How to check if input number is unique [Python] - python

I want to check if the user's 3-digit number input has no repeat digits (eg. 122, 221, 212, 555).
num = 0
while True:
try:
num = int(input("Enter a 3-Digit number: "))
if (num % 10) == 2:
print("ensure all digits are different")
This somewhat works, tells me that the numbers 122 or 212 are have repeats, but not for 221, or any other 3-digit number

num = input()
if len(set(num)) != len(num):
print("The number has repeat digits!")
else:
print("No repeat digits")

A %b gives the remainder if a is divided by b. Rather than doing this just take the number as string and check if any of the two characters present in the string are same.
num = 0
while True:
try:
num = (input("Enter a 3-Digit number: "))
if (num[0] == num[1] or num[1]==num[2] or num[2]==num[0])
print("ensure all digits are different")

You can also make use of a dictionary to check whether digits are repeated or not.
Take the remainder as you did in your code (taking mod 10)and add that remainder into the dictionary.
Everytime we take the remainder, we check inside the dictionary whether the number is present or not because if the number is present inside the dictionary, then it is not unique.
Code :
num = int(input())
dictionary = {}
while num > 0:
remainder = num % 10
if remainder in dictionary.keys():
print("Not unique")
break
else:
dictionary[remainder] = True
num = num // 10

Related

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)

Multiplication in a variable

I am writing a piece of code that needs to multiply numbers by different values, all the code for the entering and validation of the 7 digit number works however the multiplication doesn't work. This is my code.
while True:
try:
num = int(input("enter a 7 digit number: "))
check = len(str(num))
if check == 7:
print("This code is valid")
break
else:
num = int(input("enter a number that is only 7 digits: "))
except ValueError:
print("you must enter an integer")
num = int(num)
def multiplication():
num[0]*3
num[1]*1
num[2]*3
num[3]*1
num[4]*3
num[5]*1
num[6]*3
return total
multiplication()
When I run it, I get the following error:
Traceback (most recent call last):
File "\\hpdl3802\stuhomefolders$\12waj066\Year 10\Computing\A453\Code\Test v2.py", line 29, in <module>
multiplication()
File "\\hpdl3802\stuhomefolders$\12waj066\Year 10\Computing\A453\Code\Test v2.py", line 20, in multiplication
num[0]*3
TypeError: 'int' object is not subscriptable
Any feedback is welcome
Of course, your code might be written in a number of ways, optimized (check #Kasravand answer, it's awesome) or not, but with a minimal effort this is what I get:
while True:
try:
num = input("enter a 7 digit number: ")
check = len(num)
int(num) # will trigger ValueError if not a number
if check == 7:
print("This code is valid")
break
else:
print("bad length, try again")
except ValueError:
print("you must enter an integer")
def multiplication(num):
total = int(num[0])*3
total += int(num[1])*1
total += int(num[2])*3
total += int(num[3])*1
total += int(num[4])*3
total += int(num[5])*1
total += int(num[6])*3
return total
print("Answer: ", multiplication(num))
If you're bound to use an integer instead of a list for the input, you can do one of the following:
You could access the individual digits using a combination of integer division and modulo, for example:
first_digit = num // 1000000 * 3
second_digit = num // 100000 % 10 * 1
# and so on
Or you could get the input as a string and access and convert the individual digits:
# [...]
num = input("enter a number that is only 7 digits: ")
# [...]
first_digit = int(num[0]) * 3
second_digit = int(num[1]) * 3
When you convert the input number to an integer you can not use indexing on that object since integers don't support indexing. If you want to multiply your digits by a specific number you better do this before converting to integer.
So first off replace the following part:
num = int(input("enter a number that is only 7 digits: "))
with:
num = input("enter a number that is only 7 digits: ")
The you can use repeat and chain functions from itertools module in order to create your repeated numbers, then use a list comprehension to calculate the multiplication:
>>> from itertools import repeat, chain
>>> N = 7
>>> li = list(chain.from_iterable(repeat([3, 1], N/2 + 1)))
>>> num = '1290286'
>>> [i * j for i, j in zip(map(int, num), li)]
[3, 2, 27, 0, 6, 8, 18]
This code will works:
while True:
try:
num = int(input("enter a 7 digit number: "))
except ValueError:
print("you must enter an integer")
else:
if len(str(num)) != 7:
print("enter a number that is only 7 digits")
else:
break
num = str(num)
def multiplication():
total = 0
for i,m in enumerate([3,1,3,1,3,1,3]):
total += int(num[i])*m # transform the index of text into a integer
return total
print(multiplication())
This should help
num = ''
check = 0
while True:
try:
num = raw_input("enter a 7 digit number: ")
check = len(num)
if check == 7:
print("This code is valid")
break
else:
print "enter a number that is only 7 digits"
except ValueError:
print("you must enter an integer")
def multiplication():
total = 0
for i in range(check):
if i % 2 == 0:
total += int(num[i]) * 3
else:
total += int(num[i]) * 1
print total
multiplication()

ISBN final digit finder

I am working in python 3 and I am making a program that will take in a 10 digit ISBN Number and applying a method to it to find the 11th number.
Here is my current code
ISBN=input('Please enter the 10 digit number: ')
while len(ISBN)!= 10:
print('Please make sure you have entered a number which is exactly 10 characters long.')
ISBN=int(input('Please enter the 10 digit number: '))
continue
else:
Digit1=int(ISBN[0])*11
Digit2=int(ISBN[1])*10
Digit3=int(ISBN[2])*9
Digit4=int(ISBN[3])*8
Digit5=int(ISBN[4])*7
Digit6=int(ISBN[5])*6
Digit7=int(ISBN[6])*5
Digit8=int(ISBN[7])*4
Digit9=int(ISBN[8])*3
Digit10=int(ISBN[9])*2
Sum=(Digit1+Digit2+Digit3+Digit4+Digit5+Digit6+Digit7+Digit8+Digit9+Digit10)
Mod=Sum%11
Digit11=11-Mod
if Digit11==10:
Digit11='X'
ISBNNumber=str(ISBN)+str(Digit11)
print('Your 11 digit ISBN Number is ' + ISBNNumber)
I want to create some kind of loop so that the number after "Digit" for the variable name increases starting from 1 (or zero if it makes life easier), the number in the square brackets increases starting from 0 and the multiplication number to decrease from 11 to 2.
Is there any way of doing this code in a more efficient way?
I think this should do what you want.
def get_isbn_number(isbn):
digits = [(11 - i) * num for i, num in enumerate(map(int, list(isbn)))]
digit_11 = 11 - (sum(digits) % 11)
if digit_11 == 10:
digit_11 = 'X'
digits.append(digit_11)
isbn_number = "".join(map(str, digits))
return isbn_number
EXAMPLE
>>> print(get_isbn_number('2345432681'))
22303640281810242428
>>> print(get_isbn_number('2345432680'))
2230364028181024240X
Explanation of second line:
digits = [(11 - i) * num for i, num in enumerate(map(int, list(isbn)))]
Could be written out like:
isbn_letters = list(isbn) # turn a string into a list of characters
isbn_numbers = map(int, isbn_letters) # run the function int() on each of the items in the list
digits = [] # empty list to hold the digits
for i, num in enumerate(isbn_numbers): # loop over the numbers - i is a 0 based counter you get for free when using enumerate
digits.append((11 - i) * num) # If you notice the pattern, if you subtract the counter value (starting at 0) from 11 then you get your desired multiplier
Terms you should look up to understand the one line version of the code:
map,
enumerate,
list conprehension
ISBN=int(input('Please enter the 10 digit number: ')) # Ensuring ISBN is an integer
while len(ISBN)!= 10:
print('Please make sure you have entered a number which is exactly 10 characters long.')
ISBN=int(input('Please enter the 10 digit number: '))
continue
else:
Sum = 0
for i in range(len(ISBN)):
Sum += ISBN[i]
Mod=Sum%11
Digit11=11-Mod
if Digit11==10:
Digit11='X'
ISBNNumber=str(ISBN)+str(Digit11)
print('Your 11 digit ISBN Number is ' + ISBNNumber)

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