[英]How to group items in one list based on consecutive elements in another list?
I have a simple Data sorting problem:我有一个简单的数据排序问题:
I have two lists:我有两个列表:
List1: [1,1,1,2,3,3,4,5,5]
List2: [objectA, objectB, objectC, objectD, ... objectI]
列表 1:
[1,1,1,2,3,3,4,5,5]
列表 2: [objectA, objectB, objectC, objectD, ... objectI]
The lists have the same length.列表具有相同的长度。 I'd like to sort List2 into a nested List as follows:
我想将 List2 排序为嵌套列表,如下所示:
List3: [[objectA, objectB, objectC], [objectD], ...]
列表 3:
[[objectA, objectB, objectC], [objectD], ...]
So basically sort list2 to into sublists according to list1.所以基本上根据 list1 将 list2 排序为子列表。 Is this a list or a dictionary problem and how do I solve it?
这是列表问题还是字典问题,我该如何解决?
I think that the best way to solve it is using a dictionary for grouping the elements of List2
according to the elements in List1
我认为解决它的最好方法是使用字典根据
List1
中的元素对List2
的元素进行分组
For your example that should be like this对于你的例子应该是这样的
List1 = [1,1,1,2,3,3,4,5,5]
List2 = ['objectA', 'objectB', 'objectC', 'objectD', ... 'objectI']
grouping = {}
for key, value in zip(List1, List2):
if key not in grouping:
grouping[key] = []
grouping[key].append(value)
List3 = [grouping[key] for key in sorted(grouping.keys())]
print(List3)
You can use something like that:你可以使用类似的东西:
from collections import Counter
list1=[1,1,1,2,2,3,4]
list2=["objectA", "objectB", "objectC", "objectD","objectE","objectF","objectG"]
to_dict=dict(Counter(list1).items()) #{1: 3, 2: 2, 3: 1, 4: 1}
final=[]
i=0
for k,v in to_dict.items():
final.append(list2[i:v+i])
i+=v
Output(final) :输出(最终) :
[
['objectA', 'objectB', 'objectC'],
['objectD', 'objectE'],
['objectF'],
['objectG']
]
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