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我想编写一个函数来构建需要使用具有不同值的 f 字符串的字节类型字符串。我能想到的唯一方法就是下面的代码。有人有更好的建议吗？在代码中，我有字符串，就像我有级别一样，但在我的实际代码中，字符串大约是 600 个字符

def get_level_string(x):
    size = dict(v1=1, v2= 200, v3= 30000)
    s = size.get('v1')
    name = lambda x: f"I have level value as {x} in the house"
    return {
            'level1': b'%a' % (name(size['v1'])),
            'level2': b'%a' % (name(size['v2'])),
            'level3': b'%a' % (name(size['v3'])),
            }[x]

a = get_level_string('level1')
b = get_level_string('level2')
c = get_level_string('level3')
print(a, type(a))
print(b, type(b))
print(c, type(c))
=> #b"'I have level value as 1 in the house'" <class 'bytes'>
=> #b"'I have level value as 200 in the house'" <class 'bytes'>
=> #b"'I have level value as 30000 in the house'" <class 'bytes'>
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1 回答 1

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您可以通过生成字符串然后调用它们的encode方法使它们成为bytes对象来使这变得更简单。请注意，您的函数实际上只是构建一个字典，然后在其中查找内容。只构建一次字典然后__getitem__以不同的名称提供绑定方法要简单得多。

template = "I have level value as {} in the house"
size_list = (1, 200, 30000)
sizes = {f"level{i}": template.format(x).encode() for i, x in enumerate(size_list, start=1)}
get_level_string = sizes.__getitem__


# tests
a = get_level_string('level1')
b = get_level_string('level2')
c = get_level_string('level3')
print(a, type(a))
print(b, type(b))
print(c, type(c))

印刷

b'I have level value as 1 in the house' <class 'bytes'>
b'I have level value as 200 in the house' <class 'bytes'>
b'I have level value as 30000 in the house' <class 'bytes'>

为您的测试

于 2018-08-10T18:00:52.230 回答