我已经尝试解决这个问题一段时间了,我不能让它通过pep8。 这是我的代码:
1
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
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')
3
if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')'
and regex.count('(') > 1):
print('hi')
我在3个if语句中得到以下PEP8错误:
E125 continuation line does not distinguish itself from next logical line
知道它有什么问题吗?这些线条用第一个支架缩进,所以我真的没有线索。
答案 0 :(得分:21)
1
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
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')
3
if (len(regex) > 2 and regex[0] == '(' and regex[-1] == ')'
and regex.count('(') > 1):
print('hi')
答案 1 :(得分:2)
我使用PyCharm(非常适合指出PEP8错误)进行编辑,它说这个版本没问题:
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')
答案 2 :(得分:0)
我并不是说我喜欢这种解决方案,但我认为在if
之后删除空格比将第二行与len
调用的胆量排成一行不太妥协。其他答案在这里建议:
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')