I mostly have experience programming in visual basic, and am trying to learn python. I am trying to make a conditional loop in the form of a do until loop. I am not sure how to do it in python and i could use some help. I have a do until loop with an if statement within it. this is what i have.
number = 18
do while number = 1
if number%2==0 then
number = number/2
else number = (number*3)+1
loop
print(number)
Any help would be great. Thanks
There is no do … while loop in Python. I believe somewhere in the FAQ it explains why, and how to work around it.
But that doesn't matter because what you've written isn't a do … while loop, it's just a plain while loop. The whole point of a do … while in every language that has one is that you put the test condition after the loop body, instead of before, which guarantees that the loop will run at least once, and which allows you to avoid doing pre-loop setup that you'd have to duplicate inside the loop.
So, to translate your code to Python:
number = 18
while number == 1:
if number%2==0:
number = number/2
else:
number = (number*3)+1
print(number)
However, it's worth noting that since number = 18 before the loop, it will never be == 1 the first time through, so this is just going to skip the whole thing and print out 18.
You may have wanted while number != 1 or while number > 1, but you have to figure out what you actually want to write before anyone can write it.
There is no 'do/while' in Python. The closest is a loop that is guaranteed to loop once then get an exit test after 1 iteration.
while True: # kinda like 'do'
# do something at least once
if fail_condition(): # here is your 'while' test
break # end the loop if 'fail_condition' is True
# loop again if 'fail_condition' is not True
Roughly your code (Collatz/Hailstone?) would look like this:
number, i = 18, 0
while number>1:
i+=1
if number%2==0:
number = number/2
else:
number = (number*3)+1
print 'Reached {} after {} loops'.format(number,i)
# prints 'Reached 1 after 20 loops'
is that above code even in python? instead of do while, have it just be while. and after the if statement, dont have then. a quick reformatting would look something like this:
number = 18
while number == 18:
if number%2==0:
number = number/2
else:
number = (number*3)+1
print(number)
do while is not a usual construct in python. Simply use a while loop.
https://wiki.python.org/moin/WhileLoop
Your code could be modified as follows for instance.
number = 18
while number == 1:
if ( number % 2 ) == 0:
number = number / 2
else:
number = ( number * 3 ) + 1
print number
One of the biggest differences between these languages is that python is all about tab delimitation. You need not specify the start and end of a loop, the tabs do that for you. Additionally not the ':'s after conditional statements.
I would note however that the above code would not yield any valid answer. Perhaps you meant for the loop to run while number was not equal to 1?
Related
Hey so im pretty new to programming in general and I was having a crack at a question I found for the collatz function,
The code I wrote after some trial and error is as follows:
def collatz(number):
if number % 2 == 0:
number = number//2
print(number)
return number
elif number%2 != 0:
number = 3*number + 1
print(number)
return number
n = int(input("plz enter the number:"))
while n != 1:
n = collatz(n)
Output:
plz enter the number:3
10
5
16
8
4
2
1
This code works but im not sure how the variable values are being alloted, cuz after running this program I can see that in the shell "number = 3" but "n = 1", why is this the case? Shouldnt "number" also equal to 1? Because I am returning the value of number within the function?
Also just to clear my concepts, at the initial moment when I input n = 3, at that moment n = number = 3, then does this returned value of "number" automatically become the new value of n, when i call it in the while loop?
Just wanted to check cuz im a little weak when it comes to doing stuff that needs to pass parameters.
edit:
Why is this case diff then what was just answered?
def testfile(number):
number = number -1
print(number)
return number
n = int(input("enter:"))
while n != 2:
n = testfile(n)
Output:
enter:5
4
3
2
When the input is given as n = 5, then why does number = 3 instead of 5 as was just explained below?
Here's how your program works.
You ask for a number and store it in variable n.
You open a loop which continues until n is 1
Every time the loop repeats, you're calling your function and passing a COPY of n. If you add one to the copy inside the function, your original n will not change.
The COPY is called number. You perform your little tricks with number, output it to the screen, and here's the confusing part: you return a copy of number right back to your loop. And where does it go? It goes right back to n. This overwrites whatever was in n previously.
I'm in the middle of my homework. And i can't find out how to do this solution.
I have tried to used the break under for-statement but nothing return.
The problem is "Complete the following program so that the loop stops when it has found the smallest positive integer greater than 1000 that is divisible by both 33 and 273."
This is my code that i have tried to do it
n = 1001 #This one is required
while True: #This one too
for i in range(n,___): # I don't know what should i put in the blank
if i%33 == 0 and i%273 == 0: # I really confused about this line
break # Should i break it now?, or in the other lines?
print(f"The value of n is {n}") #This one is also required
I don't know that i should put break in which lines (or i don't have to used it?) or i should created a function that called a minimum number of the list?
I'm sorry about my language and how silly i am at my programming skill
I would accept every comment. Thank you
You already have a while True: loop, you don't need the inner for loop to search for your number, just keep incrementing n in the while loop instead of adding a new counter, when the number you're looking for is found, the infinite while True: loop will stop (using break), and so your print statement will be executed:
n = 1001 # start at 1001
while True: # start infinite loop
if n % 33 == 0 and n % 273 == 0: # if `n` found
break # exit the loop
n += 1 # else, increment `n` and repeat
print(f"The value of n is {n}") # done, print the result
Output:
The value of n is 3003
Thanks for saying it's homework! Makes it better to explain things in more detail than just giving an answer.
There are few things to explain:
1) n%33 is the remainder of dividing n by 33. So 66%33 is 0 and 67%33 is 1.
2) For loops are generally when you need to loop over a defined range (not always, but usually). E.g. "add up the first 100 integers". A while loop makes more sense here. It will definitely terminate, because at some point you'll get to 33 * 237.
3) if i%33 == 0 and i%237 == 0: means we want to do something when the number can be divided evenly (no remainder) by both 37 and 237.
n=1001
while True:
if n%33==0 and n%237==0:
print(n)
break
n+=1
A for loop will not help you here, because you don't know when to end the loop. You usually use for loops when the range of things you want to loop over is already known.
Instead, do the following:
before starting your while: True loop: set i to 0,
then increase i with 1 every time to the loop
also, don't forget to stop the loop when i>1000!
Well you could still use a for loop, as long as the upper limit is at least as high as the maximum possible result. The result would be in i, not in n, and the for loop will suffice, not an additional while loop. The for loop will break when the remainder when dividing by both 33 and 237 is zero (i.e. they are both factors).
n = 1001 #This one is required
for i in range(n, 33 * 237 + 1): # I don't know what should i put in the blank
if i % 33 == 0 and i % 237 == 0: # I really confused about this line
break #
print(f"The value of i is {i}") #This one is also required
You could also use a while loop and use the same logic for the condition. In this case we test that at least one is not a factor and continue the loop until both 33 and 237 are evenly divisible into i.
n = 1001 #This one is required
i = n
while i % 33 or i % 237:
i += 1
print(f"The value of i is {i}")
I am trying to create a program that prints out a list of numbers starting at 0 and leading up to a number the user inputs (represented by the variable "number"). I am required to use "while" loops to solve this problem (I already have a functional "for" loop version of the assignment). The program should mark anything in that list divisible by 3 with the word "Fizz," divisible by 5 with the word "Buzz," and anything divisible by both with "FizzBuzz" while also including unlabeled numbers outside of those specifications.
Every time I run this program, it ignores the conditions and just prints the word "FizzBuzz" however many times is represented by the number inputted. (I typically use 15 because it has at least one example of each condition, so that means I get 15 "FizzBuzz"s in a row).
To find out why it was doing that, I used print(i) instead of the rest of the program under the first conditional and it gave me 15 counts of the number 0, so there is reason to believe the program is completely ignoring the range I gave it and just outputting copies of i based on the user number input.
Any help would be appreciated!
number = int(input("Enter a Number"))
i = 0
while(i < number + 1):
if number % 3 == 0 and number % 5 == 0:
print("Fizzbuzz")
elif number % 5 == 0:
print("Buzz")
elif number % 3 == 0:
print("Fizz")
else:
print(number)
i += 1
print ("Done!")
You meant to check the divisibility of i, which increments every loop, not of number which doesn't change.
You also meant to print(i) in the else clause.
Have a little problem. I'm writing a simple program that takes an input of numbers (for example, 1567) and it adds the odd numbers together as well as lists them in the output. Here is my code:
import math
def oddsum(n):
y=n%10
if(y==0):
return
if(y%2!=0):
oddsum(int(n/10))
print (str(y),end="")
print (" ",end="")
else:
oddsum(int(n/10))
def main():
n=int(input("Enter a value : "))
print("The odd numbers are ",end="")
oddsum(n)
s = 0
while n!=0:
y=n%10
if(y%2!=0):
s += y
n //= 10
print("The sum would be ",end=' ')
print("=",s)
return
main()
It outputs just fine, in the example it will print 1 5 and 7 as the odd numbers. However, when it calculates the sum, it just says "7" instead of 13 like it should be. I can't really understand the logic behind what I'm doing wrong. If anyone could help me out a bit I'd appreciate it :)
I understand it's an issue with the "s += y" as it's just adding the 7 basically, but I'm not sure how to grab the 3 numbers of the output and add them together.
As #Anthony mentions, your code forever stays at 156 since it is an even num.
I would suggest you directly use the string input and loop through each element.
n = input("Enter a value : ") #'1567'
sum_of_input = sum(int(i) for i in n if int(i)%2) #1+5+7=13
[print(i, end="") for i in n if int(i)%2] #prints '157'
Note that int(i)%2 will return 1 if it is odd.
1567 % 10 will return 7. You might want to add the numbers you printed in oddsum to a list, and use the sum function on that list to return the right answer.
The immediate issue is that n only changes if the remainder is odd. eg 1,567 will correctly grab 7 and then n=156. 156 is even, so s fails to increment and n fails to divide by 10, instead sitting forever at 156.
More broadly, why aren't you taking advantage of your function? You're already looping through to figure out if a number is odd. You could add a global parameter (or just keep passing it down) to increment it.
And on a even more efficient scale, you don't need recursion to do this. You could take advantage of python's abilities to do lists. Convert your number (1567) into a string ('1567') and then loop through the string characters:
total = 0
for c in '1567':
c_int = int(c)
if c_int%2!= 0:
total += c_int
print(c)
print(total)
Here is the problem from leetcode:
Write an algorithm to determine if a number is "happy".
A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.
Example: 19 is a happy number
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
question 1: from my understanding, if 1 happens,return True else it will infinitely runs the loop where i understand from this sentence "or it loops endlessly in a cycle which does not include 1."
However,after i found some answer to this questions from internet, i found my understanding was wrong.It is supposed to be if 1 happens return True,**also if any number repeats in the set, then end the loop and return False.**Well, be honestly, the problem says "or it loops endlessly in a cycle which does not include 1." why we just let the program run endlessly in a cycle as it states??
questions 2: For the second question, i think it is about understanding the codes.This is what i found from internet:
def isHappy(n):
stop = {1}
while n not in stop:
stop.add(n)
n = sum(int(d)**2 for d in str(n))
return n == 1
From this codes, i can understand that if 1 happens, it will stop running the while loop and return True.However, if repeat number happens in the set, i think it also will stop running the while loop and return True,becuase next line followed by while statement is still return n==1.However, in fact it will output False.e.g
since 89 repeats again in the set, output False.
Sorry for my wordy description.Simply speaking, my questions is how this false comes out? in the coding, there is no place explicitly return False
The isHappy function you posted is correct.
This is how it works -
It will calculate the squares of the digits and save it in n.
If n is 1 it stops. Else, it appends n to change list and loops. But as soon as n is any number which is there in the change list, the loop breaks.
In Python, when control comes out of the loop, it has the last values stored in variables. So, n has the last calculation stored.
Now, if n is 1, n == 1 returns True else False.
Hope that helps.
This is how I did , even though it is a bit messy
#Python program to check if a number is a happy number
#Creating an user defined function which returns True if the number is happy else False
def happy(num) :
#Creating a copy of the number in a string and a iterating point
copy = str(num)
square = num
#Proceeding with while loop due to unknown iterations
while True : #inifinte loop
square = sum(int(j) ** 2 for j in list(str(square))) #Sum of the number's digit squares
if square == 1:
test = True
break
elif square == 4:
test = False
break
return test
check = int(input('Enter the number: '))
print('The number {} {} a happy number'.format(check, ('is') if happy(check) == True else ('is not')))
below is a popular solution to the happy number problem. this key is the sum of square of each digit did not appear in the past, otherwise, there will be an endless loop.
class Solution:
# #param {integer} n
# #return {boolean}
def isHappy(self, n):
s=set()
while n!=1 and n not in s:
s.add(n)
n= sum([int(x)**2 for x in str(n)])
return n==1