I must create two functions. One that can tell whether one number is odd or even by returning t/f, and the other will call the first function then return how many even numbers there are.
This is my code so far:
Even = [0,2,4,6,8]
IsEvenInput = int(input("Please enter a number: "))
def IsEvenDigit(a):
if a in Even:
return True
else:
return False
y = IsEvenDigit(IsEvenInput)
print(y)
def CountEven(b):
count = 0
for a in b:
if IsEvenDigit(a):
count+=1
return count
d = input("Please enter more than one number: ")
y = CountEven(d)
print(y)
This keeps outputting 0 and doesn't actually count. What am I doing wrong now?
d = input("Please enter more than one number: ")
This is going to return a string of numbers, perhaps separated by spaces. You'll need to split() the string into the sequence of text digits and then turn those into integers.
There's a general approach to determining whether a number is odd or even using the modulus / remainder operator, %: if the remainder after division by 2 is 0 then the number is even.
Here is another approach:
def is_even(number):
return number % 2 == 0
def even_count(numbers_list):
count = 0
for number in numbers_list:
if is_even(number): count += 1
return count
raw_numbers = input("Please enter more than one number: ")
numbers_list = [int(i) for i in raw_numbers.split()]
count = even_count(numbers_list)
print(count)
This will take care of all other numbers too.
So by calling CountEvent(d) outside the scope of the function CountEven, you aren't using recursion, you're simple calling the function after it's been defined.
Try reducing the amount of code outside of your functions.
#Start by declaring your functions:
def isEven(n):
return n % 2 == 0
def countEven():
count = 0
string_of_numbers = input("Please enter numbers separated by spaces: ")
list_of_number_characters = string_of_numbers.split(' ')
for number in list_of_number_characters:
number_as_int = int(number)
if isEven(number_as_int) == True:
count = count + 1
print("There were " + str(count) + " even numbers found.")
countEven() #Call the function that runs your program
You are counting whether the integers - [0, 2, 4, 6, 8] etc. - are characters in a string - "0", "2", "4", "6", "8" etc. Currently, IsEvenDigit(a) will never be true, because a character in a string will not be in the list of even integers, so the code beneath the if statement will never be executed. You need IsEvenDigit(int(a)) in the CountEven function.
On another topic, a commenter to your post suggested reading PEP 8. Your code is actually formatted pretty well, its just in Python, CamelCase is used just for classes, and words_seperated_by_underscores is used for variables and function names.
Or if you want brevity and unreadability, some code:
main = lambda: sum(map(lambda x: int(x) % 2 == 0, (i for i in input("Enter a number: "))))
main()
It does define 2 (anonymous) functions!
A possible solution:
def is_even(n):
return not n % 2
def count_even(numbers)
return sum(map(is_even, numbers))
nums = input("Enter numbers separated by spaces: ")
nums = map(int, nums.split())
print(count_even(nums))
Related
I'm writing a program that takes a number from input and generates it's palindrome number. My program only prints the first half not the second. I tried reverse, didnt work.I have incluede those lines as comment.
My Code:
def show_palindrome(maximum):
maximum = int(input("Enter Number : "))
for number in range(1, maximum + 1):
temporary = number
reverse = 0
while (temporary > 0):
reminder = temporary % 10
reverse = (reverse * 10) + reminder
temporary = temporary //10
if(number == reverse):
#number2 = number[::-1]
#print(number,number2, end = '')
print(number, end = '')
show_palindrome(3)
My output:
123
The output I need:
12321
I believe you're looking for something like this:
def show_palindrome(maximum = None):
if not maximum:
maximum = input("Enter Number : ")
output = str(maximum)
for number in range(1, int(maximum)):
output = str(int(maximum) - number) + str(output) + str(int(maximum) - number)
return output
print(show_palindrome(3))
this returns 12321 for instance
A couple of things I would do differently in your function:
If you're going to require the input in the def (The way you make it optional is to set it equal to something when you declare it like I have it set to None (maximum=None), then you don't need the input() statement.
Since you already know how long you want it to be ( you require it declared when you initialize the function, it's just 2*maximum - 1) there's really no need to use a while loop.
Other than that good job! Keep it up!
You can try something simpler like this:
def show_palindrome():
num = input("Enter Number : ")
print(num + num[:-1][::-1])
show_palindrome()
Input:
123
12
1
Output:
12321
121
1
Apart from the string slice method (as used by #PApostol), you could also use the reversed method :
def show_palindrome():
value = input("Enter Number : ")
print(value[:-1] + "".join(reversed(value)))
Input:
123
Output:
12321
def palindrome(maximum):
number = ''.join([str(num) for num in range(1, maximum + 1)])
return int(number + number[-2::-1])
print(palindrome(3))
Output:
12321
Your code seems over complicated, and I don't even know what to fix in it. What I can tell is that passing a parameter maximum to overwrite it just after with maximum = int(input("Enter Number : ")) is useless, don't pass the parameter.
So let's back to easy things
Build palindrome using strings method : slicing backwards from index -2, and reverse it with -1 increment
def show_palindrome():
value = input("Enter Number : ")
print(value + value[-2::-1])
Build palindrome using math operations : save the remainder, except the first one
def show_palindrome():
value = input("Enter Number : ")
result = value
value = int(value) // 10 # remove last char which would be redundant
while value > 0:
result += str(value % 10)
value = value // 10
print(result)
show_palindrome()
print("\tWELCOME TO DRY_RUN CLASS ASSAIGNTMENT!\t")
userList = []
def mainSystem():
number = 1
userInput = int(input("Enter the size of the List: "))
if userInput > 0:
for x in range(0, userInput):
variable = int(input(print("Enter number",number )))
number = number + 1
userList.append(variable)
else:
print("Number should not be less than or equal to '0'!")
def maxAll():
maxofall = 0
for element in userList:
if element > maxofall:
maxofall = element
print("The maximum number is:", element)
while True:
mainSystem()
askUser = int(input("What do you want to do with the numbers?\n1.Max All\n2.Average\n3.Quit\nYour answer: "))
if askUser == 1:
maxAll()
this is the code i am using right now...
what do i need to fix i am getting an error like this wheni am executing line no. 9
:-
Enter Number whatever the number is
Noneinput starts here
???
def mainSystem():
number = 1
userInput = int(input("Enter the size of the List: "))
if userInput > 0:
for x in range(0, userInput):
variable = int(input("Enter number {} : ".format(number) ))
number = number + 1
userList.append(variable)
else:
print("Number should not be less than or equal to '0'!")
Change your function to this.
print() is a function. Which returns nothing (None) in your case.
input() function thinks this None object is worth displaying on console.
Hence None appears on screen before taking any input.
There are a couple of problems:
1)You have a misplaced print() inside input() in your mainSystem() function
2)You probably want to use f-strings to print the right number instead of the string literal 'number'
def mainSystem():
number = 1
userInput = int(input("Enter the size of the List: "))
if userInput > 0:
for x in range(0, userInput):
variable = int(input(f"Enter number {number} : "))
number = number + 1
userList.append(variable)
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)
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 display the number of one's in any given integer number?
I am very new to python so keep this in mind.
I need the user to input a whole number.
Then I want it to spit out how many one's are in the number they input.
i am using python 3.3.4
How would I be able to do this with this following code?
num = int(input ("Input a number containing more than 2 digits"))
count = 0
for i in range (0):
if num(i) == "1":
count = count + 1
print (count)
I don't know what i'm doing wrong
it gives me 'int' object is not callable error
Something like this:
Int is not iterable so you may need to convert into string:
>>> num = 1231112
>>> str(num).count('1')
4
>>>
str(num).count('1') works just fine, but if you're just learning python and want to code your own program to do it, I'd use something like this, but you're on the right track with the code you supplied:
count = 0
for i in str(num):
if i == "1":
count = count + 1 # or count += 1
print(count)
Just to help you, here is a function that will print out each digit right to left.
def count_ones(num):
while num:
print(num % 10) # last digit
num //= 10 # get rid of last digit
num = 1112111
answer = str(num).count("1")
num = int(input (" Input a number to have the number of ones be counted "))
count = 0
for i in str(num):
if i == "1":
count = count + 1 # or count += 1
print (' The number of ones you have is ' + str(count))
So i took the user input and added it to the correct answer since when i tried the answer from crclayton it didn't work. So this works for me.