Question

这看起来真的很难看，有没有办法让它看起来更像pythonic？

if self.cleaned_data['string1_val'] == '' and latestSI.string1_val is None :
    if self.cleaned_data['string2_val'] == '' and latestSI.string2_val is None :
        return False
    elif self.cleaned_data['string2_val'] == latestSI.string2_val :
        return False
    else:
        return True
elif self.cleaned_data['string1_val'] == latestSI.string1_val :
    if self.cleaned_data['string2_val'] == '' and latestSI.string2_val is None :
        return False
    elif self.cleaned_data['string2_val'] == latestSI.string2_val :
        return False
    else:
        return True
else:
    return True

Answer 1

def eq(x,y):
    return x == ('' if y is None else y)

if eq(self.cleaned_data['string1_val'],latestSI.string1_val):
    return not eq(self.cleaned_data['string2_val'],latestSI.string2_val)

嗯。看起来问题已经改变了。通过添加最终else: True，逻辑可以更改为

return not (eq(self.cleaned_data['string1_val'],latestSI.string1_val)
            and eq(self.cleaned_data['string2_val'],latestSI.string2_val))

Answer 2

您的所有问题都来自None。清理它们，你的逻辑变得微不足道。

cd1 = self.cleaned_data['string1_val']
lsi1 = latestSI.string1_val
cd2 = self.cleaned_data['string2_val']
lsi2 = latestSI.string2_val
if lsi1 is None:
  lsi1 = ''
if lsi2 is None:
  lsi2 = ''
return not (cd1 == lsi1 and cd2 == lsi2)

Answer 3

我认为这个表达式是等价的（我无法测试它，因为我无权访问你的其余代码）。但是，这真的很长很难理解，如果我是你，我宁愿保持不变：

if (self.cleaned_data['string1_val'] == latestSI.string1_val) or (not self.cleaned_data['string1_val'] and not latestSI.string1_val):
    return (not self.cleaned_data['string2_val'] or not latestSI.string2_val) and self.cleaned_data['string2_val'] != latestSI.string2_val
else:
    return True

Answer 4

内部if / elif / else链可以用and-operator和or-operator替换。只要您看到return True和return False代替每个替代方案的正文，这种替换就应该是自动的。

此外，还有多个对self.cleaned_data的引用可以用以下方法解决：

cleaned = self.cleaned_data

寻找更多pythonic的东西

4 个答案: