关于python：无法通过不合逻辑的行pep8错误

if-statementpep8python

Can't get past illogical line pep8 error

我已经试着解决这个问题有一段时间了，我就是不能让它通过佩普8。这是我的代码：

1。

1
2
3
4
5

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')' and
sum(regex.count(char) for char in splitter) == 1 and
regex.count('(') == 1 and regex.count(')') == 1):

print('hi')

2。

1
2
3
4
5

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')' and
sum(regex.count(char) for char in splitter) == 1 and
regex.count('(') == 1 and regex.count(')') == 1):

print('hi')

三。

1
2
3
4

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')'
and regex.count('(') > 1):

print('hi')

我在3条if语句中的每一条上都得到以下PEP8错误：

1	E125 continuation line does not distinguish itself from next logical line

你知道怎么回事吗？行与第一个括号缩进，所以我真的没有线索。

相关讨论

1。

1
2
3
4
5

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')' and
sum(regex.count(char) for char in splitter) == 1 and
regex.count('(') == 1 and regex.count(')') == 1):

print('hi')

2。

1
2
3
4
5

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')' and
sum(regex.count(char) for char in splitter) == 1 and
regex.count('(') == 1 and regex.count(')') == 1):

print('hi')

三。

1
2
3
4

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')'
and regex.count('(') > 1):

print('hi')

相关讨论

我正在使用Pycharm(这非常适合指出PEP8错误)进行编辑，它说这个版本可以：

1
2
3
4
5

if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')' and
sum(regex.count(char) for char in splitter) == 1 and
regex.count('(') == 1 and regex.count(')') == 1):

print('hi')

相关讨论

我不是说我喜欢这个解决方案，但我认为在if之后移除空间比用len呼叫的勇气排第二行要少一些妥协，就像这里的其他答案所建议的那样：

1
2
3
4
5

if(len(regex) > 2 and regex[0] == '(' and regex[-1] == ')' and
sum(regex.count(char) for char in splitter) == 1 and
regex.count('(') == 1 and regex.count(')') == 1):

print('hi')