This code works but it is not very efficient is there any help on a faster code in python to find n knowing that n is an integer above 0 and that n has no upper bound, how(x) will return you 1 if x>n, 0 if x = n, and -1 if x
def how(x):
if x > n:
return 1
elif x < n:
return -1
else:
return 0
def find(how):
if how(1) == 1:
return 1
x = 2
while how(x) != 1:
x = x**x
v = x
while how(x) != 0:
if how(x) == 1:
v = x
x = (x+1)//2
else:
x += (v-x+1)//2
return x
Rebecca, I've added some print statements so you can see where goes what wrong. As Patrick Artner said... its a bit confusing which way to go so I've tried to clean-up some things that enable you to continue exploring comparison of two variables against each other (and fake error catching (0).
Lets start and remove the confusing lingo and produce something workable code. With current below script it runs and with value = 1, reference = 1 you get the below print result in a continues loop until YOU stop the script manually:
v1 = n: error
loop1 1
def selector(v1, n):
if v1 > n:
print 'v1 > n', v1, n
return 1
elif v1 < n:
print 'v1 < n', v1, n
return -1
else:
print 'v1 == n: error'
return 0
def find(value, reference):
if selector(value, reference) == 1:
return 1
while selector(value, reference) != 1:
x = value**value
print 'loop1', x
v = x
while selector(value, reference) != 0:
print 'loop2'
if selector(value, reference) == 1:
v = value
x = (value+1)/2
print 'loop2-if', v, x
else:
x += (v-(value+1))/2
print 'loop2-else', x
print ' Almost done...'
return x
if __name__ == '__main__':
n = 1
print find(1, 1)
Happy exploring,....
Related
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've tested to see if the list is empty and I'm sure it's not empty. So why this error continues?
The code is to find the smallest amount of factorial to a sum is equal to N.
def fat(n):
if n == 0 or n == 1:
return 1
else:
return n * fat(n - 1)
n = int(input())
ns = [x+1 for x in range(n)]
listaFat = []
l = []
for x in ns:
if fat(x) < n:
listaFat.append(x)
maior = max(listaFat)
listaFat.remove(maior)
print(listaFat)
print(len(listaFat))
while (fat(listaFat[-1]) + fat(maior)) < n:
l.append(listaFat[-1])
if len(listaFat) > 0:
listaFat.remove(listaFat[-1])
continue
else: break
print(l)
print(listaFat)
I have created the below script to find out the prime factors of a number :
def check_if_no_is_prime(n):
if n <= 3:
return True
else:
limit = int(math.sqrt(n))
for i in range(2,limit + 1):
if n % i == 0:
return False
return True
def find_prime_factors(x):
prime_factors = []
if check_if_no_is_prime(x):
prime_factors.append(1)
prime_factors.append(x)
else:
while x % 2 == 0 and x > 1:
prime_factors.append(2)
x = x // 2
for i in range(3,x+1,2):
while x % i == 0 and x > 1:
if check_if_no_is_prime(i):
prime_factors.append(i)
x = x // i
if x <= 1:
return prime_factors
return prime_factors
no = int(input())
check = find_prime_factors(no)
print (check)
I am not sure whether this is the best and efficient way to do this ?
Can someone please point out any better way to do this ?
using sieve of erathnostanes to get all prime numbers from 2 to whatever limit inputted
def sieve(N):
from math import floor,sqrt
A=[1 for x in range(N+1)]
for count in range(2):
A[count]=0
for i in range(floor(sqrt(N))+1):
if A[i]==1:
for k in range(i*i,N+1,i):
A[k]=0
ans=list(enumerate(A))
res=[]
for (i,j) in ans:
if j==1:
res+=[i]
return res
print(sieve(100))
#my code
I wrote the python code below that solves and prints each possible solution for anything under 6 unit fractions, but given how I programmed it, it takes infinitely long to check for 7 fractions. Any ideas on how to modify the code to find all the possible solutions more efficienty?
import sys
from fractions import Fraction
import os
#myfile = open('7fractions.txt', 'w')
max = 7 #>2 #THIS VARIABLE DECIDES HOW MANY FRACTIONS ARE ALLOWED
A = [0] * max
A[0] = 1
def printList(A):
return str(A).strip('[]')
def sumList(A):
sum = 0
for i in A:
if i != 0:
sum += Fraction(1, i)
return sum
def sumTest(A):
sum = 0
v = 0
for i in range(0, len(A)):
if A[i] == 0 and v == 0:
v = Fraction(1,A[i-1])
if v != 0:
sum += v
else:
sum += Fraction(1, A[i])
return sum
def solve(n, A):
if n == max - 2:
while (sumTest(A) > 1):
print(A)
if sumList(A) < 1:
e = 1 - sumList(A)
if e.numerator == 1 and e.denominator>A[n-1]:
A[n+1] = e.denominator
#myfile.write(printList(A) + '\n')
print(A)
A[n+1] = 0
A[n] += 1
else:
while (sumTest(A) > 1):
if sumList(A) < 1:
A[n+1] = A[n] + 1
solve(n+1, A)
A[n+1] = 0
A[n] += 1
#execute
solve(0, A)
def primetest(x):
if x < 2:
return False
if x == 2:
return True
if x % 2 == 0:
return False
for i in range(3,(x**0.5)+1):
if x % i == 0:
return False
return True
def nthprime(n):
primes = []
x = 2
while len(primes) < n:
if primetest(x) == True:
primes.append(x)
x = x + 1
return list(-1)
print nthprime(10001)
Whenever I try to run this it says that "print nthprime(10001)" is invalid syntax.
-prime test is to test wether a number is prime and nthprime creates a list of prime numbers a certain lengths and then return the last element of the list.
print is a function in Python 3, not a statement. You should change your last line of code to:
print(nthprime(10001))
In your code:
def nthprime(n):
primes = []
x = 2
while len(primes) < n:
if primetest(x) == True:
primes.append(x)
x = x + 1
return list(-1) // this is the error
I think you meant primes[-1], like this:
def nthprime(n):
primes = []
x = 2
while len(primes) < n:
if primetest(x) == True:
primes.append(x)
x = x + 1
return primes[-1] // this is now correct
You're also going to need to specify a range in integers, not float. So this:
for i in range(3,(x**0.5)+1):
Becomes this:
for i in range(3,int((x**0.5)+1)): // note the "int"