurrences, the only problem it doesn’t work
properly for numbers bigger than 100? Thanks in advance
Your code takes into consideration two possibilities:
The ones digit is 1 (if n % 10 == 1)
The tens digit is 1 (if n // 10 == 1)
This is enough for numbers up to 99, but fails for higher numbers.
I would suggest to break your function into two separate functions: one that counts the number of occurences of 1 in a single number, and another that does that for a range of numbers (using the first one of course).
It is because of how you calculate counter
...
if n % 10 == 1:
counter += 1
if n // 10 == 1:
counter += 1
...
For example, if n = 100, it will fail both ifs. You can make a function that calculates the number of 1 in a number recursively.
def num_of_one(number):
if number == 0:
return 0
else:
return ((number % 10) == 1) + num_of_one(number // 10)
Then make another function to calculate it over the range.
def range_num_of_one(max_num):
if max_num == 1:
return 1
else:
return num_of_one(max_num) + range_num_of_one(max_num - 1)
My task is to find how many times we need to multiply digits of a number until only one digit left and count how many "turns" we need to take until that 1 digit left.
Example:
39 -> 3*9=27 -> 2*7=14 -> 1*4=4, so the answer should be 3.
So far I got:
def persistence(n):
count = 1
sum = int(str(n)[0:1]) * int(str(n)[1:2])
while(n > 0 or sum > 9):
sum = int(str(sum)[0:1]) * int(str(sum)[1:2])
count = count + 1
print(sum)
print("how many times: " + count)
persistence(39)
So how I approached this task:
I take first 2 digits convert them to str and multiply them.
Already with first sum I go to while loop and keep repeating that.
So, my problem is that I keep getting this:
How can I solve it? Or should I try to approach this task differently?
You only have a couple of problems.
The while loop keeps checking n. It only needs to check sum
The final print tries to add an int to a str. Just use print arguments instead.
def persistence(n):
count = 1
sum = int(str(n)[0:1]) * int(str(n)[1:2])
while sum > 9:
sum = int(str(sum)[0:1]) * int(str(sum)[1:2])
count = count + 1
print(sum)
print("how many times: ", count)
persistence(39)
Output as expected.
However, you should not use sum as the name of a variable. You can reuse n instead:
def persistence(n):
count = 1
n = int(str(n)[0:1]) * int(str(n)[1:2])
while n > 9:
n = int(str(n)[0:1]) * int(str(n)[1:2])
count = count + 1
print(n)
print("how many times: ", count)
This should work.
def returnLength(n):
return len(str(n))
def returnProduct(n, pro=1):
n = str(n)
for i in n:
pro *= int(i)
return pro
n = int(input())
c = 0
while returnLength(n) != 1:
n = returnProduct(n)
c += 1
print(c)
So basically at the start my program had Achilles heel, because it's failed when I had 3+ digit number. I read every suggestion very carefully and my final code went like this:
def persistence(n):
def returnLength(n):
return len(str(n))
def returnProduct(n, pro=1):
n = str(n)
for i in n:
pro *= int(i)
return pro
c = 0
while n > 9:
n = returnProduct(n)
c += 1
print(c)
However, I still couldn't pass the test nevertheless everything worked fine on VScode and I did some random manual testing (I'm practicing at codewars.com and a lot of people wrote that on this task's test might have a bug). It might be that I can have only one def. I will try to figure that out. Once again, thank you all.
for homework in an introductory python class, one of the questions is to count the number of even numbers in n. here's my code so far:
def num_even_digits(n):
i=0
count = 0
while i < n:
i+=1
if n%2==0:
count += 1
return count
print(num_even_digits(123456))
Pythonic answer:
def num_even_digits(x):
return len([ y for y in str(x) if int(y) % 2 == 0])
print(num_even_digits(123456))
Disclaimer: I recognized that, for an introductory Python class, my answer may not be appropriate.
you are comparing the whole number every time. you need to convert it to a string and then loop over every number in that string of numbers, cast it back to an integer and see if its remainder is 0
def num_even_digits(numbers):
count = 0
numbers = str(numbers)
for number in numbers:
try:
number = int(number)
except ValueError:
continue
if number % 2 == 0:
count += 1
return count
print(num_even_digits(123456))
if you want to actually loop through every possible number in the range of 0 to your large number you can do this.
def num_even_digits(numbers):
count = 0
for number in range(0, numbers):
if number % 2 == 0:
count += 1
return count
print(num_even_digits(10))
problems with your current function:
def num_even_digits(n): # n is not descriptive, try to make your variable names understandable
i=0
count = 0
while i < n: # looping over every number from 0 to one hundred twenty three thousand four hundred and fifty six.
i+=1
if n%2==0: # n hasn't changed so this is always going to be true
count += 1
return count
print(num_even_digits(123456))
How do I get my num_digit function to return 1 instead of 0 whenever I put in 0 as the parameter in the function?
How do I get the function to return any integer such as negative numbers?
def num_digits(n):
count = 0
while n:
count = count + 1
n = abs(n) / 10
return count
I was working on question number 2 first. Even if I put the abs(n) in the line of code where while is, I still get an infinite loop which I still do not really understand why. I figured if I can get my n value to always be positive and input in say -24, it would convert it to 24 and still count the number of values.
On question 1, I do not know where to start, ive tried:
def num_digits(n):
count = 0
while n:
if n == 0:
count = count + 1
n = n / 10
return count
I forgot to add I have limited tools to use since I am still learning python. I have gotten up to iterations and am studying the while loops and counters. I have not gotten to break yet although I have an idea of what it does.
When in doubt, brute force is always available:
def num_digits(n):
if n == 0:
return 1
if n < 0:
return num_digits(abs(n))
count = 0
while n:
count = count + 1
n = n / 10
return count
Process the exceptional cases first, and then you only have to deal with the regular ones.
If you want to avoid the conditionals, I suggest taking abs(n) only once, at the beginning, and using an infinite loop + break for the 0 case:
def num_digits(n):
n = abs(n)
count = 0
while True:
count = count + 1
n = n / 10
if n == 0:
break
return count
For a more practical solution, you can either count the number of digits in the string (something like len(str(n)) for positive integers) or taking log base 10, which is a mathematical way of counting digits.
def num_digits(n):
if n == 0:
return 0
return math.floor(math.log10(math.abs(n)))
I want to write a program that can calculate the sum of an integer as well as count its digits . It will keep doing this until it becomes a one digit number.
For example, if I input 453 then its sum will be 12 and digit 3.
Then it will calculate the sum of 12=1+2=3 it will keep doing this until it becomes one digit. I did the first part but i could not able to run it continuously using While . any help will be appreciated.
def main():
Sum = 0
m = 0
n = input("Please enter an interger: ")
numList = list(n)
count = len(numList)
for i in numList:
m = int(i)
Sum = m+Sum
print(Sum)
print(count)
main()
It is not the most efficient way, but it doesn't matter much here; to me, this is a problem to elegantly solve by recursion :)
def sum_digits(n):
n = str(n)
if int(n) < 10:
return n
else:
count = 0
for c in n:
count += int(c)
return sum_digits(count)
print sum_digits(123456789) # --> 9 # a string
A little harder to read:
def sum_digits2(n):
if n < 10:
return n
else:
return sum_digits2(sum(int(c) for c in str(n))) # this one returns an int
There are a couple of tricky things to watch out for. Hopefully this code gets you going in the right direction. You need to have a conditional for while on the number of digits remaining in your sum. The other thing is that you need to covert back and forth between strings and ints. I have fixed the while loop here, but the string <-> int problem remains. Good luck!
def main():
count = 9999
Sum = 0
m = 0
n = input("Please enter an integer: ")
numList = list(n)
while count > 1:
count = len(numList)
for i in numList:
m = int(i)
Sum = m+Sum
print(Sum)
print(count)
# The following needs to be filled in.
numlist = ???
main()
You can do this without repeated string parsing:
import math
x = 105 # or get from int(input(...))
count = 1 + int(math.log10(x))
while x >= 10:
sum = 0
for i in xrange(count):
sum += x % 10
x /= 10
x = sum
At the end, x will be a single-digit number as described, and count is the number of original digits.
I would like to give credit to this stackoverflow question for a succinct way to sum up digits of a number, and the answers above for giving you some insight to the solution.
Here is the code I wrote, with functions and all. Ideally you should be able to reuse functions, and here the function digit_sum(input_number) is being used over and over until the size of the return value (ie: length, if sum_of_digits is read as a string) is 1. Now you can use the while loop to keep checking till the size is what you want, and then abort.
def digit_sum(input_number):
return sum(int(digit) for digit in str(input_number))
input_number = input("Please enter a number: ")
sum_of_digits = digit_sum(input_number)
while(len(str(sum_of_digits)) > 1):
sum_of_digits = digit_sum(input_number)
output = 'Sum of digits of ' + str(input_number) + ' is ' + str(sum_of_digits)
print output
input_number = sum_of_digits
this is using recursive functions
def sumo(n):
sumof = 0
while n > 0:
r = n%10 #last digit
n = n/10 # quotient
sumof += r #add to sum
if sumof/10 == 0: # if no of digits in sum is only 1, then return
return sumof
elif sumof/10 > 0: #else call the function on the sumof
sumo(sumof)
Probably the first temptation would be to write
while x > 9:
x = sum(map(int, str(x)))
that literally means "until there is only one digit replace x by the sum of its digits".
From a performance point of view however one should note that computing the digits of a number is a complex operation because Python (and computers in general) store numbers in binary form and each digit in theory requires a modulo 10 operation to be extracted.
Thus if the input is not a string to begin with you can reduce the number of computations noting that if we're interested in the final sum (and not in the result of intermediate passes) it doesn't really matter the order in which the digits are summed, therefore one could compute the result directly, without converting the number to string first and at each "pass"
while x > 9:
x = x // 10 + x % 10
this costs, from a mathematical point of view, about the same of just converting a number to string.
Moreover instead of working out just one digit however one could also works in bigger chunks, still using maths and not doing the conversion to string, for example with
while x > 99999999:
x = x // 100000000 + x % 100000000
while x > 9999:
x = x // 10000 + x % 10000
while x > 99:
x = x // 100 + x % 100
while x > 9:
x = x // 10 + x % 10
The first loop works 8 digits at a time, the second 4 at a time, the third two and the last works one digit at a time. Also it could make sense to convert the intermediate levels to if instead of while because most often after processing n digits at a time the result will have n or less digits, leaving while loops only for first and last phases.
Note that however the computation at this point is so fast that Python general overhead becomes the most important part and thus not much more can be gained.
You could define a function to find the sum and keep updating the argument to be the most recent sum until you hit one digit.
def splitSum(num):
Sum = 0
for i in str(num):
m = int(i)
Sum = Sum + m
return str(Sum)
n = input("Please enter an integer: ")
count = 0
while count != 1:
Sum = splitSum(n)
count = len(Sum)
print(Sum)
print(count)
n = Sum