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。它应该是这样的：

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

我想与每个人的身份证进行比较，这是独一无二的。

编辑：结果将是我的函数比较。它打印差异。我不知道如何启动

Answer 1

您可以将列表转换为集合，并获取差异

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 2

edit：我写这篇文章之前，我对集合有很多了解。现在我推荐使用下面给出的集合的解决方案。

一种解决方案：

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，如果您将数据作为字典呈现：

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()}

两者都变成了函数，返回ys中的项目，但不返回xs:中的项目

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()}

Answer 3

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!")

它是比较整个元素(id，name)的方法，所以如果只想比较id，就应该做一些修改(例如使用dicts)。