为什么这个python逻辑语句的行为与我预期的行为相反？

我在Python解释器中运行以下命令：

  some_list = []
  methodList = [method for method in dir(some_list) if (callable(getattr(some_list, method)) and (not method.find('_')))]

我想得到的是一个特定对象的所有方法名称的列表，除了用下划线命名的方法，即

\uuuuuuuuizeof\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

这就是为什么我在上面的代码中嵌套了if语句：

 if (callable(getattr(some_list, method)) and (not method.find('_')))

但是

methodList

的内容是：

“UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU"初始","迭代","新","减少","减少","减少","减少","报告","反转"__“、”、“、”、“、”、“、”、“、”、“、”、“、”、“、”、“、”、“、”、“、”、“、”、“、”、”、“、”、“、”、”、“、”子类钩子”]

事实上，这与我的预期正好相反

不应

而不是方法。只有当
方法
字符串未能包含字符串
'
？
时，find（'''
）才会返回true以获取详细信息

返回找到子字符串sub的字符串中的最低索引，例如sub包含在切片s[start:end]中。可选参数start和end解释为切片表示法。如果未找到sub，则返回-1

表达式
method.find（“'）
如果未找到下划线，则返回-1，如果以下划线开头，则返回0。应用
not
意味着只有以下划线开头的方法才会给出
True
（因为
not 0
是
True
）

使用
“not in method”
。
最好是
“not in method”
@StevenRumbalski:好建议。谢谢！澄清了问题。从perl迁移到python是一个奇怪的世界。我认为如果你试图找到神奇的方法，那么
startswith（“''.'''
将比
in
好。或者
not method.startswith（“”“）
如果要排除魔术方法和私有方法。