python - 在 python 中的循环中编写循环的更简洁的方法

Question

我有一个对象列表，这些对象是我代码中的特定类，

[object1, object2, object3, object4, object5, object6]

即这个类有两个属性:class.score和class.id

我可能有具有相同 ID 的对象。例如:

[object1.id, object2.id, object3.id, object4.id, object5.id, object6.id] = [1, 2, 3, 4, 2, 3]

但分数不同。例如:

[object1.score, object2.score, object3.score, object4.score, object5.score,
object6.score] = [0.25, 0.55, 0.6, 0.4, 0.30, .33]

我想要做的是让这个对象在 id-wise 上没有重复的列表，但是 添加分数。因此对于前面的示例，输出将是:

[object1.id, object2.id, object3.id, object4.id] = [1, 2, 3, 4]
[object1.score, object2.score, object3.score, object4.score] = [.25, .85, .93, .4]

我已经设法用两个 for 循环做到了这一点:

k = 1
    for object in list_of_objects:
        j = 1
        for object2 in list_of_objects:
            if object.id == object2.id and j > k:
                object.score = object.score + object2.score
                list_of_objects.remove(object2)
            j += 1
        k += 1

但我希望以一种更像 python 的方式来实现它:

newlist[:] = [ x for x in list_of_objects if certain_condition(x)]

谢谢。

Answer 1

itertools.groupby 正是为这种情况而制作的 https://docs.python.org/2/library/itertools.html#itertools.groupby

from itertools import groupby
# object.id is our key:
keyfunc = lambda obj: obj.id
list_of_objects = sorted(list_of_objects, key=keyfunc)

scores = [sum(score_list) for id, score_list in groupby(list_of_objects, keyfunc)]
ids = [id for id, score_list in groupby(list_of_objects, keyfunc)]