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.
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'm currently trying to use simple while loops to count down from a given positive number to zero and then stop. If the number isn't a positive number then it should loop and ask again. I've gotten most of it down, but I can't make it count down to zero. Right now it just stops at 1.
For this problem, zero is not counted as a positive number. If zero is entered, it should loop back to the initial prompt.
This is the code I have so far:
# set loop variable
looking_for_positive_number = True
# get positive number
while (looking_for_positive_number == True):
reply = input("Enter positive number: ")
int_reply = int(reply)
# count down from given positive number to 0, loop back to reply if not
while int_reply > 0:
print(int_reply)
int_reply = int_reply - 1
looking_for_positive_number = False
This is what it returns (I used 5 as the positive number):
Enter positive number: 5
5
4
3
2
1
Ideally, it should return something like this (also used 5 here):
Enter positive number: 5
5
4
3
2
1
0
I don't know what I'm missing/what I'm doing wrong. Any help would be appreciated. Thanks in advance!
EDIT: I figured it out, I just had to include an if statement in the nested while loop. Final code looks like this:
# set loop variable
looking_for_positive_number = True
# get positive number
while (looking_for_positive_number == True):
reply = input("Enter positive number: ")
int_reply = int(reply)
# count down from given positive number to 0, loop back to reply if not
# ZERO DOES NOT COUNT AS POSITIVE NUMBER -- SHOULD LOOP BACK IF 0 IS ENTERED
while int_reply > 0:
print(int_reply)
int_reply = int_reply - 1
# want to print 0 after we get down to 1 (but not include 0 in our initial loop)
if int_reply == 0:
print(0)
# finish and exit loop
looking_for_positive_number = False
# get positive number
while (looking_for_positive_number == True):
reply = input("Enter positive number: ")
int_reply = int(reply)
if int_reply <= 0:
print( "Not a positive number." )
continue
# count down from given positive number to 0, loop back to reply if not
while int_reply > 0:
print(int_reply)
int_reply = int_reply - 1
looking_for_positive_number = False
The Python range function could simplify your logic (slightly). Using a negative step value it will count down instead of counting up.
Example:
while True:
number = int(input("Enter a postive integer value: "))
if number <= 0:
print("Not a positive number.")
continue
for i in range(number, -1, -1):
print(i)
break
Output:
Enter a postive number: 5
5
4
3
2
1
0
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")
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.
What this is supposed to do is:
To ask the user for numbers using while loop until they enter 0
when 0 is pressed we need to print the average of the numbers entered so far.
(Kindly help)
import statistics as st
provided_numbers = []
while True:
number = int(input('Write a number'))
if number == 0:
break
else:
provided_numbers.append(number)
print(f'Typed numbers: {provided_numbers}')
print(f'The average of the provided numbers is {st.mean(provided_numbers)}')
Take each number and print the average so far and stop it when input is zero, right?
s = 0
count = 0
while True:
num = int(input('Write a number: '))
if num == 0:
break
s += num
count += 1
print("Average so far:",(s/count))
Could you add a piece of code you have tried or sample input and output too?