Syntax error in a while loop

[calvin dong]

In my program to brute force all the factors (not including 1) of a number, there is a syntax error in my while loop. The code looks like this.

number = raw_input(" Enter a number ")
x = 2
coollist = []
while x <= sqrt(int(number):
    if number % loop == 0:
        coollist.append(x)
    else:
        x += 1
print coollist

The syntax error reads


admin-trs-imac-2:programmingst uff yandong$ python factors.py
File "factors.py ", line 4
while x <= sqrt(int(number ):
^
SyntaxError: invalid syntax


When I take out the colon, I get a syntax error in my if statement, again saying that the colon is invalid syntax.When I take that out, i get another syntax error saying that


admin-trs-imac-2:programmingst uff yandong$ python factors.py
File "factors.py ", line 5
x += 1
^
SyntaxError: invalid syntax

Am i doing something majorly wrong? I cannot figure out what.

(by the way there are indents in the correct places but they do not show up)
(Also the arrow in the syntax error in the while loop should be pointing to the colon)

[dwblas]

Count the parenthesis.

[dwblas]

if number % loop == 0:

loop is not declared any where. Also, if this condition is True, it will be an infinite loop, since the value of "x" won't change. Can't tell anything else without code tags (the "#" symbol) at the top of the post box.

[calvin dong]

So, what would you recommend in order to fix the syntax error?

[bvdet]

This would fix your syntax error (missing closing parenthesis):while x <= sqrt(int(number )):
Your while loop could be a for loop similar to this:for i in xrange(2, int(num**0.5)+1 )

Then if not num%i, add the number i to your factors list.

raw_input returns a string, so you need to convert number to an integer.

[Glenton]

Hi. Are you looking for *all* factors? Because even if you fix your code you won't find them this way. You'd need to include number/x also to find the factors that are bigger than sqrt(number).

I think sympy has some functions to find all factors.

[dwblas]

Any factors larger than the square root will have a corresponding factor less than. For example, 20 and 5 are factors of 100, but 20 was found when 5 was tested, so anything greater than 10 is redundant.

[bvdet]

Hey Glenton. It's been a while. Glad to have you back.

This will compile the factors up to num**0.5:

factors = [i for i in xrange(2, int(num**0.5)+1) if not num%i]

This will extend the list the rest of the way:

factors.extend([num/outList[i] for i in range(len(outList)-1, -1, -1)])

If the last number in factors was the sqrt of num:

factors.extend([num/outList[i] for i in range(len(outList)-2, -1, -1)])

Anyone have a better way?

[Glenton]

Thanks, bvdet. It has been a while. Too little python in my life at the moment...

If you're allowed to use numpy you could use a mask. It's more brute force in the sense that it just checks everything:

import number
factors=numpy.arange(2,n+1)
factors[n%factors==0]

where n is the number.

if you want to be a bit more clever (as per OP) you could do this:

import numpy
factors=numpy.arange(2,int(n**0.5)+1)
f1=factors[n%factors==0]
s=set(f1).union(set(n/f1))

[calvin dong]

i still get an error(when adding the extra parenthese at the end of sqrt in my while loop) it says that "sqrt" is not defined. I used bvdet's code exactly.(His first revision)

[bvdet]

You need this:from math import sqrt

[calvin dong]

I changed my code, but for some reason it still doesn't work. There isn't an error displayed, just nothing happens.

number = int(input(" Enter a number "))
x = 1
coollist = []
while x <= pow(int(number),0.5):
    if (number % x) == 0:
        coollist.append(x)
    else:
        x += 1
coollist.append(number)
print coollist

[Glenton]

How are you running your script?

[bvdet]

In the while loop, if (number % x) == 0 is True, x never increments and your loop will never end.

[calvin dong]

I feel stupid for not having caught this obvious error...thank you all.