Local variable referenced before assignment in Python

Question

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.

Ashwini Chaudhary · Accepted Answer · 2013-07-06 20:57:59Z

41

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.

edited Jul 6, 2013 at 20:57

answered Jul 6, 2013 at 20:45

Ashwini Chaudhary

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1 Comment

Or Halimi Over a year ago

thanks, i though that define the var outside of the function will solve this problem. so every time i will use global var in function i will have to define it as global?

rvf0068 · Accepted Answer · 2018-10-08 18:38:56Z

2

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

edited Oct 8, 2018 at 18:38

rvf0068

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answered Oct 29, 2015 at 23:19

Colin Schoen

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