Reputation: 671
Truel=""
count = 0
finle_touch=False #true after it find the first 3 upperletter
# check if there is 1 lower letter after three upper letter
def one_lower(i):
count=0
if i == i.lower:
finle_touch=True
Truel=i
# check for 3 upper letter
def three_upper(s):
for i in s:
if count == 3:
if finle_touch==True:
break
else:
one_lower(i)
elif i == i.upper:
count +=1
print(count) #for debug
else:
count ==0
finle_touch=False
stuff="dsfsfFSfsssfSFSFFSsfssSSsSSSS......."
three_upper(stuff)
print(Truel)
So I have a lot of string on 'stuff' and I like to find 1 lowercase letter that's surrounded by 3 uppercase letter.
But when I run this code I get:
Traceback (most recent call last):
File "C:\Python33\mypy\code.py", line 1294, in <module>
three_upper(stuff)
File "C:\Python33\mypy\code.py", line 1280, in three_upper
if count == 3:
UnboundLocalError: local variable 'count' referenced before assignment
I don't understand why.
Upvotes: 23
Views: 74850
Reputation: 2592
It is actually better to use nonlocal in this case. Use global as sparingly as possible. More information about nonlocal here docs.python.org/3/reference/simple_stmts.html#nonlocal
Upvotes: 2
Reputation: 250891
Due to this line count +=1
python thinks that count
is a local variable and will not search the global scope when you used if count == 3:
. That's why you got that error.
Use global
statement to handle that:
def three_upper(s): #check for 3 upper letter
global count
for i in s:
From docs:
All variable assignments in a function store the value in the local symbol table; whereas variable references first look in the local symbol table, then in the global symbol table, and then in the table of built-in names. Thus, global variables cannot be directly assigned a value within a function (unless named in a global statement), although they may be referenced.
Upvotes: 41