is there a way to keep the counter going without counting the negatives and only to stop when the input is zero?
count = 0
total = 0
n = input()
while n != '0':
count = count + 1
total = total + int(n) ** 2
n = input()
print(total)
Here is an example of execution result.
Input: -1 10 8 4 2 0
Output: 184
Since you want only number to enter the loop you can use isnumeric() built in function to check that.
You need if() : break here.
num = input()
...
while(isnumeric(num)):
...
if(num == "0"):
break;
The response you wait for is:
ignore negative number
count positive numbers
stop when input is 0
count = 0
total = 0
n = int(input())
while (n != 0):
count += 1
if (n > 0):
total = total + n**2
num = int(input())
print(total)
Your code was already OK except that you did not cast the number n into int and you did not test n to take away negative values.
Execution:
When you enter -1 10 8 4 2 0, it should show 184
You can parse your Input to an integer (number) and check if it's larger than zero:
count = 0
total = 0
num = int(input())
while number != 0:
if number < 0:
continue
count += 1
total = total + num**2
num = int(input())
print(total)
The difference between pass, continue, break and return are:
pass = ignore me an just go on, usefull when you create a function that has no purpose yet
continue = ignore everything else in the loop and start a new loop
break = break the loop
return = end of a function - a return statement can be used to give an output to a function but also as a way to break out of the function like the break statement does in loops.
Related
n = int(input())
counter = 0
while n > 0:
if (n // 2) > 1:
counter = counter +1
print (counter)
Hi,
I am a python learner and I am having problems with this homework I was given.
Read a natural number from the input.
Find out how many times in a row this number can be divided by two
(e.g. 80 -> 40 -> 20 -> 10 -> 5, the answer is 4 times)
And I should use while loop to do it.
Any Ideas, because I really don't have any idea how to do it.
This is my best try
You are not changing n. I would write it like this:
while (n % 2) == 0:
n //= 2
counter += 1
Try this, we take the value from "n" and check whether it is divisible by two or not. If it is divisible by two, we increment the counter and then divide that number by 2. If not, it will print the output.
n = int(input("Input your number: "))
counter = 0
while n % 2 != 1:
counter = counter + 1
n = n/2
print(counter)
Your while loop condition is wrong.
While the number is evenly divisible by 2, divide it by 2 and increment counter
num = int(input('Enter a number: '))
counter = 0
while num % 2 == 0 and num != 0:
num = num / 2
counter = counter + 1
print(counter)
The code above will not work as intended. The intended functionality is to take an input natural number and find out how many times in a row the number can be divided by 2. However, the code will only divide the number by 2 once.
To fix this, you can change the code to the following:
n = int(input())
counter = 0
while n > 0:
if (n % 2) == 0:
counter = counter +1
n = n // 2
print (counter)
You need to test whether a number is divisible by 2. You can do this in one of two ways...
x % 2 == 0 # will be True if it's divisible by 2
x & 1 == 0 # will be True if it's divisible by 2
So, you need a loop where you test for divisibility by 2, if True divide your original value by 2 (changing its value) and increment a counter
N = int(input())
counter = 0
if N != 0:
while N % 2 == 0:
counter += 1
N //= 2
print(counter)
Or, if you prefer more esoteric programming, then how about this:
N = int(input())
b = bin(N)
print(0 if (o := b.rfind('1')) < 0 else b[o:].count('0'))
I have a double while loop, and it does not seem to be working because of some logic I'm doing incorrectly. I'm not sure what is wrong exactly, but I feel the code may be too complicated and somewhere, there is an error.
enter code here
import math
print("How many numbers am I estimating John?")
count = int(input("COUNT> "))
print("Input each number to estimate.")
better_guess = 0
initial_guess = 10
i = 0
j = 0
t = 1
list = []
for j in range(count):
num = float(input("NUMBER> "))
list.append(num)
j = j + 1
if j == count:
print("The square roots are as follows:")
while i <= len(list):
while t != 0 :
initial_guess = 10
better_guess = (initial_guess + (list[i])/initial_guess) / 2
if initial_guess == better_guess:
print(f"OUTPUT After {t} iterations, {list[i]}^0.5 = {better_guess}")
i = i + 1
break
initial_guess = better_guess
i = i + 1
There are some errors in your code, #x pie has pointed out some of them but not all. The most important is that you are need to initalize t for every number in the list, so you can get the iterations for the numbers separately. Also, t needs to be incremented in the while loop, not inside the if block.
You can also clean up the code considerably, for example the j variable is not being used, list comprehension can be used to shorten the code (pythonic way), and iterating over lists can be done with for num in list.
Putting this altogether produces this:
count = int(input('How many numbers am I estimating John? \nCOUNT> '))
print("Input each number to estimate.")
list = [float(input(f'NUMBER {i+1}> ')) for i in range(count)]
print("The square roots are as follows:")
for num in list:
initial_guess = 10
t = 0
while True:
better_guess = (initial_guess + num/initial_guess) / 2
t += 1
if initial_guess == better_guess:
print(f"OUTPUT After {t} iterations, {num}^0.5 = {better_guess}")
break
initial_guess = better_guess
Sample run:
How many numbers am I estimating John?
COUNT> 4
Input each number to estimate.
NUMBER 1> 1
NUMBER 2> 9
NUMBER 3> 16
NUMBER 4> 4
The square roots are as follows:
OUTPUT After 9 iterations, 1.0^0.5 = 1.0
OUTPUT After 7 iterations, 9.0^0.5 = 3.0
OUTPUT After 7 iterations, 16.0^0.5 = 4.0
OUTPUT After 8 iterations, 4.0^0.5 = 2.0
#viggnah is right, and I just ignored the num of iterations. But I think #viggnah 's num of iterations are 1 bigger than the actual num of iterations. E.g., if input is 4 and initial guess is 2, the iteration should be 0 rather than 1. Also I add except in case of illegal input.
I suppose the following code works as you expect.
import math
print("How many numbers am I estimating John?")
count = int(input("COUNT> "))
print("Input each number to estimate.")
better_guess = 0
initial_guess = 10
i = 0
j = 0
t = 1
list = []
while True:
try:
num = float(input("NUMBER> "))
list.append(num)
j = j + 1
if j == count:
print("The square roots are as follows:")
break
except:
print("Invalid input! Try again.")
while i < len(list):
initial_guess = 10
t = 0
while True:
better_guess = (initial_guess + (list[i])/initial_guess) / 2
if initial_guess == better_guess:
print(f"OUTPUT After {t} iterations, {list[i]}^0.5 = {better_guess}")
break
t = t + 1
initial_guess = better_guess
i = i + 1
You need to understand:
We only need initialize guess once for each number. So do it in the first while loop;
tneeds to be updated when initial_guess==better_guess rather than i, I believe this is a clerical error;
initial_guessneeds to be updated in the second loop;
i have to write a hailstone program in python
you pick a number, if it's even then half it, and if it's odd then multiply it by 3 and add 1 to it. it says to continue this pattern until the number becomes 1.
the program will need methods for the following:
accepting user input
when printing the sequence, the program should loop until the number 1.
print a count for the number of times the loop had to run to make the sequence.
here's a sample run:
prompt (input)
Enter a positive integer (1-1000). To quit, enter -1: 20
20 10 5 16 8 4 2 1
The loop executed 8 times.
Enter a positive integer (1-1000). To quit, enter -1: 30
30 15 46 23 70 35 106 53 160 80 40 20 10 5 16 8 4 2 1
The loop executed 19 times.
Enter a positive integer (1-1000). To quit, enter -1: -1
Thank you for playing Hailstone.
right now i have this:
count = 0
def hailstone(n):
if n > 0
print(n)
if n > 1:
if n % 2 == 0:
hailstone(n / 2)
else:
hailstone((n * 3) + 1)
count = count + 1
i don't know what to do after this
Try to think in a modular way, make two functions: check_number() and user_call(). Check_number will verify if the current number in the loop is odd or even and the user_call() just wraps it to count how many times the loop did iterate.
I found the exercise in a great book called Automate Boring Stuff with Python, you have to check it out, if you don't know it already.
Here's my code. Try to use what serves you the best.
from sys import exit
def check_number(number):
if number % 2 ==0:
print(number // 2)
return(number // 2)
else:
print(number*3+1)
return number*3+1
def user_call(number):
count = 0
while number != 1:
count += 1
number = check_number(number)
return count
if __name__ == "__main__":
try:
number = int(input('Give a number \n'))
count = user_call(number)
print('count ',count)
except Exception as e:
exit()
you can use global
visit https://www.programiz.com/python-programming/global-keyword to learn more
import sys
res = []
def hailstone(number):
global res
if number > 1:
if number % 2 == 0:
res.append( number // 2 )
hailstone(res[len(res)-1])
else:
res.append(number * 3 + 1)
hailstone(res[len(res)-1])
return res
number = int(input('Enter a positive integer. To quit, enter -1: '))
if number <= 0 or number == 0:
print('Thank you for playing Hailstone.')
sys.exit()
else:
answers = hailstone(number)
for answer in answers:
print(answer)
print('The loop executed {} times.'.format(len(answers) + 1))
I used recursion to solve the problem.
Heres my code:
Edit: All criteria met
count = 0
list_num = []
def input_check():
number = int(input("Enter a positive integer (1-1000). To quit, enter -1: "))
if number >= 1 and number <= 1000:
hailstone_game(number)
elif number == -1:
return
else:
print("Please type in a number between 1-1000")
input_check()
def hailstone_game(number):
global count
while number != 1:
count += 1
list_num.append(number)
if number % 2 == 0:
return hailstone_game(int(number/2))
else:
return hailstone_game(int(number*3+1))
list_num.append(1) # cheap uncreative way to add the one
print(*list_num, sep=" ")
print(f"The loop executed {count} times.")
return
input_check()
Additional stuff that could be done:
- Catching non-integer inputs using try / except
Keep in mind when programming it is a good habit to keep different functions of your code separate, by defining functions for each set of 'commands'. This leads to more readable and easier to maintain code. Of course in this situation it doesn't matter as the code is short.
Your recursive function is missing a base/terminating condition so it goes into an infinite loop.
resultArray = [] #list
def hailstone(n):
if n <= 0: # Base Condition
return
if n > 0:
resultArray.append(n)
if n > 1:
if n % 2 == 0:
hailstone(int(n/2))
else:
hailstone((n * 3) + 1)
# function call
hailstone(20)
print(len(resultArray), resultArray)
Output
8 [20, 10, 5, 16, 8, 4, 2, 1]
Here's a recursive approach for the problem.
count=0
def hailstone(n):
global count
count+=1
if n==1:
print(n)
else:
if n%2==0:
print(n)
hailstone(int(n/2))
else:
print(n)
hailstone(3*n+1)
hailstone(21)
print(f"Loop executed {count} times")
My task is to:
"Write a program that will keep asking the user for some numbers.
If the user hits enter/return without typing anything, the program stops and prints the average of all the numbers that were given. The average should be given to 2 decimal places.
If at any point a 0 is entered, that should not be included in the calculation of the average"
I've been trying for a while, but I can't figure out how to make the programs act on anything I instruct when the user hits 'enter' or for it to ignore the 0.
This is my current code:
count = 0
sum = 0
number = 1
while number >= 0:
number = int(input())
if number == '\n':
print ('hey')
break
if number > 0:
sum = sum + number
count= count + 1
elif number == 0:
count= count + 1
number += 1
avg = str((sum/count))
print('Average is {:.2f}'.format(avg))
You're very close! Almost all of it is perfect!
Here is some more pythonic code, that works.
I've put comments explaining changes:
count = 0
sum = 0
# no longer need to say number = 1
while True: # no need to check for input number >= 0 here
number = input()
if number = '': # user just hit enter key, input left blank
print('hey')
break
if number != 0:
sum += int(number) # same as sum = sum + number
count += 1 # same as count = count + 1
# if number is 0, we don't do anything!
print(f'Average is {count/sum:.2f}') # same as '... {:.2f} ...'.format(count/sum)
Why your code didn't work:
When a user just presses enter instead of typing a number, the input() function doesn't return '\n', rather it returns ''.
I really hope this helps you learn!
Try this:
amount = 0 # Number of non-zero numbers input
nums = 0 # Sum of numbers input
while True:
number = input()
if not number: # Breaks out if nothing is entered
break
if int(number) != 0: # Only add to the variables if the number input is not 0
nums+=int(number)
amount += 1
print(round(nums/amount,2)) # Print out the average rounded to 2 digits
Input:
1
2
3
4
Output:
2.5
Or you can use numpy:
import numpy as np
n = []
while True:
number = input()
if not number: # Breaks out if nothing is entered
break
if int(number) != 0: # Only add to the variables if the number input is not 0
n.append(int(number))
print(round(np.average(n),2)) # Print out the average rounded to 2 digits
A list can store information of the values, number of values and the order of the values.
Try this:
numbers = []
while True:
num = input('Enter number:')
if num == '':
print('Average is', round(sum(numbers)/len(numbers), 2)) # print
numbers = [] # reset
if num != '0' and num != '': numbers.append(int(num)) # add to list
Benefit of this code, it does not break out and runs continuously.
I'm doing this assignment:
Write a program that prints all even numbers less than the input
number using the while loop.
The input format:
The maximum number N that varies from 1 to 200.
The output format:
All even numbers less than N in ascending order. Each number must be
on a separate line.
N = int(input())
i = 0
while 200 >= N >= 1:
i += 1
if i % 2 == 0 and N > i:
print(i)
and its output like:
10 # this is my input
2
4
6
8
but there is an error about time exceed.
The simple code would be:
import math
N = int(input(""))
print("1. " + str(N))
num = 1
while num < math.ceil(N/2):
print (str(num) + ". " + str(num * 2))
num += 1
The problem is that the while loop never stops
while 200 >= N >= 1 In this case because you never change the value of N the condition will always be true. Maybe you can do something more like this:
N = int(input())
if N > 0 and N <= 200:
i = 0
while i < N:
i += 2
print(i)
else
print("The input can only be a number from 1 to 200")