初始化（复杂）静态数据成员的Python方法 [英] Pythonic Way to Initialize (Complex) Static Data Members

问题描述

我有一个包含复杂数据成员的类，我想保持其静态。我想使用函数将其初始化一次。 Pythonic是这样的：

I have a class with a complex data member that I want to keep "static". I want to initialize it once, using a function. How Pythonic is something like this:

def generate_data():
    ... do some analysis and return complex object e.g. list ...

class Coo:
    data_member = generate_data()
    ... rest of class code ...

函数 generate_data 需要很长时间才能完成，并且返回在a范围内保持不变的数据正在运行的程序。我不希望每次实例化Coo类时都运行它。

The function generate_data takes a long while to complete and returns data that remains constant in the scope of a running program. I don't want it to run every time class Coo is instantiated.

此外，只要我不为在 __ init __ 中，它将保持静态吗？如果Coo中的方法将一些值附加到 data_member （假设它是一个列表），该添加项将可用于其余实例吗？

Also, to verify, as long as I don't assign anything to data_member in __init__, it will remain "static"? What if a method in Coo appends some value to data_member (assuming it's a list) - will this addition be available to the rest of the instances?

谢谢

推荐答案

在所有方面您都是正确的。 data_member 将创建一次，并且可用于 coo 的所有实例。如果有任何实例对其进行修改，则该修改将对所有其他实例可见。

You're right on all counts. data_member will be created once, and will be available to all instances of coo. If any instance modifies it, that modification will be visible to all other instances.

下面是一个示例，展示了所有这些内容，并在最后显示了输出：

Here's an example that demonstrates all this, with its output shown at the end:

def generate_data():
    print "Generating"
    return [1,2,3]

class coo:
    data_member = generate_data()
    def modify(self):
        self.data_member.append(4)

    def display(self):
        print self.data_member

x = coo()
y = coo()
y.modify()
x.display()

# Output:
# Generating
# [1, 2, 3, 4]

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