I have been coding this problem for HackerRank and I ran into so many problems. The problem is called "Plus Minus" and I am doing it in Python 3. The directions are on https://www.hackerrank.com/challenges/plus-minus/problem. I tried so many things and it says that "there is no response on stdout". I guess a none-type is being returned. Here is the code.:
def plusMinus(arr):
p = 0
neg = 0
z = arr.count(0)
no = 0
for num in range(n):
if arr[num] < 0:
neg+=1
if arr[num] > 0:
p+=1
else:
no += 1
continue
return p/n
The following are the issues:
1) variable n, which represents length of the array, needs to be passed to the function plusMinus
2) No need to maintain the extra variable no, as you have already calculated the zero count. Therefore, we can eliminate the extra else condition.
3) No need to use continue statement, as there is no code after the statement.
4) The function needs to print the values instead of returning.
Have a look at the following code with proper naming of variables for easy understanding:
def plusMinus(arr, n):
positive_count = 0
negative_count = 0
zero_count = arr.count(0)
for num in range(n):
if arr[num] < 0:
negative_count += 1
if arr[num] > 0:
positive_count += 1
print(positive_count/n)
print(negative_count/n)
print(zero_count/n)
if __name__ == '__main__':
n = int(input())
arr = list(map(int, input().rstrip().split()))
plusMinus(arr, n)
The 6 decimals at the end are needed too :
Positive_Values = 0
Zeros = 0
Negative_Values = 0
n = int(input())
array = list(map(int,input().split()))
if len(array) != n:
print(f"Error, the list only has {len(array)} numbers out of {n}")
else:
for i in range(0,n):
if array[i] == 0:
Zeros +=1
elif array[i] > 0:
Positive_Values += 1
else:
Negative_Values += 1
Proportion_Positive_Values = Positive_Values / n
Proportion_Of_Zeros = Zeros / n
Proportion_Negative_Values = Negative_Values / n
print('{:.6f}'.format(Proportion_Positive_Values))
print('{:.6f}'.format(Proportion_Negative_Values))
print('{:.6f}'.format(Proportion_Of_Zeros))
I am trying to write a function that returns the number of prime numbers that exist up to and including a given number.
Initially this was my code:
def count_primes(num):
prime = [2]
x = 3
if num < 2:
return 0
while x <= num:
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
return len(prime)
How ever I realise this code will run forever because of the following line of code:
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
Can anyone help to explain to me why will this end up with an infinite loop compared to the following code?
for y in prime:
if x%y == 0:
print('not prime')
x+=2
break
else:
prime.append(x)
x += 2
Have tried searching for this, but can't find exactly what I'm looking for.
I want to make a function that will recursively find the factors of a number; for example, the factors of 12 are 1, 2, 3, 4, 6 & 12.
I can write this fairly simply using a for loop with an if statement:
#a function to find the factors of a given number
def print_factors(x):
print ("The factors of %s are:" % number)
for i in range(1, x + 1):
if number % i == 0: #if the number divided by i is zero, then i is a factor of that number
print (i)
number = int(input("Enter a number: "))
print (print_factors(number))
However, when I try to change it to a recursive function, I am getting just a loop of the "The factors of x are:" statement. This is what I currently have:
#uses recursive function to print all the letters of an integer
def print_factors(x): #function to print factors of the number with the argument n
print ("The factors of %s are:" % number)
while print_factors(x) != 0: #to break the recursion loop
for i in range(1,x + 1):
if x % i == 0:
print (i)
number = int(input("Enter a number: "))
print_factors(number)
The error must be coming in either when I am calling the function again, or to do with the while loop (as far as I understand, you need a while loop in a recursive function, in order to break it?)
There are quite many problems with your recursive approach. In fact its not recursive at all.
1) Your function doesn't return anything but your while loop has a comparision while print_factors(x) != 0:
2) Even if your function was returning a value, it would never get to the point of evaluating it and comparing due to the way you have coded.
You are constantly calling your function with the same parameter over and over which is why you are getting a loop of print statements.
In a recursive approach, you define a problem in terms of a simpler version of itself.
And you need a base case to break out of recursive function, not a while loop.
Here is a very naive recursive approach.
def factors(x,i):
if i==0:
return
if x%i == 0:
print(i)
return factors (x,i-1) #simpler version of the problem
factors(12,12)
I think we do using below method:
def findfactor(n):
factorizeDict
def factorize(acc, x):
if(n%x == 0 and n/x >= x):
if(n/x > x):
acc += [x, n//x]
return factorize(acc, x+1)
else:
acc += [x]
return acc
elif(n%x != 0):
return factorize(acc, x+1)
else:
return acc
return factorize(list(), 1)
def factors(x,i=None) :
if i is None :
print('the factors of %s are : ' %x)
print(x,end=' ')
i = int(x/2)
if i == 0 :
return
if x % i == 0 :
print(i,end=' ')
return factors(x,i-1)
num1 = int(input('enter number : '))
print(factors(num1))
Recursion is a functional heritage and so using it with functional style yields the best results. This means avoiding things like mutations, variable reassignments, and other side effects. That said, here's how I'd write factors -
def factors(n, m = 2):
if m >= n:
return
if n % m == 0:
yield m
yield from factors(n, m + 1)
print(list(factors(10))) # [2,5]
print(list(factors(24))) # [2,3,4,6,8,12]
print(list(factors(99))) # [3,9,11,33]
And here's prime_factors -
def prime_factors(n, m = 2):
if m > n:
return
elif n % m == 0:
yield m
yield from prime_factors(n // m, m)
else:
yield from prime_factors(n, m + 1)
print(list(prime_factors(10))) # [2,5]
print(list(prime_factors(24))) # [2,2,2,3]
print(list(prime_factors(99))) # [3,3,11]
def fact (n , a = 2):
if n <= a :
return n
elif n % a != 0 :
return fact(n , a + 1 )
elif n % a == 0:
return str(a) + f" * {str(fact(n / a , a ))}"
Here is another way. The 'x' is the number you want to find the factors of. The 'c = 1' is used as a counter, using it we'll divide your number by 1, then by 2, all the way up to and including your nubmer, and if the modular returns a 0, then we know that number is a factor, so we print it out.
def factors (x,c=1):
if c == x: return x
else:
if x%c == 0: print(c)
return factors(x,c+1)
While I was running the code, I got a "maximum recursion depth exceeded in comparison" error. I'm not exactly sure which part of the code to look at to fix this problem. This numToBinary function is basically supposed to convert a number n to a binary number with bit size k. I would greatly appreciate any input on how I can resolve this issue!
def numToBinary(k, n):
''' converts number to binary number bit size k'''
def binary(n):
if n == 0:
return ''
elif n%2 == 1:
return binary(n/2)+'1'
else:
return binary(n/2)+ '0'
temp = binary(n)
if len(temp) <= k:
answer = '0' * (k - len(temp)) + temp
elif len(temp) > k:
answer = temp[-k:]
return answer
print (numToBinary(6, 10))
You need floor division, double /, in python3 / does truediv so you are getting floats from n/2:
def binary(n):
if n == 0:
return ''
elif n%2 == 1:
return binary(n//2) + '1' # // floor
else:
return binary(n//2)+ '0' # // floor
Once you make the change, it will work fine:
In [50]: numToBinary(6, 10)
Out[50]: '001010'
You can also use else in place of the elif, if the len of temp is not <= then it has to be greater than:
def numToBinary(k, n):
''' converts number to binary number bit size k'''
def binary(n):
if n == 0:
return ''
elif n % 2 == 1:
return binary(n//2)+'1'
else:
return binary(n//2) + '0'
temp = binary(n)
if len(temp) <= k:
answer = '0' * (k - len(temp)) + temp
else:
answer = temp[-k:]
return answer
If you wanted to see exactly what was happening you should put a print in you own code, if you added a print(n) in binary you would see a lot of output like:
5.125332723668738e-143
2.562666361834369e-143
1.2813331809171846e-143
6.406665904585923e-144
3.2033329522929615e-144
Which meant you eventually hit the recursion limit.
from random import randint
stats = {'correct': 0, 'wrong': 0}
correct = 0
for i in range(10):
n1 = randint(1, 10)
n2 = randint(1, 10)
prod = n1 * n2
ans = input("What's %dx%d?" % (n1, n2))
if ans == prod:
print ('Your answer is correct. Well done.')
stats['correct'] += 1
else:
print (('Your answer is wrong. I am afraid the answer is %d.') % prod)
stats['wrong'] += 1
print (("\nI asked you 10 questions. You got %d of them right.") % correct)
print ("Well done!")
The answer received from the input is a string.
if ans == str(prod):
Should do the trick...