I'm trying to solve this problem but I'm just stuck at the very end.
I need to make a function that cycles the 5 digit number around. So for example, the input is 12345 and then it should be 51234, 45123 etc. Then when it cycles it out, until it's at the beginning again which is 12345, it should print out the biggest number of all of the cycled ones.
def number_cycle(number):
if number >= 99999 or number < 9999:#this is for limiting the number to 5 digits
print('Error,number isnt in 5 digit format')
else:
e = number%10 #5th digit
d = int(((number-e)/10)%10)#4th digit
c = int(((((number - e)/10)-d)/10)%10)#3rd digit
b = int(((((((number - e)/10)-d)/10)-c)/10)%10)#2nd digit
a = int(((((((((number - e)/10)-d)/10)-c)/10)-b)/10)%10)#1st digit
print(e,a,b,c,d)
print(d,e,a,b,c)
print(c,d,e,a,b)
print(b,c,d,e,a)
print(a,b,c,d,e)
number = eval(input('Input number:'))
I can't figure out how to get the biggest number out of all these.
Can you help?
You can try this solution:
def get_largest(number):
num_str = str(number)
num_len = len(num_str)
if num_len != 5:
print(f"Error: {number} is not a 5 digit number")
return
largest = 0
for i in range(num_len):
cycled_number = int(num_str[i:] + num_str[:i])
print(cycled_number)
if cycled_number > largest:
largest = cycled_number
print("Largest: ", largest)
number = int(input("Input number: "))
get_largest(number)
It will basically convert your 5 digit number to a string using a for loop, and then rotate the string, compare the rotated numbers and print the largest one.
Note:
In your original code snippet I'd suggest not to use the built-in eval method as it can be quite dangerous if you don't know what you are doing so avoid it as much as you can except you really have no other way to deal with a problem. More here:
https://nedbatchelder.com/blog/201206/eval_really_is_dangerous.html
https://medium.com/swlh/hacking-python-applications-5d4cd541b3f1
Related
So for my assignment I have to start with these parameters.
"Write a program that will average all of a numbers digits. You must use % for this lab to access the right most digit of the number. You will use / to chop off the right most digit."
Now I'm kinda confused where to start here's the code provided we must follow
def go( num ):
return 0
while ( True ):
# enter a number
val = int(input("Enter a number :: "))
print( go( val ) )
As a beginner I'm pretty damn confused where to start.It's supposed to average the digits in a number, and this is confusing me right now.
You can do something like this, i followed Crytofool's method
number = int(input("Enter a number"))
count = 0
result = 0
while number:
result, number = result + number % 10, number // 10 #grabbing induvidual digits of base 10, chopping off right most digit
count = count +1
print(result/count)
I wrote a simple code for:
enter any number and a digit and count how many times the digit is in the number.
the code i wrote is:
num= int(input("enter a number"))
n=num
digit = int(input("enter the digit"))
times=0
while n > 0 :
d = n%10
if d==digit :
times += 1
continue
else:
continue
n=n//10
print ("no. of times digit gets repeated is ", times)
When I tried this code, somehow it gave me nothing
Remove the else and continue statements because the loop always hits the continue and never goes to n=n//10
num= int(input("enter a number"))
n=num
digit = int(input("enter the digit"))
times=0
while n > 0 :
d = n%10
if d==digit :
times += 1
n=n//10
print ("no. of times digit gets repeated is ", times)
Output:
enter a number1111222233344567433232222222
enter the digit2
no. of times digit gets repeated is 12
if d==digit :
times += 1
continue
else:
continue
n=n//10
There is no way to reach the code line above that divides n by ten because both the true and false branches restart the loop with continue, hence n will never change value and you'll loop forever (for non-zero number input).
You should remove continue from both branches and, in fact, you don't need the else part since it doesn't do anything:
if d == digit:
times += 1
n = n // 10
The line n=n//10 never gets executed because of the continues before it. You don't need continue if you don't intend to skip the remaining of this loop iteration.
The other answers point out your continue misuse, but there're a couple Pythonic ways to do this.
divmod() neatly does division and modulus in one operation:
num = int(input("enter a number"))
digit = int(input("enter the digit"))
times = 0
while num > 0:
num, d = divmod(num, 10)
if d == digit:
times += 1
print("no. of times digit gets repeated is ", times)
You can also more simply not do anything with numbers, but with strings, and use str.count:
num = input("enter a number")
digit = input("enter the digit")
print("no. of times digit gets repeated is ", num.count(digit))
So im new to python and this question should be fairly easy for anybody in here except me obvisouly haha so this is my code
for c in range(0,20):
print("number",c+1)
number = input ("Enter number:\n")
number = int(number)
if (number < 1 or number > 9):
print("try again by inputing a positive number < 10")
c -=1
so as you may think if the number someone inputs is bigger than 9 or smaller than 0 i just want my c to stay where it is so i get 20 positive numbers from 1-9 instead it doesnt do that and it keeps increasing even tho i have the c-=1 thingy down there
First, do not use range(0,20) but range(20) instead.
Second, range returns an iterator. This means, that when you do c-=1 you do not go back in the iterator, you decrease the number returned by the iterator. Meaning that if c=5 and the number you entered in input is 20, c will become 4, but when returning to the start of the loop, c be be 6.
You probably want something of this sort:
c = 0
while c < 20:
print("number",c+1)
number = input ("Enter number:\n")
number = int(number)
if (number < 1 or number > 9):
print("try again by inputing a positive number < 10")
continue
c += 1
an alternative and simple solution to this would be:
i=20
while(i!=0):
number = input("Enter number:\n")
number = int(number)
if (number <1 or number >9):
print("Invalid no try within 0-9 ")
else
print(number)
i-=1
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
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