I am new to Python and currently reading Python 3 for absolute beginner and face following problem.
I would like to calculate factorial with procedure.
request user to input an non negative number n
then use for loop to calculate factorial
and the code is like that:
N = input("Please input factorial you would like to calculate: ")
ans = 1
for i in range(1,N+1,1):
ans = ans*i
print(ans)
while i would like to add a feature to check whether input number N is non-negative number. like:
if N != int(N) and N < 0:
I want the user to input N again if it is NOT non-negative number.
Thanks for your gentle help.
The construct could look like this:
while True:
N = input("Please input factorial you would like to calculate: ")
try: # try to ...
N = int(N) # convert it to an integer.
except ValueError: # If that didn't succeed...
print("Invalid input: not an integer.")
continue # retry by restarting the while loop.
if N > 0: # valid input
break # then leave the while loop.
# If we are here, we are about to re-enter the while loop.
print("Invalid input: not positive.")
In Python 3, input() returns a string. You have to convert it to a number in all cases. Your N != int(N) thus makes no sense, as you cannot compare a string with an int.
Instead, try to convert it to an int directly, and if that doesn't work, let the user enter again. That rejects floating point numbers as well as everything else which is not valid as an integer.
In Python's math library, there is a factorial function. You can use it like so:
import math
...
ans = math.factorial(N)
Since you want to calculate using a loop however, have you considered the following?
ans = -1
while ans < 0:
N = input("Please enter a positive integer: ")
if N.isdigit() == True:
n = int(N)
if n >= 0:
ans = n
for x in range (n-1, 1, -1):
ans *= x
print (ans)
Note, the second solution doesn't work for N = 0, where ans = 1 is correct by definition of factorial.
Number = int(input("Enter the number to calculate the factorial: "))
factorial = 1
for i in range(1,Number+1):
factorial = i*factorial
print("Factorial of ",Number," is : ", factorial)
def factorial(a):
if a == 1:
return 1
else:
return a * factorial(a - 1)
print('factorial of number', factorial(5))
Start
Declare Integer n,i,n!
Display “Enter a nonnegative integer.”
Input n
For i=1 to n-1 Step1,
Display “n!=i*n”
End for
Stop
You can check math module for python.
# math.factorial(x)
Return x factorial.
Raises ValueError if x is not integral or is negative.
Related
I know how to do this with a while loop and know how to use a for-loop in other languages like Java and C++. I want to use a for-loop in place of where I have written the while loop asking for the user input.
# You are required to use for-loop to solve this and round your answer to 2 decimal places. Write
# a program that takes n ∈ N (i.e., any positive integer including zero) from the user and use the
# input value to compute the sum of the following series:
n = -1
while n < 0:
n = int(input("Enter a value to compute: "))
# keep asking for user input until a whole number (0, 1, 2, 3, etc...) has been entered
k = 0
sum = 0
# To hold the sum of the fraction to be displayed
lastTerm = 0
# This variable represents the last term to be added to the fraction sum before the while loop below terminates
if n == 0:
sum = 0
elif n == 1:
sum = 1
else:
while lastTerm != 1 / n:
lastTerm = (n - k) / (k + 1)
sum = sum + (n - k) / (k + 1)
k += 1
print("{:.2f}".format(sum))
# Print the sum to two decimal places
One option is to catch the exception which is thrown when you cannot convert the input to an int, i.e.
while(True):
try:
# read input and try and covert to integer
n = int(input("Enter a value to compute: "))
# if we get here we got an int but it may be negative
if n < 0:
raise ValueError
# if we get here we have a non-negative integer so exit loop
break
# catch any error thrown by int()
except ValueError:
print("Entered value was not a postive whole number")
Alternative, slightly cleaner but I'm not 100% sure isdigit() will cover all cases
while(true):
n = input("Enter a value to compute: ")
if value.isdigit():
break
else:
print("Entered value was not a postive whole number")
How about this? It uses the for loop and sums all the values in the list.
x=[1,2,3,4] #== test list to keep the for loop going
sum_list=[]
for i in x:
j=float(input("Enter a number: "))
if not j.is_integer() or j<0:
sum_list.append(j)
x.append(1) #=== Add element in list to keep the cyclone going
else:
break
sums=sum(sum_list)
print("The Sum of all the numbers is: ",round(sums,2))
Use this to check for whole numbers -
if num < 0:
# Not a whole number
elif num >= 0:
# A whole number
for a for loop:
import itertools
for _ in itertools.repeat([]): # An infinite for loop
num = input('Enter number : ')
if num < 0:
# Not a whole number
pass # This will ask again
elif num >= 0:
# A whole number
break # break from for loop to continue the program
Easier Way -
mylist = [1]
for i in mylist : # infinite loop
num = int(input('Enter number : '))
if num < 0:
mylist.append(1)
pass # This will ask again
elif num >= 0:
# A whole number
break
i have to write a program with a loop that asks the user to enter a series of positive numbers. the user should enter a negative number to signal the end of the series, and after all positive numbers have been entered, the program should display their sum.
i don't know what to do after this, or whether this is even right (probably isn't):
x = input("numbers: ")
lst = []
for i in x:
if i > 0:
i.insert(lst)
if i < 0:
break
you should use input in the loop to enter the intergers.
lst = []
while True:
x = int(input('numbers:'))
if x > 0:
lst.append(x)
if x < 0:
print(sum(lst))
break
def series():
sum1 = 0
while True:
number = int(input("Choose a number, -1 to quit:"))
if number>=0:
sum1+=number
else:
return sum1
series()
while True means that the algorithm has to keep entering the loop until something breaks it or a value is returned which ends the algorithm.
Therefore, while True keep asking for an integer input , if the given integer is positive or equal to zero, add them to a sum1 , else return this accumulated sum1
You can use a while loop and check this condition constantly.
i, total = 0, 0
while i>=0:
total+=i
i = int(input('enter number:'))
print(total)
Also, don't use:
for loop for tasks you don't know exactly how many times it is going to loop
while loop as while True unless absolutely necessary. It's more prone to errors.
variable names that have the same name as some built-in functions or are reserved in some other ways. The most common ones I see are id, sum & list.
do you want that type of code
sum_=0
while True:
x=int(input())
if x<0:
break
sum_+=x
print(sum_)
Output:
2
0
3
4
5
-1
14
if you want a negative number as a break of input loop
convert string to list; input().split()
convert type from string to int or folat; map(int, input().split())
if you want only sum, it is simple to calculate only sum
x = map(int, input("numbers: ").split())
ans = 0
for i in x:
if i >= 0:
ans += i
if i < 0:
break
print(ans)
I have this following that I can't seem to solve correctly.
I have to write a function that takes a positive integer n as an input and checks if it is a palindrome (a number that's the same forward and backward). If it is, I have to return this number. If it isn't, I have to calculate the sum of the original number with the reversed number. This step is repeated until the result is a palindrome. This is the code I have so far:
x = input("Enter a positive integer: ")
def generate_palindrome(n):
if n == n[::-1]:
return n
else:
while n != n[::-1]:
r = n[::-1]
int(n) += int(r)
return n
generate_palindrome(x)
Of course this returns an error, since int(n) += int(r) is not possible. However, when not using int, it adds it as a string and that's also incorrect.
I'm not sure how far I'm off with the solution. How can I implement this correctly?
You are close. You just need another variable for the integer version of your string.Then just be clear which holds a string and which holds an integer:
def generate_palindrome(n):
if n == n[::-1]:
return n
else:
while n != n[::-1]: # n needs to be a string for this to work
r = int(n[::-1]) # r and s need to be numbers for correct addition
s = int(n)
n = str(r + s) # n needs to still be a string for the next loop iteration
return n
generate_palindrome("989")
# '989'
generate_palindrome("98")
# '8813200023188'
FWIW, I don't know that this is always guaranteed to return a reasonably sized number for every input. For example, I gave up on generate_palindrome("5798").
This code is for asserting values into the list until the user inputs a negative number.
Is there a way better to implement it?
Below is my tried version.
i = -1
while i > -1:
x = raw_input("Enter array limit")
for i in limit(x):
listArr = list()
y = raw_input("Enter number")
y.append(listArr)
print (listArr)
Something like this should meet the requirements:
odds = [] # similiar to `listArr` but we only want odd numbers
limit = 5
for _ in range(limit): # Input 5 numbers into an array
y = int(input('Enter number: '))
if y > -1 and y % 2 !=0: # if odd number
odds.append(y) # add to array
else:
break # break after a negative number as input
if odds:
print(min(odds)) # and display minimum odd number
# and if no negative integer input then also display minimum odd number
else:
print(0) # and if no odd number display zero
If Python2 use raw_input() as used in question code, otherwise use input().
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