python 比较两个列表之间的差异

Question

我有两个这样的列表：

newList = (
    (1546, 'John'),
    (8794, 'Michael'),
    (892416, 'Dave'),
    (456789, 'Lucy'),
    )

oldList = (
    (1546, 'John'),
    (8794, 'Michael'),
    (892416, 'Dave'),
    (246456, 'Alexander')
    )

我想要一个比较两个列表的函数。它会是这样的：

def compare(new, old):
    print('Alexander is not anymore in the new list !')
    print('Lucy is new !')
    return newList

我想比较每个人的id是唯一的。

编辑：结果将是我的函数比较。它打印差异。我不知道如何开始

Answer 1

您可以使用 set:

def compare(old, new):
    oldSet = set(old)
    newSet = set(new)
    removedElements =  oldSet - newSet
    newElements = newSet - oldSet
    for element in removedElements:
        print(element[1] + " is not anymore in the new list!")
    for element in newElements:
        print(element[1] + " is new!")

这是比较整个元素（id、name）的方法，所以如果你只想比较id，你应该做一些修改（例如使用dicts）。

Answer 2

您可以将列表转换为集合并取差

n = set(l[1] for l in newList)
o = set(l[1] for l in oldList)
print n - o # set(['Lucy'])
print o - n # set(['Alexander'])

Answer 3

编辑：我在对集合了解很多之前就写了这篇文章。现在我会推荐使用下面给出的集合的解决方案。

一个解决方案：

removed = [o for o in old if o[0] not in [n[0] for n in new]]
added = [n for n in new if n[0] not in [o[0] for o in old]]

或者，如果您将数据显示为字典：

old = dict(old) # if you do go for this approach be sure to build these
new = dict(new) # variables as dictionaries, not convert them each time

removed = {k:old[k] for k in old.keys() - new.keys()}
added = {k:new[k] for k in new.keys() - old.keys()}

两者都变成函数，返回 ys 中的项目，但不在 xs 中：

def tuple_list_additions(xs,ys):
  return [y for y in ys if y[0] not in [x[0] for x in xs]]

def dict_additions(xs,ys):
  return {k:ys[k] for k in ys.keys() - xs.keys()}

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