UnboundLocalError: local variable 'i' referenced before assignment. That is the error I got from my code, but my 'i' is in local range still.
the fuction factor2(n) works by itself. but when put in a for loop, it throw the error. I know there is something behind this, but I can not figure out.
def tri_num(n):
return n * (n + 1) // 2
def factors2(n):
f_ = 1
for i in range(2, int(n ** .5) + 1):
count_ = 0
while n % i == 0:
count_ += 1
n /= i
if count_ != 0:
f_ *= count_ + 1
else:
if n > i:
f_ *= 2
return f_
import itertools
for n in itertools.count(1):
m = tri_num(n)
q= factors2(m)
if q > 500:
print(m,q)
break
*
---------------------------------------------------------------------------
UnboundLocalError Traceback (most recent call last)
<ipython-input-52-07bebaa60ddd> in <module>()
2 for n in itertools.count(1):
3 m = tri_num(n)
----> 4 q= factors2(m)
5 if q > 500:
6 print(m,q)
<ipython-input-51-214ce9ab60ad> in factors2(n)
9 f_ *= count_ + 1
10 else:
---> 11 if n > i:
12 f_ *= 2
13 return f_
UnboundLocalError: local variable 'i' referenced before assignment
*
'else:' is part of the for loop, it should not throw an error
s = factors2(81)
print(s)
give me my answer '5', no error for that.
my script should work and give me two numbers
UPDATE: Thank you, I figured out. It is a bug in my factors2(), it fail to function when n = 0, 1 and 2. it will be fixed
You are iterating over an empty range, which means i is never assigned to before you enter the for loop's else clause.
For instance, the first value yielded by count is 1. Then tri_num(1) == 1, so factors2 gets called with n == 1. This results in a call to range(2, int(1**.5)+1), which results in range(2,2) which is indeed empty. Since the iterator never yields a value, i is never assigned to, and the else clause is entered, where you assume i does have a value. QED.