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How to print only the last result of an array of sum in python

I want to calculate the sum of the natural numbers from 1 up to an input number. I wrote this code:
number=int(input("enter a natural number"))
if number<0:
print("The number is not positive")
else:
n=0
for i in range (1,number+1):
n+=i
print(n)
But it prints multiple numbers instead. For example, if the user puts five, the program should print 15, but I get this:
1
3
6
10
15
How can I fix the code so that only 15 appears?

You have all the steps because your print statement is in your for loop.
Change it like this:
number = int(input("Enter a positive natural number: "))
if number < 0:
print("The number needs to be positive")
exit() # Stops the program
result = 0
for i in range(1, number + 1):
result += i
print(result) # We print after the calculations
There's also a mathematical alternative (see here):
number = int(input("Enter a positive natural number: "))
if number < 0:
print("The number needs to be positive")
exit() # Stops the program
print(number * (number + 1) / 2)

As I've pointed out and suggested earlier in comments, you could move the print statement out of for-loop to print the final sum.
Or you could try to use generator expression to get all number's total (sum), because we don't care the intermediate sums.
This simple sum of all up to the number in one shot.
number=int(input("enter a natural number"))
if number < 0:
print("The number is not positive")
# exit or try again <---------
else:
print(sum(range(1, number + 1))) # given 5 -> print 15

Something like this?
number = int(input("enter a natural number"))
if number < 0:
print("The number is not positive")
else:
n = 0
for i in range (1,number + 1):
n += i
print(n)

The answer to your question is that you are printing the n every time you change it. You are looking for the last answer when you run the code. This code should solve it.
number = int(input("enter a natural number"))
if number < 0:
print("The num < 0")
else:
n = 0
l = []
for i in range (0, number+1):
n+=i
l.append(n)
print(l[len(l)-1])

How to loop a function def in python until I write the number 0

I'm trying to do a def function and have it add the digits of any number entered and stop when I type the number "0", for example:
Enter the number: 25
Sum of digits: 7
Enter the number: 38
Sum of digits: 11
Enter the number: 0
loop finished
I have created the code for the sum of digits of the entered number, but when the program finishes adding, the cycle is over, but what I am looking for is to ask again for another number until finally when I enter the number "0" the cycle ends :(
This is my code:
def sum_dig():
s=0
num = int(input("Enter a number: "))
while num != 0 and num>0:
r=num%10
s=s+r
num=num//10
print("The sum of the digits is:",s)
if num>0:
return num
sum_dig()

Use list() to break the input number (as a string) into a list of digits, and sum them using a list comprehension. Use while True to make an infinite loop, and exit it using return. Print the sum of digits using f-strings or formatted string literals:
def sum_dig():
while True:
num = input("Enter a number: ")
if int(num) <= 0:
return
s = sum([int(d) for d in list(num)])
print(f'The sum of the digits is: {s}')
sum_dig()

In order to get continuous input, you can use while True and add your condition of break which is if num == 0 in this case.
def sum_dig():
while True:
s = 0
num = int(input("Enter a number: "))
# Break condition
if num == 0:
print('loop finished')
break
while num > 0:
r=num%10
s=s+r
num=num//10
print("The sum of the digits is:",s)
sum_dig()

A better approach would be to have sum_dig take in the number for which you want to sum the digits as a parameter, and then have a while loop that takes care of getting the user input, converting it to a number, and calling the sum_digit function.
def sum_dig(num): # takes in the number as a parameter (assumed to be non-zero)
s=0
while num > 0: # equivalent to num != 0 and num > 0
r = num % 10
s = s + r
num = num // 10
return s
while True:
num = int(input("Enter a number: "))
if num == 0:
break
print("The sum of the digits is: " + sum_dig(num))
This enables your code to adhere to the Single-Responsibility Principle, wherein each unit of code has a single responsibility. Here, the function is responsible for taking an input number and returning the sum of its digits (as indicated by its name), and the loop is responsible for continuously reading in user input, casting it, checking that it is not the exit value (0), and then calling the processing function on the input and printing its output.

Rustam Garayev's answer surely solves the problem but as an alternative (since I thought that you were also trying to create it in a recursive way), consider this very similar (recursive) version:
def sum_dig():
s=0
num = int(input("Enter a number: "))
if not num: # == 0
return num
while num>0:
r= num %10
s= s+r
num= num//10
print("The sum of the digits is:",s)
sum_dig()

Python Program to check if a number is armstrong or not is not working, what am I doing wrong?

n=int(input("Enter a Number: "))
x=0
y=0
z=0
while(n>0):
x=n%10
y=x**3
z=z+y
n=n//10
print (z)
#The z here is the same value which I enter, yet it doesn't work.
#If I enter 407 as n, z becomes (4^3)+(0^3)+(7^3) which is 407
if (z==n):
#But even when 407==407, it just wont print the bottom statement
print ("The number is Armstrong")
else:
print ("The number isn't Armstrong")
#it prints that it isn't an Armstrong number

After the while loop, n already became 4//10 which is 0, so it'll never equal z which is 407.
You will want to keep a copy of the original input for comparison.
As a general advice, use a debugger or at least print() your objects to see where the assignments went wrong.

Without using any built-in method
Armstrong number is 371 because 3**3 + 7**3 + 1**3 = 371. according this rule 123 is not Armstrong number because 1**3 + 2**3 + 3**3 is not equal to 123
def count_digit(n):
count = 0
while n > 0:
count += 1
n //= 10
return count
def is_armstrong(n):
given = n
result = 0
digit = count_digit(n)
while n > 0:
reminder = n % 10
result += reminder ** digit
n //= 10
return given == result
is_armstrong(371)
>> True
is_armstrong(123)
>> False

You can take in your initial number as a string so we can more easily convert it to a list. We can then map to create that list of ints. After we can use list comprehension to raise all int in that list to the power that is the len of our list. If the sum of this list equals our input, then we have an Armstrong number.
n = input('Enter a number: ')
nums = list(map(int, n))
raised = [i**len(nums) for i in nums]
if sum(raised) == int(n):
print('The number is Armstrong')
else:
print('The number is not Armstrong')
Expanded list comprehension:
raised = []
for i in nums:
i = i**len(nums)
raised.append(i)
print(raised)
Alternate for map:
nums = []
for i in n:
i = int(i)
nums.append(int(i))

I corrected your code:
n = int(input("Enter a Number: "))
x = 0
y = 0
z = 0
num = n
while n > 0:
x = n % 10
y = x**len(str(num))
z = z+y
n = n//10
print(z)
if (z == num):
print ("The number is Armstrong")
else:
print ("The number isn't Armstrong")
But you can still do it in many ways better. Look at the code of vash_the_stampede and ggorlen.
Or:
def isArmstrong(n):
print(f"{n} is {'' if int(n) == sum(int(i)**len(n) for i in n) else 'not '}an Armstrong number")
isArmstrong(input("Please enter a number: "))

Definition: a number n is an Armstrong number if the sum of each digit in n taken to the power of the total digits in n is equal to n.
It's important to keep track of the original number n, because it'll be needed to compare against the result of z (your variable representing the sum). Since you're mutating n in your while loop, there's no grounds for comparison against your original input, so if (z==n): isn't working like you expect. Save n in another variable, say, original, before reducing it to 0.
Additionally, your code has arbitrarily chosen 3 as the number of digits in the number. For your function to work correctly for any number, you'll need a way to count its digits. One way is to convert the number to a string and take the length.
I strongly recommend using descriptive variable names which reduces the chance of confusing yourself and others. It's only apparent that z represents your sum and x your remainder by virtue of reading through the code. If the code was any longer or more complex, it could be a nightmare to make sense of.
Lastly, Python is not a particularly flexible language from a style standpoint. I recommend adhering to the style guide as best as possible to keep your code readable.
Here's a working example:
def armstrong(n):
total = 0
original = n
digits = len(str(n))
while n > 0:
total += (n % 10) ** digits
n //= 10
return total == original
if __name__ == "__main__":
while 1:
print(armstrong(int(input("Enter a Number: "))))
Output:
Enter a Number: 407
True
Enter a Number: 1234
False
Enter a Number: 23
False
Enter a Number: 8
True
Enter a Number: 371
True
Try it!

total=0
def Armstrong(n):
m=list(n)
global total
for i in m:
total+=pow(int(i),len(n))
if total==int(n):
print ("it is Armstrong number")
else:
print("it is not Armstrong number")
Armstrong(input("enter your number"))
print(total)

Use enter in python 3

I'm writing for creating a list of number from input and get the average of the list. The requirement is: when the user enters a number, the number will be appended to the list; when the user press Enter, the input section will stop and conduct and calculation section.
Here is my code:
n = (input("please input a number"))
numlist = []
while n != '':
numlist.append(float(n))
n = float(input("please input a number"))
N = 0
Sum = 0
for c in numlist:
N = N+1
Sum = Sum+c
Ave = Sum/N
print("there are",N,"numbers","the average is",Ave)
if I enter numbers, everything works fine. But when I press Enter, it shows ValueError. I know the problem is with float(). How can I solve this?

You don't need the float() around the input() function inside your loop because you call float() when you append n to numlist.

this should solve ur prob ,by adding a try,catch block around ur print statement
n = (input("please input a number"))
numlist = []
while True :
numlist.append(float(n))
#####cath the exception and break out of
try :
n = float(input("please input a number"))
except ValueError :
break
N = 0
Sum = 0
for c in numlist:
N = N+1
Sum = Sum+c
Ave = Sum/N
print("there are",N,"numbers","the average is",Ave)

Check if numbers are in a certain range in python (with a loop)? [duplicate]

This question already has answers here:
Determine whether integer is between two other integers
(16 answers)
Closed 4 years ago.
Here's my code:
total = int(input("How many students are there "))
print("Please enter their scores, between 1 - 100")
myList = []
for i in range (total):
n = int(input("Enter a test score >> "))
myList.append(n)
Basically I'm writing a program to calculate test scores but first the user has to enter the scores which are between 0 - 100.
If the user enters a test score out of that range, I want the program to tell the user to rewrite that number. I don't want the program to just end with a error. How can I do that?

while True:
n = int(input("enter a number between 0 and 100: "))
if 0 <= n <= 100:
break
print('try again')
Just like the code in your question, this will work both in Python 2.x and 3.x.

First, you have to know how to check whether a value is in a range. That's easy:
if n in range(0, 101):
Almost a direct translation from English. (This is only a good solution for Python 3.0 or later, but you're clearly using Python 3.)
Next, if you want to make them keep trying until they enter something valid, just do it in a loop:
for i in range(total):
while True:
n = int(input("Enter a test score >> "))
if n in range(0, 101):
break
myList.append(n)
Again, almost a direct translation from English.
But it might be much clearer if you break this out into a separate function:
def getTestScore():
while True:
n = int(input("Enter a test score >> "))
if n in range(0, 101):
return n
for i in range(total):
n = getTestScore()
myList.append(n)
As f p points out, the program will still "just end with a error" if they type something that isn't an integer, such as "A+". Handling that is a bit trickier. The int function will raise a ValueError if you give it a string that isn't a valid representation of an integer. So:
def getTestScore():
while True:
try:
n = int(input("Enter a test score >> "))
except ValueError:
pass
else:
if n in range(0, 101):
return n

You can use a helper function like:
def input_number(min, max):
while True:
n = input("Please enter a number between {} and {}:".format(min, max))
n = int(n)
if (min <= n <= max):
return n
else:
print("Bzzt! Wrong.")