Here is my code, where I get the syntax error:
def cube(number):
return number*number*number
def by_three(number):
if number % 3==0:
cube(number)
return number
else:
return False
Please note that indentation in Python is extremely important since it defines where blocks start and end. I guess your code should be:
def by_three(number):
if number % 3==0:
cube(number)
return number
else:
return False
This should work:
First, indent your code correctly.
def by_three(number):
if number % 3==0:
cube(number)
return number
else:
return False
For cube(number), you can use base**exponent
def cube(number):
return number**3
Alternatively, number^3 is the same as number*number*number, so
def cube(number):
return number*number*number
In addition to indentation, you might want to use mathematical order of precedence and indicating a * for the multiplication.
This works for me with python 3
#!/usr/local/bin/python3
def cube(number):
return (number * number * number)
def by_three(number):
if (number % 3) == 0:
cubed = cube(number)
return cubed
else:
return False
def main():
x = 3
output = '%d' % by_three(x)
print(output)
if __name__ == "__main__":
main()
Related
My perfect number function is not working as intended :(. It prints false even though it should print true :(
def perfect_check(number):
z = []
for i in range(1, number):
if number % i == 0:
z.append(i)
if sum(z) == number:
return True
else:
return False
print(perfect_check(6))
def perfect_check(number):
z = []
for i in range(1, number):
if number % i == 0:
z.append(i)
if sum(z) == number:
return True
else:
return False
print(perfect_check(6))
You have put the if-else statement inside your for loop. It should be outside the for loop. Then, your code will work correctly.
I have this code for checking if a number is a palindrome and for some reason it returns false for number=1 even though it is palindrome. Why is that? The code works for other cases such as 12321.
def palindrome_integer(number):
if number != int:
return False
elif str(number) == str(number)[::-1]:
return True
else:
return False
If you want to check if number is integer, you should use isistance.
def palindrome_integer(number):
if not isinstance(number, int):
return False
elif str(number) == str(number)[::-1]:
return True
else:
return False
The rest of your code seems to work fine.
One-liner:
return isinstance(n, int) and str(n) == str(n)[::-1]
Or slightly more contrived:
import re
x = str(n)
return re.match(r”\d+“, x) and x == x[::-1]
solution without string
def palindrome_integer(num):
copy = num
rev = 0
while num!= 0:
rev = rev*10+num%10
num = num//10
return rev == copy
def palindrome_integer(number):
return type(number) == int and str(number)[::-1] == str(number)
Don't hesitate to comment if you have problem in understanding the solution.
we need to define a function that compares two strings and if they are different we want to know the index. the problem is that no matter what insert we use we always get -1 even when they are not the same.
def mutation_detector(seq1,seq2):
if DNAval(seq1) and DNAval(seq2) == True:
if len(seq1) == len(seq2):
for i in range(0, len(seq1)) and range(0, len(seq2)):
if seq1[i] != seq2[i]:
return(i)
else:
return(-1)
else:
return('Wrong input')
else:
return('Wrong input')
print(mutation_detector('ATCGGGTA','ATCGGCTA'))
Basically you're using and wrong and have some basic Logic errors I think:
and doesn't say "do A and B" but rather"if A is True and B is True"
My attempt at fixing it is as follows:
if DNAval(seq1) and DNAval(seq2):
if len(seq1) == len(seq2):
for (i, (elem1, elem2)) in enumerate(zip(seq1, seq2)):
if elem1 != elem2:
return i
return -1
Your if-else in the loop always returns in the First loop iteration: it takes the first two chars and compares them; are they equal? No -> return -1
If you follow the logic, it begins comparing strings. If the two letters are the same, it goes for the ELSE (because they are not different) and ends the routine after checking only the first letter.
You only want the routine to return -1 if it makes it all the way through the for loop without returning an index number. So,
Change as below:
def test(seq1, seq2):
if len(seq1) == len(seq2):
for i in range(0, len(seq1)):
if seq1[i] != seq2[i]:
return(i)
return(-1)
else:
return('Wrong input')
print( test('Hello1', 'Hello2'))
print('done')
The problem is, that you are returning -1 on the first time you run through the for loop, because the else-clause is immediately entered.
Use the else clause with the for loop itself instead.
For example:
def compare_strings(seq1, seq2):
if len(seq1) == len(seq2):
for i in range(0, len(seq1)):
if seq1[i] != seq2[i]:
return i
else:
return -1
else:
return 'Wrong input'
(mind you, raising a custom exception might be better than returning "Wrong input" here...)
Here's how I would do it
def mutation_detector(seq1, seq2):
if not DNAval(seq1) or not DNAval(seq2) or len(seq1) != len(seq2):
return "Wrong input" # maybe raise ValueError if DNAval fails instead?
for index, (base1, base2) in enumerate(zip(seq1, seq2)):
if base1 != base2:
return index
return -1
Try this,
It will return index when unable to compare otherwise it will compare two character and to print equal again compare it with length of string.
def mutation_detector(seq1,seq2):
count=0
if len(seq1) == len(seq2):
for i in range(0, len(seq1)) and range(0, len(seq2)):
if seq1[i] != seq2[i]:
return i
else:
count=count+1
if count==len(seq1):
return 'Equal'
print(mutation_detector('ATCGGCTA','ATCGGCTA'))
def comp_string(string1,string2):
if len(string1)!=len(string2):
return "length not equal"
else:
for i in range(len(string1)):
if string1[i] != string2[i]:
return i
return "equal"
Can anyone tell me why when I input 1, 2, 3, and 4 into this code, my output is 6, 2, 3.00? I thought that every time my while loop evaluated to true it would increment the count by one, but the output is not making sense. It's taking the total of 3 of the numbers, but only 2 for the count? I'm probably just overlooking something so an extra pair of eyes would be awesome.
def calcAverage(total, count):
average = float(total)/float(count)
return format(average, ',.2f')
def inputPositiveInteger():
str_in = input("Please enter a positive integer, anything else to quit: ")
if not str_in.isdigit():
return -1
else:
try:
pos_int = int(str_in)
return pos_int
except:
return -1
def main():
total = 0
count = 0
while inputPositiveInteger() != -1:
total += inputPositiveInteger()
count += 1
else:
if count != 0:
print(total)
print(count)
print(calcAverage(total, count))
main()
The error with your code is that on this piece of code...
while inputPositiveInteger() != -1:
total += inputPositiveInteger()
You first call inputPositiveInteger and throw out the result in your condition. You need to store the result, otherwise one input out of two is ignored and the other is added even if it is -1.
num = inputPositiveInteger()
while num != -1:
total += num
count += 1
num = inputPositiveInteger()
Improvements
Although, note that your code can be significantly improved. See the comments in the following improved version of your code.
def calcAverage(total, count):
# In Python3, / is a float division you do not need a float cast
average = total / count
return format(average, ',.2f')
def inputPositiveInteger():
str_int = input("Please enter a positive integer, anything else to quit: ")
# If str_int.isdigit() returns True you can safely assume the int cast will work
return int(str_int) if str_int.isdigit() else -1
# In Python, we usually rely on this format to run the main script
if __name__ == '__main__':
# Using the second form of iter is a neat way to loop over user inputs
nums = iter(inputPositiveInteger, -1)
sum_ = sum(nums)
print(sum_)
print(len(nums))
print(calcAverage(sum_, len(nums)))
One detail worth reading about in the above code is the second form of iter.
def fizz_buzz(i):
if i % 15 == 0:
return ("FizzBuzz")
elif i % 5 == 0:
return ("Buzz")
elif i % 3 == 0:
return ("Fizz")
else:
return (i)
for i in range(1, 21):
print(fizz_buzz(i))
Where and how would do a new line command here with commas?
Trying to get an output like this: 1,2,Fizz,4,Buzz,Fizz,7,8,Fizz,Buzz,11,Fizz,13,14,FizzBuzz,16,17,Fizz,19, Buzz
but sideways and with commas.
Pass end=',' to print()
def fizz_buzz(i):
if i % 15 == 0:
return ("FizzBuzz")
elif i % 5 == 0:
return ("Buzz")
elif i % 3 == 0:
return ("Fizz")
else:
return (i)
for i in range(1, 21):
print(fizz_buzz(i), end=',')
# prints
1,2,Fizz,4,Buzz,Fizz,7,8,Fizz,Buzz,11,Fizz,13,14,FizzBuzz,16,17,Fizz,19,Buzz,
If you do not want the trailing comma, end the range at 20 and follow with print(fizz_buzz(20))
Consider making a list & joining the elements with a comma. There are some great examples here.
For example:
def fizz_buzz(i):
# your code
my_list = []
for i in range(1,21):
my_list.append( str(fizz_buzz(i)) )
print ",".join(my_list)
There are more elegant ways of doing this -- using generators &c, as in the linked answer --, but this simple code will do what you want. Note that the join() method accepts string only, hence the str() in the list.append(); alternatively you could ensure that your fizz_buzz function returns strings regardless.