[英]How to change default value of optional function parameter
I need to change the global variable S
at a.py
from b.py
, but it is used as a default value in a function at a.py
.我需要从
b.py
更改a.py
的全局变量S
,但它用作a.py
函数中的默认值。
a.py一个.py
S = "string"
def f(s=S):
print(s)
print(S)
b.py b.py
import a
def main():
a.S = "another string"
a.f()
if __name__ == "__main__":
main()
python b.py
outputs python b.py
输出
string
another string
instead of expected而不是预期
another string
another string
If I call af
in b.py
like this如果我像这样在
b.py
调用af
a.f(a.S)
this works as expected, but is there any way to change default variable value?这按预期工作,但是有什么方法可以更改默认变量值吗?
The short answer is: You can't.简短的回答是:你不能。
The reason for this is that the function default arguments are created at function definition time, and the defaults are not meant to be re-defined.这样做的原因是函数默认参数是在函数定义时创建的,默认值并不意味着要重新定义。 The variable name is bound once to a value and that is all, you can't re-bind that name to another value.
变量名称与一个值绑定一次,仅此而已,您不能将该名称重新绑定到另一个值。 First, let's look at variables in global scope:
首先,让我们看看全局范围内的变量:
# create a string in global scope
a = "string"
# b is "string"
b = a
a += " new" # b is still "string", a is a new object since strings are immutable
You've now just bound a new name to "string", and "string new" is a completely new value bound to a, it does not change b because str += str
returns a new str
, making a
and b
refer to different objects.您现在刚刚将一个新名称绑定到“string”,而“string new”是一个绑定到 a 的全新值,它不会改变 b 因为
str += str
返回一个新的str
,使a
和b
指代不同对象。
The same happens with functions:函数也会发生同样的情况:
x = "123"
# this expression is compiled here at definition time
def a(f=x):
print(f)
x = "222"
a()
# 123
The variable f
was defined with the default of "123"
at definition time.变量
f
在定义时使用默认值"123"
定义。 This can't be changed.这是无法改变的。 Even with mutable defaults such as in this question:
即使使用可变默认值,例如在这个问题中:
x = []
def a(f=x):
print(x)
a()
[]
# mutate the reference to the default defined in the function
x.append(1)
a()
[1]
x
[1]
The default argument was already defined, and the name f
was bound to the value []
, that cannot be changed.默认参数已经定义,名称
f
绑定到值[]
,不能更改。 You can mutate the value associated with f
, but you cannot bind f
to a new value as a default.您可以改变与
f
关联的值,但不能将f
绑定到新值作为默认值。 To further illustrate:进一步说明:
x = []
def a(f=x):
f.append(1)
print(f)
a()
x
[1]
# re-defining x simply binds a new value to the name x
x = [1,2,3]
# the default is still the same value that it was when you defined the
# function, albeit, a mutable one
a()
[1, 1]
It might be better to either A) pass the global variable in as an argument to the function or B) use the global variable as global
. A) 将全局变量作为参数传递给函数或 B) 使用全局变量作为
global
可能会更好。 If you are going to change the global variable you wish to use, don't set it as a default parameter and choose a more suitable default:如果要更改要使用的全局变量,请不要将其设置为默认参数并选择更合适的默认值:
# some global value
x = "some default"
# I'm choosing a default of None here
# so I can either explicitly pass something or
# check against the None singleton
def a(f=None):
f = f if f is not None else x
print(f)
a()
some default
x = "other default"
a()
other default
a('non default')
non default
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