Python 基于选择的列表理解
基本上,如果我要编写一个具有可变返回元素的函数,如下所示:Python 基于选择的列表理解,python,list-comprehension,Python,List Comprehension,基本上,如果我要编写一个具有可变返回元素的函数,如下所示: def func(elem1=True, elem2=True, elem3=True, elem4=False): x = MyClass() ret = [] if elem1: ret.extend([x.func1()]) if elem2: ret.extend([x.obj1]) if elem3: ret.extend([x.func2
def func(elem1=True, elem2=True, elem3=True, elem4=False):
x = MyClass()
ret = []
if elem1:
ret.extend([x.func1()])
if elem2:
ret.extend([x.obj1])
if elem3:
ret.extend([x.func2().attr1])
if elem4:
ret.extend(x.list_obj3)
return ret
事情变得相当漫长和多风。有可能这样做吗?也许:
def func(elem1=True, elem2=True, elem3=True, elem4=False):
x = MyClass()
return [x.func1() if elem1,
x.obj1 if elem2,
x.func2().attr1 if elem3,
x.list_obj3 if elem4]
多么整洁啊
我知道这是可以做到的:
def func(elem1=True, elem2=True, elem3=True, elem4=False):
x = MyClass()
ret = [x.func1(), x.obj1, x.func2().attr1, x.list_obj3]
choices = [elem1, elem2, elem3, elem4]
return [r for i, r in enumerate(ret) if choices[i]]
但是如果用户不需要,我不想计算元素;计算其中的一些操作有点昂贵。如果在lambdas中隐藏操作,则可以使用惰性计算:
def func(elem1=True, elem2=True, elem3=True, elem4=False):
x = MyClass()
return [L() for inc,L in (
(elem1, lambda: x.func1()),
(elem2, lambda: x.obj1),
(elem3, lambda: x.func2().attr1),
(elem4, lambda: x.list_obj3),
) if inc]
问一个稍微不同的问题,你能得到像matlab/octave这样的行为吗?在matlab/octave中,如果你分配给两个变量,你只计算前两个结果,而不计算结果3和4 例如:
a, b = func()
Python不能完全做到这一点,因为func()不知道它需要多少返回值,但您可以使用:
from itertools import islice
def func():
x = MyClass()
yield x.fun c1()
yield x.obj1
yield x.func2().attr1
yield x.list_obj3
a, b = islice(func(), 2)
我不确定它是否更好,但您可以使用decorator添加数组索引语义,这将允许您编写:
@sliceable
def func():
...
a, b = func()[:2]
这很容易实现:
from itertools import islice
class SlicedIterator(object):
def __init__(self, it):
self.it = it
def __iter__(self):
return self.it
def __getitem__(self, idx):
if not isinstance(idx, slice):
for _ in range(idx): next(self.it)
return next(self.it)
return list(islice(self.it, idx.start, idx.stop, idx.step))
def sliceable(f):
def wraps(*args, **kw):
return SlicedIterator(f(*args, **kw))
return wraps
测试:
@sliceable
def f():
print("compute 1")
yield 1
print("compute 2")
yield 2
print("compute 3")
yield 3
print("compute 4")
yield 4
print("== compute all four")
a, b, c, d = f()
print("== compute first two")
a, b = f()[:2]
print("== compute one only")
a = f()[0]
print("== all as a list")
a = f()[:]
给出:
== compute all four
compute 1
compute 2
compute 3
compute 4
== compute first two
compute 1
compute 2
== compute one only
compute 1
== all as a list
compute 1
compute 2
compute 3
compute 4