将整数解释为枚举标志 [英] interpret an integer as enum flags

问题描述

假设我有一个这样的Flag枚举:

Suppose I have a Flag enum like this:

From enum import Flag
class suspicion(Flag):
     TOT_PCNT_LOW = 1
     IND_PCNT_LOW = 2
     IND_SCORE_LOW = 4

假设我有一个使用这些标志构造的整数

And suppose I have an integer that was constructed using these flags

i = suspicion.IND_PCNT_LOW.value | suspicion.TOT_PCNT_LOW.value      # i == 3

如何获取整数中设置的标志名称的列表?例如:

How do I get a list of the names of the flags that are set in the integer? For example:

print(some_method(i))   # prints ["IND_PCNT_LOW", "TOT_PCNT_LOW"]

推荐答案

设置代码:

from enum import Flag
class Suspicion(Flag):
     TOT_PCNT_LOW = 1
     IND_PCNT_LOW = 2
     IND_SCORE_LOW = 4

a_flag = Suspicion.TOT_PCNT_LOW | Suspicion.IND_PCNT_LOW

在3.10之前的Python中

In Python before 3.10

def flag_names(flag):
    return [f.name for f in flag.__class__ if f & flag == f]

---

在3.10及更高版本中，它会更容易

---

It's slightly easier in 3.10+

def flag_names(flag):
    # Python 3.10+
    return [f.name for f in flag]

---

无论哪种情况:

---

In either case:

print(flag_names(a_flag))

得到你

['TOT_PCNT_LOW', 'IND_PCNT_LOW']

---

如果传入的 flag 参数可以是整数而不是 Suspicion 成员，则您也可以处理该问题:

---

If the incoming flag argument may be an integer instead of a Suspicion member, you can handle that as well:

def flag_names(flag, enum=None):
    """
    returns names of component flags making up flag
    if `flag` is an integer, `enum` must be enumeration to match against
    """
    if enum is None and not isinstance(flag, Flag):
        raise ValueError('`enum` must be specifed if `flag` is an `int`')
    enum = enum or flag.__class__
    return [f.name for f in enum if f & flag == f]

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