Question

好的，这是我的代码：

tick = 1
week = 1
month = 1
year = 1

def new_week(week, tick):
    week = week + 1
    tick = tick + 1
    if tick == 4:
        new_month()
        tick = 1

new_week(week, tick)
print week, tick
new_week(week, tick)
print week, tick

无论后来多少次我告诉它运行new_week（）函数并打印周＆＃39;周的值。并且＆＃39; tick＆＃39;，它总是为两者打印1 ...为什么它忽略了'周=周+ 1＆＃39; ＆＃39; tick＆＃39; ??

我正在运行Python 2.7 btw

Answer 1

全局名称week和tick被函数参数的名称所遮蔽。

您需要删除这些参数（因为它们是不必要的），并且还将week和tick声明为函数内的全局（这将允许您修改它们的值）：

tick = 1
week = 1
month = 1
year = 1

def new_week():
    global week, tick  # Declare that week and tick are global
    week = week + 1    # Modify the value of the global name week
    tick = tick + 1    # Modify the value of the global name tick
    if tick == 4:
        new_month()
        tick = 1

new_week()
print week, tick
new_week()
print week, tick

输出：

2 2
3 3

但是请注意，许多Python程序员认为使用全局声明的函数非常不优雅。更加pythonic的方法是使用值的字典，并有一个函数来递增它们。

以下是一个非常简单的示例，可以帮助您入门：

dct = {
    'tick': 1,
    'week': 1,
    'month': 1,
    'year': 1
}

def increment(key):
    dct[key] += 1

increment('week')
print dct['week']
increment('tick')
print dct['tick']

输出：

2
2

Answer 2

这是关于范围的。

# here you are setting your variables, there scope is essentially global
tick = 1
week = 1
month = 1
year = 1

# we enter the function definition
def new_week(week, tick):
# you passed in your variables, and increment them
# however, they are only changed within the local scope of this function
# and since you didn't RETURN them as shown below, you are leaving the
# global variables from above, unchanged
week = week + 1
tick = tick + 1
if tick == 4:
    new_month()
    tick = 1

# when you print them out below, you are printing the unchanged globally scoped variables
new_week(week, tick)
print week, tick 
new_week(week, tick)
print week, tick

为了获得理想的结果，您可以这样做：

tick = 1
week = 1
month = 1
year = 1

def new_week(week, tick):
    week += 1
    tick += 1
    if tick == 4:
        new_month()
        tick = 1
    return (week, tick)  #### return the variables

>>> week, tick = new_week(week, tick) ### capture the returned variables
>>> print week, tick
(2, 2)

每次致电tick时，week和new_week都会继续增加。显然，tick一旦达到4就会重置。如果您想将变量存储在list或dict中，那么您可以执行以下操作：

使用dict：

vars = {'tick': 1, 'week': 1, 'month': 1, 'year': 1}

现在改变了函数处理数据的方式：

    def new_week(lst=None):
        if lst:
            for key in lst:
                vars[key] += 1
            if vars['tick'] == 4:
                new_month()
                vars['tick'] = 1

>>> new_week(['week', 'tick'])
>>> print vars['week'], vars['tick']
2 2

在此处阅读范围：

https://www.inkling.com/read/learning-python-mark-lutz-4th/chapter-17/python-scope-basics

http://www.python-course.eu/python3_global_vs_local_variables.php

Answer 3

首先，week = week + 1只是将（本地）名称week重新绑定为新值;它不会影响调用范围中的变量。但是，由于int是不可变的，因此即使week+=1也不会影响调用范围;它仍然只是更新函数本地week变量。

一种选择是使用dict而不是单个变量。由于dict是一个可变对象，因此您可以将引用传递给dict作为参数，并且对该对象的更改将在调用范围中可见。

times = { 'tick': 1, 'week': 1, 'month': 1, 'year': 1}

def new_week(t):
   t['week'] += 1
   t['tick'] += 1
   if t['tick'] == 4:
       new_month(t)
       t['tick'] = 1

new_week(times)

值不加

3 个答案: