如何在 Python 中增加 lambda 中的数字？

Question

可以在 lambda 函数中增加像

int

这样的数字对象吗？

想象一下有一个窥视功能，例如：

def _peek(cb, iter):
    for i in iter:
        cb(i)

如何查看这些值并将其添加到总和中，如以下简单示例所示：

numbers = (1, 2, 3)
s = 0

# Doesn't work, because __add__ doesn't add inline
_peek(s.__add__, numbers)

# Doesn't work, because s is outside of scope (syntax error)
_peek(lambda x: s += x, numbers)

# Does work, but requires an additional function
def _sum(var):
  nonlocal s
  s += var

_peek(_sum, numbers)

# Does work, but reduces numbers
sum = reduce(lambda x, y: x+y, numbers)

这是一个现实世界的例子：

@dataclass
class Vote:
    count = 0
    def add_count(self, count: int):
      self.count += count

vote = Vote()

# Doesn't work work
_peek(lambda x: vote.count += x, map(lambda x: x['count'], data))

# Does work, but requires additional function
_peek(vote.add_count, map(lambda x: x['count'], data))

在Java中，我可以轻松编写：

@Test
public void test_numbers() {
    class Vote {
        int count = 0;
    }

    var vote = new Vote();

    var count = Stream.of(1,2,3).peek(i -> vote.count+=i).filter(i -> i > 1).count();
    assert vote.count == 6;
    assert count == 2;
}

Answer 1

您可以使用可调用类。

def _peek(cb, iter):
    for i in iter:
        cb(i)

class Adder:
    def __init__(self, *args, **kwargs):
        self.value = 0
    def __call__(self, *args, **kwargs):
        self.value += args[0]

adder = Adder()
_peek(adder, (1, 2, 3))
print(adder.value)

结果是

6

。

Answer 2

如果我们将总和放入容器（例如列表）中，那么我们可以引用该容器而不必担心范围；我们可以使用

.append

和

.pop

:

等方法来改变它的内容（即总和）

def _peek(cb, iter):
    for i in iter:
        cb(i)

numbers = (1, 2, 3)
s_container = [0]

_peek(lambda x: s_container.append(s_container.pop() + x), numbers)

# Extract the result
s = s_container.pop()