Write a loop that reads positive integers from standard input and that terminates when it reads an integer that is not positive. After the loop terminates, it prints out, on a line by itself, the sum of all the even integers read.
number =int(input())
even_number = 0
while number >= 0:
if number % 2 == 0:
even_number += number
print(even_number)
number = int(input())
it says:
Exception occurred(, EOFError('EOF when reading a line',), )
Exception occurred(, EOFError('EOF when reading a line',), )
The value of _stdout is incorrect.
We think you might want to consider using: >
We think you might want to consider using: sum
We think you might want to consider using: +
Solutions with your approach don't usually use: +=
After the loop terminates, it prints out, on a line by itself, the sum of all the even integers read.
You seem to be printing each time the sum is updated, instead of just once at the end of the loop. Try moving the print to after the while block.
Related
I need to identify the count of numbers with non-repeating digits in the range of two numbers.
Suppose n1=11 and n2=15.
There is the number 11, which has repeated digits, but 12, 13, 14 and 15 have no repeated digits. So, the output is 4.
Wrote this code:
n1=int(input())
n2=int(input())
count=0
for i in range(n1,n2+1):
lst=[]
x=i
while (n1>0):
a=x%10
lst.append(a)
x=x//10
for j in range(0,len(lst)-1):
for k in range(j+1,len(lst)):
if (lst[j]==lst[k]):
break
else:
count=count+1
print (count)
While running the code and after inputting the two numbers, it does not run the code but still accepts input. What did I miss?
The reason your code doesn't run is because it gets stuck in your while loop, it can never exit that condition, since n1 > 0 will never have a chance to be evaluated as False, unless the input itself is <= 0.
Anyway, your approach is over complicated, not quite readable and not exactly pythonic. Here's a simpler, and more readable approach:
from collections import Counter
n1 = int(input())
n2 = int(input())
count = 0
for num in range(n1, n2+1):
num = str(num)
digit_count = Counter(num)
has_repeating_digits = any((True for count in digit_count.values() if count > 1))
if not has_repeating_digits:
count += 1
print(count)
When writing code, in general you should try to avoid nesting too much stuff (in your original example you have 4 nested loops, that's readability and debugging nightmare), and try using self-describing variable names (so a, x, j, k, b... are kind of a no-go).
If in a IPython session you run import this you can also read the "Zen of Python", which kind of sums up the concept of writing proper pythonic code.
I'm writing a program to check if a given user input is a palindrome or not. if it is the program should print "Yes", if not "no". I realize that this program is entirely too complex since I actually only needed to check the whole word using the reversed() function, but I ended up making it quite complex by splitting the word into two lists and then checking the lists against each other.
Despite that, I'm not clear why the last conditional isn't returning the expected "Yes" when I pass it "racecar" as an input. When I print the lists in line 23 and 24, I get two lists that are identical, but then when I compare them in the conditional, I always get "No" meaning they are not equal to each other. can anyone explain why this is? I've tried to convert the lists to strings but no luck.
def odd_or_even(a): # function for determining if odd or even
if len(a) % 2 == 0:
return True
else:
return False
the_string = input("How about a word?\n")
x = int(len(the_string))
odd_or_even(the_string) # find out if the word has an odd or an even number of characters
if odd_or_even(the_string) == True: # if even
for i in range(x):
first_half = the_string[0:int((x/2))] #create a list with part 1
second_half = the_string[(x-(int((x/2)))):x] #create a list with part 2
else: #if odd
for i in range(x):
first_half = the_string[:(int((x-1)/2))] #create a list with part 1 without the middle index
second_half = the_string[int(int(x-1)/2)+1:] #create a list with part 2 without the middle index
print(list(reversed(second_half)))
print(list(first_half))
if first_half == reversed(second_half): ##### NOT WORKING BUT DONT KNOW WHY #####
print("Yes")
else:
print("No")
Despite your comments first_half and second_half are substrings of your input, not lists. When you print them out, you're converting them to lists, but in the comparison, you do not convert first_half or reversed(second_half). Thus you are comparing a string to an iterator (returned by reversed), which will always be false.
So a basic fix is to do the conversion for the if, just like you did when printing the lists out:
if list(first_half) == list(reversed(second_half)):
A better fix might be to compare as strings, by making one of the slices use a step of -1, so you don't need to use reversed. Try second_half = the_string[-1:x//2:-1] (or similar, you probably need to tweak either the even or odd case by one). Or you could use the "alien smiley" slice to reverse the string after you slice it out of the input: second_half = second_half[::-1].
There are a few other oddities in your code, like your for i in range(x) loop that overwrites all of its results except the last one. Just use x - 1 in the slicing code and you don't need that loop at all. You're also calling int a lot more often than you need to (if you used // instead of /, you could get rid of literally all of the int calls).
I am programming a program that will make a list of all numbers from 1 to 200 that:
are not divisible by 7 or 11
do not contain the digit 7 or 11 in their number.
I want to use the pass function so when the condition is not met, it will continue with the next number. I don't really know how to do it. The pass function is probably the wrong one. I know the break function does also not work because it will end the whole loop.
Please explain me how to make this program work on this way. There are probably plenty other ways to calculate this, but the point is that i want to understand how to use the for loops better :).
n = 200 #all digits till and including 200
numbers = [] #empty list for accumulation values
for i in range(1,(n+1)):
if i%7 == 0 or i%11 == 0: #if one of these 3 conditions are met
pass #it should continue to the next number (i)
if str(7) in str(i):
pass
if str(11) in str(i):
pass
numbers.append(i)
print(numbers)
print(sum(numbers)) # for my assignment i need to sum the list
use continue in place of pass.
pass does nothing
continue skips to the next loop
so i had to use continue in my example.
I am a novice trying to learn python from Automate The Boring Stuff with Python by Al Sweigart and I came across his block of code to answer a math problem: "What is the sum of all the numbers from 0 to 100?" Apparently, this was a question Gauss got when his teacher wanted to keep him busy.
Sweigart used a for loop and range() function to get the answer:
total = 0
for num in range(101):
total=total+num
print(total)
A page later he states that "you can actually use a while loop to do the same thing as a for loop; for loops are more concise."
How would this statement be rendered in a while loop?
I tried replacing the for with while but got an error: "name 'num' is not defined." I also tried to set up a summation math equation using another block of code from another forum, but completely got lost.
print('Gauss was presented with a math problem: add up all the numbers from 0 to 100. What was the total?')
a=[1,2,3,4,5,...,100]
i=0
while i< len(a)-1:
result=(a[i]+a[i+1])/2
print(result)
i +=1
Then, I tried to setup i in an equation that would loop until each number was added, but got stuck.
print('Gauss was presented with a math problem: add up all the numbers from 0 to 100. What was the total?')
i=0
while i<101:
i=i+1
a=i
Would the while statement be too complex to warrant the effort?
Your last example comes close.
A for loop of this form:
for x in range(N):
# ...
can be replaced by a while loop like this:
x = 0
while x < N:
# ...
x += 1 # equivalent to x = x + 1
Just make sure you leave the rest of the code unchanged!
The for loop is more concise. Notice how we need a "counter" variable , in this case i with the while loop. That's not to say we don't need them in a for loop, however they're integrated nicely into the syntax to make for cleaner code.
i = 0
total = 0
while i < 101:
total += i
i += 1
print(total)
Python's for loop syntax is also a foreach equivalent:
for eachItem in list:
# Do something
I'm working on my final for a class I'm taking(Python 3) im stuck at this part.
he gave us a file with numbers inside of it. we opened it and add those numbers to a list.
"Create a function called makeOdd() that returns an integer value. This function should take in any integer and reduce it down to an odd number by dividing it in half until it becomes an odd number.
o For example 10 would be cut in half to 5.
o 9 is already odd, so it would stay 9.
o But 12 would be cut in half to 6, and then cut in half again to 3.
o While 16 would be cut to 8 which gets cut to 4 which gets cut to 2 which gets cut to 1.
Apply this function to every number in the array. "
I have tried to search the internet but i have not clue where to even begin with this one. any help would be nice.
Here my whole final so far:
#imports needed to run this code.
from Final_Functions import *
#Defines empty list
myList = []
sumthing = 0
sortList = []
oddList = []
count = 0
#Starts the Final Project with my name,class, and quarter
intro()
print("***************************************************************",'\n')
#Opens the data file and reads it then places the intrager into a list we can use later.
with open('FinalData.Data', 'r') as f:
myList = [line.strip() for line in f]
print("File Read Complete",'\n')
#Finds the Sum and Adverage of this list from FinalData.Data
print("*******************sum and avg*********************************")
for oneLine in myList:
tempNum = int(oneLine)
sumthing = sumthing + tempNum
avg = sumthing /1111
print("The Sum of the List is:",sumthing)
print("The Adverage of the List is:",avg,'\n')
print("***************************************************************",'\n')
#finds and prints off the first Ten and the last ten numbers in the list
firstTen(myList)
lastTen(myList)
print("***************************************************************",'\n')
#Lest sort the list then find the first and last ten numbers in this list
sortList = myList
sortList.sort()
firstTen(sortList)
lastTen(sortList)
print("****************************************************************",'\n')
Language:Python 3
I don't want to give you the answer outright, so I'm going to talk you through the process and let you generate your own code.
You can't solve this problem in a single step. You need to divide repeatedly and check the value every time to see if it's odd.
Broadly speaking, when you need to repeat a process there are two ways to proceed; looping and recursion. (Ok, there are lots, but those are the most common)
When looping, you'd check if the current number x is odd. If not, halve it and check again. Once the loop has completed, x will be your result.
If using recursion, have a function that takes x. If it's odd, simply return x, otherwise call the function again, passing in x/2.
Either of those methods will solve your problem and both are fundamental concepts.
adding to what #Basic said, never do import * is a bad practice and is a potential source of problem later on...
looks like you are still confuse in this simple matter, you want to given a number X reduce it to a odd number by dividing it by 2, right? then ask yourself how I do this by hand? the answer is what #Basic said you first ask "X is a even number?" if the answer is No then I and done reducing this number, but if the answer is Yes then the next step dividing it by 2 and save the result in X, then repeat this process until you get to the desire result. Hint: use a while
to answer your question about
for num in myList:
if num != 0:
num = float(num)
num / 2
the problem here is that you don't save the result of the division, to do that is as simple as this
for num in myList:
if num != 0:
num = float(num)
num = num / 2