Question

如果在python中按名称（不能索引）存储结果，如何将函数调用的结果分配给多个变量。

例如（在Python 3中测试），

import random

# foo, as defined somewhere else where we can't or don't want to change it
def foo():
    t = random.randint(1,100)
    # put in a dummy class instead of just "return t,t+1" 
    # because otherwise we could subscript or just A,B = foo()
    class Cat(object):
        x = t
        y = t + 1

    return Cat()

# METHOD 1
# clearly wrong; A should be 1 more than B; they point to fields of different objects
A,B = foo().x, foo().y
print(A,B)

# METHOD 2
# correct, but requires two lines and an implicit variable
t = foo()
A,B = t.x, t.y
del t # don't really want t lying around
print(A,B)

# METHOD 3
# correct and one line, but an obfuscated mess
A,B = [ (t.x,t.y) for t in (foo(),) ][0]
print(A,B) 
print(t) # this will raise an exception, but unless you know your python cold it might not be obvious before running

# METHOD 4
# Conforms to the suggestions in the links below without modifying the initial function foo or class Cat.
# But while all subsequent calls are pretty, but we have to use an otherwise meaningless shell function
def get_foo():
    t = foo()
    return t.x, t.y

A,B = get_foo()

我们不想做什么

如果结果是可转位的（Cat扩展元组/列表，我们使用了namedtuple等），我们可以简单地编写A,B = foo()，如{上面的注释中所示{1}}课程。例如That's what's recommended here。

我们假设我们有充分的理由不允许这样做。也许我们喜欢从变量名称中分配清晰度（如果它们比Cat和x更有意义），或者对象主要不是容器。也许字段是属性，因此访问实际上涉及方法调用。我们不必承担任何人回答这个问题; y课程可以按面值进行。

This question已经讨论了如何以最佳方式设计函数/类; 如果函数的预期返回值已经很好地定义并且不涉及类似元组的访问，那么在返回时接受多个值的最佳方法是什么？

Answer 1

我强烈建议使用多个语句，或者只保留结果对象而不解压缩其属性。也就是说，您可以使用operator.attrgetter：

from operator import attrgetter
a, b, c = attrgetter('a', 'b', 'c')(foo())

在python中的一行中分配函数调用的结果

1 个答案: