I'm trying to write a code for calculating the number of factors for an arbitrary integer number but unfortunately when I run that I receive a false answer
I have tried for loop without defining function in this case and I got the result.Contrary to that, when I define a function I can't see the proper result
r = 0
def factor(a):
global r
for i in range(1, a + 1):
if a % i == 0:
r += 1
return r
a = int(input())
factor(a)
for example 18 has 6 factors but I receive just 1.
Use print to check your code. Indentation in Python matters. Also, global is not needed.
def factor(a):
r = 0
for i in range(1, a + 1):
if a % i == 0:
print('i', i)
r += 1
return r
a = int(input())
print(factor(a))
It was an indentation problem: the function should only return after the loop has finished iterating.
r = 0
def factor(a):
global r
for i in range(1, a + 1):
if a % i == 0:
r += 1
return r
a = 18 # int(input())
factor(a)
output:
6
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))
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,....
I'm a newbie with programming and I'm in trouble with this code:
def supercalcx(a, b):
n = a
while a <= b:
n = n * a
a = a + 1
print(n)
The IDE give me the error: "TypeError: can't multiply sequence by non-int of type 'str'", but I'm sure the inputs are ints or floats, can anyone explain me the problem. Thanks !
This function works:
>>> def supercalcx(a, b):
... n = a
... while a <= b:
... n = n * a
... a = a + 1
... print(n)
...
>>> supercalcx(2, 4)
48
Your function does not convert between data types. A very crude method of this is to do the following below:
def supercalcx(a,b):
n = int(a)
a = int(a)
b = int(b)
while a <= b:
n = n * a
a = a + 1
print(n)
A couple of suggestions to improve your code:
A function should rarely have the print() function inside of it; instead, use the return keyword. You can change a = a + 1 to a += 1 and n = n * a to n *= a. You can also introduce try and except which will attempt to perform whatever is tabbed under try and if something throws an error specified by the except block, it will then perform whatever is tabbed under except. A somewhat improved version is below:
def supercalcx(a, b):
try:
n = int(a)
a = int(a)
b = int(b)
except ValueError:
return "Unable to convert to integers!"
while a <= b:
n *= a
a += 1
return n
print(supercalcx("1", 2))
print(supercalcx(1, 2))
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)
Lets assume that i have 2 strings
M = "sses"
N = "assesses"
I have to count how many times string M is present into string N
I am not allowed to use any import or methods just loops and range() if needed.
M = "sses"
N = "assesses"
counter = 0
if M in N:
counter +=1
print(counter)
This isn't good enough i need loop to go trough N and count all M present
in this case it is 2.
def count(M, N):
i = 0
count = 0
while True:
try:
i = N.index(M, i)+1
count += 1
except ValueError:
break
return count
Or a one-liner without str.index:
def count(M, N):
return sum(N[i:i+len(M)]==M for i in range(len(N)-len(M)+1))
The same without using the sum function:
def count(M, N):
count = 0
for i in range(len(N)-len(M)+1):
if N[i:i+len(M)] == M:
count += 1
return count