使用对称差异获取已删除元素的索引

Question

CSV文件中有两列：

oldCol1 = [1, 2, 3, 4, 5]

oldCol2 = ['A', 'B', 'C', 'D', 'E']

现在，我更新csv并添加新行

newCol1 = [1, 2, 3, 4, 5, 6]

newCol2 = ['A', 'B', 'C', 'D', 'E', 'A']

我只想获取新添加的元素。所以，我正在尝试：

newListCol1 = list(set(oldCol1).symmetric_difference(newCol1))

现在，我的问题是如何从第二列中获取新添加的元素？

#Here, I want to get two lists: [6] and ['A']. 

谢谢你的帮助！

更新：

新添加的元素可以在列表中的任何位置（而不仅仅是结尾）-造成混乱的原因！

Answer 1

#if they can be anywhere#mehtod 1from collections import Counter oldCol1 = [1, 2, 3, 4, 5]oldCol2 = ['A', 'B', 'C', 'D', 'E']newCol1 = [1, 2, 3, 4, 5, 6]newCol1_1 = [1, 2, 3, 4, 5, 6, 6, 7, 7] #different examplenewCol2 = ['A', 'B', 'C', 'D', 'E', 'A']print(list((Counter(newCol1) - Counter(oldCol1)))) # returns a list of unique valueprint(list((Counter(newCol2) - Counter(oldCol2))))new_item_added_dict = Counter(newCol1_1) - Counter(oldCol1)print( list(new_item_added_dict.elements())) # elements() returns an iterator# if you want all the new values even duplicates like in newCol1_1 # ie if you want ans to be [6, 6, 7, 7] then use elements()# else use list() if you just want unique value updates [6,7]print( list(new_item_added_dict)) # output # [6] # ['A'] # [6, 6, 7, 7] # [6, 7]#--------------------------------------------------------------------- #method 2from collections import defaultdictoldCol1 = [1, 2, 3, 4, 5]newCol1 = [1, 2, 3, 4, 5, 6]  # -->[6]# [1, 2, 3, 4, 5, 6, 5] --> [6,5]new_item_list = []oldlist_dict = defaultdict(lambda:0) #default value of key is 0 and with defualtdict you will not key errorfor item in oldCol1:    oldlist_dict[item] += 1for item in newCol1:    if item in oldlist_dict and oldlist_dict[item] > 0:        oldlist_dict[item] -=1    else:        # its a new item         new_item_list.append(item)print(new_item_list)#--------------------------------------------------------------------- #if new items are always appended ie added to end of old listprint(newCol1[len(oldCol1):])  print(newCol2[len(oldCol2):])print(newCol1_1[len(oldCol1):])  

Answer 2

如果您知道“新添加的元素”始终附加在列表的末尾，则只需从旧列表的长度开始进行切片。IEold_til_here = len(oldCol1)list_of_new_elements_col_1 = newCol1[old_til_here:]list_of_new_elements_col_2 = newCol2[old_til_here:]

Answer 3

您将需要获取第一个索引中不存在的索引，因此仅使用不带有symmetric_difference的集合。使用enumerate（）可使索引更容易。oldCol1 = [1, 2, 3, 4, 5]oldCol2 = ['A', 'B', 'C', 'D', 'E']newCol1 = [1, 2, 3, 4, 5, 6]newCol2 = ['A', 'B', 'C', 'D', 'E', 'A']indexes = [i for i, v in enumerate(newCol1) if v not in set(oldCol1)]resultCol1 = [newCol1[i] for i in indexes]resultCol2 = [newCol2[i] for i in indexes]print(resultCol1, resultCol2)