局部變量引用時我將它的全球

Question

from random import randint 
shifts = [4, 4.2, 5, 6, 7] 
days_names = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday'] 
workers_names = ['Itai', 'Or', 'Reut', 'Kuka', 'Aviel'] 
counter = 1 

def shift_arrange(worker): 
    for day in days.values(): 
     counter+=1 
     global avilable_shifts 
     avilable_shifts = check_avilable_shifts(day) 
     if not random_shifte_selector(worker,day): soft_reset(worker) 

我設置計數器作爲一個全局變量，當我嘗試運行這段代碼我得到的局部變量錯誤：

Traceback (most recent call last): 
    File "C:\Or\mypy\shift creator\shift cretor.py", line 144, in <module> 
    for w in workers.values(): shift_arrange(w) 
    File "C:\Or\mypy\shift creator\shift cretor.py", line 105, in shift_arrange 
    counter+=1 
UnboundLocalError: local variable 'counter' referenced before assignmen 

我看到有些人在這裏問這個問題，他刪除了他的pyc文件或其他東西（我不知道它是什麼），它的工作很好。爲什麼會發生？它不會發生在程序中的其他變量。

感謝或

Answer 1

你需要聲明一個全局變量

def shift_arrange(worker): 
    global counter 
    for day in days.values(): 
     counter+=1 
     ... 

既然你在該範圍內修改counter，蟒蛇把它當作一個局部變量，除非你聲明爲global。如果你只需要閱讀它，那沒有必要。

考慮以下幾點：

這工作：雖然這確實

c = 0 
def f(): 
    print c 
    c = 1 
f() 

：

c = 0 
def f(): 
    global c 
    print c 
    c = 1 
f() 
print c # prints 1, f() modified the global value 

Answer 2

有很多值得

c = 0 
def f(): 
    print c 
f() 

雖然這並不在這裏，但是一般來說，python都有名稱和分配給它們的東西。該名稱可以在本地或全球範圍內。

對於讀取操作，python查看本地範圍，然後查看全局範圍，並使用它找到的第一個（或者如果沒有，則返回錯誤）。

對於寫入... python需要知道把它放在哪裏。通常，它會做的是查看本地範圍，如果它不在那裏，則在那裏創建一個變量並賦值。這會隱藏全局變量。你可以讓它看起來像全局變量，如果它存在的話就使用它 - 但這可能是不可取的&。所以，你需要一種方法來告訴Python使用全局變量，而不是它存在（然後，如果它不存在，它會創建）。

這會導致一些奇怪的行爲，有時也是如此。除了以前的答案...

c = 0 

# Passes. We are just looking up the global variable. 
def f1(x): 
    return x + c 

# Passes, but may not be as expected. Sets a local variable c to a value, does not 
# modify the global one. 
def f2(x): 
    c = x 

# Correct way to do the above; now it sets the global variable. 
def f3(x): 
    global c 
    c = x 

# What if you mix them? 
def f4(x): 
    c = c + x 
# This fails. What happens is that first it sees that it's writing to c, and it's 
# not marked global, so it assigns c to the local space. Then it tries to dereference 
# it. Since we've marked it local, it masks the global one, and since it doesn't 
# have a value, it throws an error. c += x works the same way and also fails. 

# However, this works, though is just as equally wrong: 
def f5(x): 
    d = c 
    c = d + x 
# This one works, because we copy the value of the global c into the local d. 
# Then, the second line assigns a local c to addition of local d and x. 
# But does not update the global. 

# Each line is separate though: 
def f6(x): 
    d = c 
    c = c + 1 
# You might think that d=c already made c local for the whole function. But it doesn't 
# work like that. The first line just looks up the value for c, which it finds 
# globally, but then forgets about it - it cares about the object c is the name of, 
# not the name c itself. The second line still fails.