Question

我在Python函数中有以下代码：

strings = ('TIRS10', 'TIRS11')
if any(string in fields for string in strings):

    def transform(row):
        '''
        Transform an input row in to a named tuple, then feed it in to a
        dictionary.
        '''
        # split row in elements
        elements = row.split('|')

        # key: 1st column, replace
        key = replace_dot_comma_space(elements[0])

        # namedtuple
        ect = namedtuple(key, [fields[0], fields[1]])

        # feed namedtuples
        ect.TIRS10 = is_number(elements[1])
        ect.TIRS11 = is_number(elements[2])

        # feed dictionary
        dictionary[key] = dictionary.get(key, ect)

strings = ('b0', 'b1', 'b2', 'b3', 'b4', 'b5', 'b6', 'b7')
if any(string in fields for string in strings):

    def transform(row):
        '''
        Transform an input row in to a named tuple, then feed it in to a
        dictionary.
        '''
        # split row in elements
        elements = row.split('|')

        # key: 1st column, replace
        key = replace_dot_comma_space(elements[0])

        # *** small modification for the CWV field ***
        fields[0] = 'cwv'

        # named tuples
        cwv = namedtuple(key,
                         [replace_dot_comma_space(fields[0]),
                          replace_dot_comma_space(fields[1]),
                          replace_dot_comma_space(fields[2]),
                          replace_dot_comma_space(fields[3]),
                          replace_dot_comma_space(fields[4]),
                          replace_dot_comma_space(fields[5]),
                          replace_dot_comma_space(fields[6]),
                          replace_dot_comma_space(fields[7]),
                          replace_dot_comma_space(fields[8]),
                          replace_dot_comma_space(fields[9])])

        # feed named tuples
        cwv.subrange = to_tuple(elements[1])
        cwv.b0 = is_number(elements[2])
        cwv.b1 = is_number(elements[3])
        cwv.b2 = is_number(elements[4])
        cwv.b3 = is_number(elements[5])
        cwv.b4 = is_number(elements[6])
        cwv.b5 = is_number(elements[7])
        cwv.b6 = is_number(elements[8])
        cwv.b7 = is_number(elements[9])
        cwv.rmse = is_number(elements[10])
        dictionary[key] = dictionary.get(key, cwv)  # feed dictionary

map(transform, rows)
return dictionary

我得到了一个从xxx [F811] 行重新定义未使用的'transform'。我如何解决这个“错误”，以便不使用两个不同的功能，但只使用一个，稍后在同一个功能内？处理这种情况的正确方法是什么？

Answer 1

一种方法是将它们命名为transform1和transform2，并将map放在if-else：

中

if condition1:
    transform = transform1
else:  # if condition2
    transform = transform2

map(transform, row)

此外，重用像字符串这样的变量并不好：它可能会引入细微的错误。隐含地更改字典变量也不好。明确地传递这些东西并且明确地返回更改的对象总是更好。

如果有如上所述的更多变换，重构更清晰的是面向对象的样式，其中.transform（）是tirs-class或b-class的方法。然后字典自然是属性，是Transformer类状态的一部分，而不是全局属性。

如何避免重新定义函数（PEP8错误F811）

1 个答案: