英文:
Removing elements of one list wth respect to the other in Python
问题
I have lists ```J``` and ```Indices```. I want to remove elements of ```J``` according to locations specified in ```Indices```. For example, ```Indices[0]=1``` and ```Indices[1]=2```. This means that ```J[0][1]``` and ```J[0][2]``` should be removed in one go. But I am getting an error. I present the expected output.Indices=[1,2]J=[[2, 3, 6, 7, 9, 10]]J=J[0].remove(Indices[0])print(J)
The error is
in <module>J=J[0].remove(Indices[0])ValueError: list.remove(x): x not in list
The expected output is
[[2, 7, 9, 10]]
英文:
I have lists J and Indices. I want to remove elements of J according to locations specified in Indices. For example, Indices[0]=1 and Indices[1]=2. This means that J[0][1] and J[0][2] should be removed in one go. But I am getting an error. I present the expected output.
Indices=[1,2]J=[[2, 3, 6, 7, 9, 10]]J=J[0].remove(Indices[0])print(J)
The error is
in <module>J=J[0].remove(Indices[0])ValueError: list.remove(x): x not in list
The expected output is
[[2, 7, 9, 10]]
答案1
得分: 2
使用这个
Indices = [1, 2]
J = [[2, 3, 6, 7, 9, 10]]
for index in sorted(Indices, reverse=True):
J[0].pop(index)
print(J)
`remove`: 移除传递给它的元素，如果元素不在列表中，则报错。`pop`: 移除传递给它的索引处的元素，如果没有提供索引，则将移除最后一个元素（[-1]索引），如果索引不正确，则报错。
英文:
Use this
Indices = [1, 2]J = [[2, 3, 6, 7, 9, 10]]for index in sorted(Indices, reverse=True):J[0].pop(index)print(J)
remove: remove the element that is passed in it, error if the element is not in the list.
pop: remove the element on the index that is passed in it, if no index provide then it will remove the last element[-1 index], if the index is not correct then error.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论