在Python中根据另一个列表的情况删除一个列表的元素

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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.

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  • 本文由 发表于 2023年6月15日 16:06:59
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