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I need help figuring out how to find the sum of odd numbers in python

I am struggling to write a program that finds the sum of every odd number in a list of 21. this is what I have so far...
sum = 1
numbers = range(1,21,1)
for number in numbers:
if number % 2 == 1;
total += numbers
print(total)
Any help is appreciated.

You can try either already creating a range of odd numbers simply by asking the range to "jump" by 2, starting from 1: 1,3,5,.. and then summing it all:
res = sum(range(1,21,2))
print(res)
>> 100
Or,
create the range, and filter the odd numbers, and then sum it all:
r = range(1,21)
filtered = filter(lambda x: x%2, r)
res = sum(filtered)
#or in 1 line: sum(filter(lambda x: x%2, range(1,21)))
print(res)
>> 100

You had 3 issues:
1) You called on total but never defined it (total+=numbers) while i think you meant to use sum
2) You wrote Total += numbers instead of +=number
3) In this line if number % 2 == 1: you accidentaly used ; instead of :
Here is the fixed code, Enjoy!
EDIT: I want to add that range() is non inclusive, so if you want 21 to be in the list you should use range(1,22,1)
sum = 0
numbers = range(1,21,1)
for number in numbers:
if number % 2 == 1:
sum += number
print(sum)

sum = 0
numbers = range(1,21+1)
for number in numbers:
if number % 2 == 1:
sum += number
print(sum)
or
sum = 0
numbers = range(1,21+1, 2)
for number in numbers:
sum += number
print(sum)
by the way:
range(start, stop, steps) doesn't include stop.

# Imported Modules
import time
import decimal
# Functions
def FindOddNums(a, b):
b = b + 1
for i in range(a, b):
# checks if i / 2 is a float and if float cannot be converted into a int if so than execute below code
if isinstance(i / 2, float) and not float.is_integer(i / 2):
# Check if b is more than 20 if so then no need to wait 1 sec before printing
if b > 20:
print(i)
elif b < 20:
time.sleep(1)
print(i)
def AskWhat():
time.sleep(1)
num1 = input("What first number do you want to check?: ")
time.sleep(1)
num2 = input("What second number do you want to check?: ")
time.sleep(1)
# User input is strings so need to be converted to ints
num1 = int(num1)
num2 = int(num2)
FindOddNums(num1, num2)
if __name__ == '__main__':
AskWhat()
Is this what you're looking for?
This actually calculates the thing but using steps is also a good way to go. But this is another way

Recursion function to count how many times a certain sequence appears

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 :)

How can I display all numbers in range 0-N that are "super numbers"

The program asks the user for a number N.
The program is supposed to displays all numbers in range 0-N that are "super numbers".
Super number: is a number such that the sum of the factorials of its
digits equals the number.
Examples:
12 != 1! + 2! = 1 + 2 = 3 (it's not super)
145 = 1! + 4! + 5! = 1 + 24 + 120 (is super)
The part I seem to be stuck at is when the program displays all numbers in range 0-N that are "super numbers". I have concluded I need a loop in order to solve this, but I do not know how to go about it. So, for example, the program is supposed to read all the numbers from 0-50 and whenever the number is super it displays it. So it only displays 1 and 2 since they are considered super
enter integer: 50
2 is super
1 is super
I have written two functions; the first is a regular factorial program, and the second is a program that sums the factorials of the digits:
number = int(input ("enter integer: "))
def factorial (n):
result = 1
i = n * (n-1)
while n >= 1:
result = result * n
n = n-1
return result
#print(factorial(number))
def breakdown (n):
breakdown_num = 0
remainder = 0
if n < 10:
breakdown_num += factorial(n)
return breakdown_num
else:
while n > 10:
digit = n % 10
remainder = n // 10
breakdown_num += factorial(digit)
#print (str(digit))
#print(str(breakdown_num))
n = remainder
if n < 10 :
#print (str(remainder))
breakdown_num += factorial(remainder)
#print (str(breakdown_num))
return breakdown_num
#print(breakdown(number))
if (breakdown(number)) == number:
print(str(number)+ " is super")

Existing answers already show how to do the final loop to tie your functions together. Alternatively, you can also make use of more builtin functions and libraries, like sum, or math.factorial, and for getting the digits, you can just iterate the characters in the number's string representation.
This way, the problem can be solved in a single line of code (though it might be better to move the is-super check to a separate function).
def issuper(n):
return sum(math.factorial(int(d)) for d in str(n)) == n
N = 1000
res = [n for n in range(1, N+1) if issuper(n)]
# [1, 2, 145]

First I would slightly change how main code is executed, by moving main parts to if __name__ == '__main__', which will execute after running this .py as main file:
if __name__ == '__main__':
number = int(input ("enter integer: "))
if (breakdown(number)) == number:
print(str(number)+ " is super")
After that it seems much clearer what you should do to loop over numbers, so instead of above it would be:
if __name__ == '__main__':
number = int(input ("enter integer: "))
for i in range(number+1):
if (breakdown(i)) == i:
print(str(i)+ " is super")
Example input and output:
enter integer: 500
1 is super
2 is super
145 is super
Small advice - you don't need to call str() in print() - int will be shown the same way anyway.

I haven't done much Python in a long time but I tried my own attempt at solving this problem which I think is more readable. For what it's worth, I'm assuming when you say "displays all numbers in range 0-N" it's an exclusive upper-bound, but it's easy to make it an inclusive upper-bound if I'm wrong.
import math
def digits(n):
return (int(d) for d in str(n))
def is_super(n):
return sum(math.factorial(d) for d in digits(n)) == n
def supers_in_range(n):
return (x for x in range(n) if is_super(x))
print(list(supers_in_range(150))) # [1, 2, 145]

I would create a lookup function that tells you the factorial of a single digit number. Reason being - for 888888 you would recompute the factorial of 8 6 times - looking them up in a dict is much faster.
Add a second function that checks if a number isSuper() and then print all that are super:
# Lookup table for single digit "strings" as well as digit - no need to use a recursing
# computation for every single digit all the time - just precompute them:
faks = {0:1}
for i in range(10):
faks.setdefault(i,faks.get(i-1,1)*i) # add the "integer" digit as key
faks.setdefault(str(i), faks [i]) # add the "string" key as well
def fakN(n):
"""Returns the faktorial of a single digit number"""
if n in faks:
return faks[n]
raise ValueError("Not a single digit number")
def isSuper(number):
"Checks if the sum of each digits faktorial is the same as the whole number"
return sum(fakN(n) for n in str(number)) == number
for k in range(1000):
if isSuper(k):
print(k)
Output:
1
2
145

Use range.
for i in range(number): # This iterates over [0, N)
if (breakdown(number)) == number:
print(str(number)+ " is super")
If you want to include number N as well, write as range(number + 1).

Not quite sure about what you are asking for. From the two functions you write, it seems you have solid knowledge about Python programming. But from your question, you don't even know how to write a simple loop.
By only answering your question, what you need in your main function is:
for i in range(0,number+1):
if (breakdown(i)) == i:
print(str(i)+ " is super")

import math
def get(n):
for i in range(n):
l1 = list(str(i))
v = 0
for j in l1:
v += math.factorial(int(j))
if v == i:
print(i)
This will print all the super numbers under n.
>>> get(400000)
1
2
145
40585

I dont know how efficient the code is but it does produce the desired result :
def facto():
minr=int(input('enter the minimum range :')) #asking minimum range
maxr=int(input('enter the range maximum range :')) #asking maximum range
i=minr
while i <= maxr :
l2=[]
k=str(i)
k=list(k) #if i=[1,4,5]
for n in k: #taking each element
fact=1
while int(n) > 0: #finding factorial of each element
n=int(n)
fact=fact*n
n=n-1
l2.append(fact) #keeping factorial of each element eg : [1,24,120]
total=sum(l2) # taking the sum of l2 list eg 1+24+120 = 145
if total==i: #checking if sum is equal to the present value of i.145=145
print(total) # if sum = present value of i than print the number
i=int(i)
i=i+1
facto()
input : minr =0 , maxr=99999
output :
1
2
145
40585

How do I get my num_digit function to give me the correct output and how do I get it to return any integer value in python?

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)))

How can I Keep the following program running until it becomes a single digit number?

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