问题描述
def isBig(x):
if x > 4:
return 'apple'
else:
return 'orange'
这有效:
if isBig(y): return isBig(y)
这不起作用:
if fruit = isBig(y): return fruit
第二个为什么不起作用！?我要一线.除此以外，第一个将调用函数TWICE.
Why doesn't the 2nd one work!? I want a 1-liner. Except, the 1st one will call the function TWICE.
如何在不调用函数两次的情况下使其成为1个衬里?
推荐答案
我看到其他人已经指出了我的旧分配并设置"食谱，该食谱的最简单版本为:
I see somebody else has already pointed to my old "assign and set" Cookbook recipe, which boils down in its simplest version to:
class Holder(object):
def set(self, value):
self.value = value
return value
def get(self):
return self.value
h = Holder()
...
if h.set(isBig(y)): return h.get()
但是，这主要是为了简化Python和直接在if或while中支持赋值的语言之间的音译.如果您有成百上千个这样的检查和返回，那么最好做一些完全不同的事情:
However, this was intended mostly to ease transliteration between Python and languages where assignment is directly supported in if or while. If you have "hundreds" of such check-and-return in a cascade, it's much better to do something completely different:
hundreds = isBig, isSmall, isJuicy, isBlah, ...
for predicate in hundreds:
result = predicate(y)
if result: return result
甚至是类似的东西
return next(x for x in (f(y) for f in hundreds) if x)
如果没有谓词得到满足，则可以获取StopIteration异常，或者
if it's OK to get a StopIteration exception if no predicate is satisfied, or
return next((x for x in (f(y) for f in hundreds) if x)), None)
如果在没有谓词得到满足时，None是正确的返回值，等等.
if None is the proper return value when no predicate is satisfied, etc.
几乎总是使用(或什至希望；-)Holder技巧/非习惯用法是一种设计气味"，这表明正在寻找一种不同的，更Python化的方法-在这种情况下，Holder是合理的恰恰是我为其设计的特殊情况，即，您想要保持Python代码与某些非Python之间的密切对应的情况(您正在用Python音译参考算法，并希望它在重构之前首先工作)转换成更加Pythonic的形式，或者您正在编写Python作为原型，一旦其有效运行，它将被音译为C ++，C#，Java等).
Almost invariably, using (or even wishing for;-) the Holder trick/non-idiom is a "design smell" which suggests looking for a different and more Pythonic approach -- the one case where Holder is justified is exactly the special case for which I designed it, i.e., the case where you want to keep close correspondence between the Python code and some non-Python (you're transliterating a reference algorithm in Python and want it working first before refactoring it into a more Pythonic form, or you're writing Python as a prototype that will be transliterated into C++, C#, Java, etc, once it's working effectively).
这篇关于如何在IF条件下分配变量，然后将其返回?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！