问题描述
这个问题在某种程度上是我在这里提出的问题的继续:，@ belisarius和@Daniel提供了非常有用的答案.
This question is in a way a continuation of the question I asked here:Simple way to delete a matrix column in Mathematica to which @belisarius and @Daniel provided very helpful answers.
我通常想做的是从矩阵中提取特定的行和列，或者除去指定的行和列之后剩下的内容.因此，可以将其正式写为，找到TakeOperator和Drop Operator这样:
What I am generally trying to do is to extract from a matrix A specific lines and columns OR what remains after what those specified are removed. So this can be formally writtewn as, find TakeOperator and Drop Operator such that:
TakeOperator [A，{i1，..，ip}，{j1，...，jq}] =(A [[ik]] [[jl]])(1 Table[A[[ik]][[jl]],{k,p},{l,q}]
TakeOperator[A,{i1,..,ip},{j1,...,jq}]=(A[[ik]][[jl]]) (1<=k<=p, 1<=l<=q) =Table[A[[ik]][[jl]],{k,p},{l,q}]
我们注意到Ic = {i'1，...，i'p'} = Complement [{1，...，Length[A]}，{i1，...，ip}]; Jc = {j'1，...，j'q'} = Complement [{1，...，Length[A]}，{j1，...，jq}];
We note Ic={i'1,...,i'p'}=Complement[{1,...,Length[A]},{i1,...,ip}];Jc={j'1,...,j'q'}=Complement[{1,...,Length[A]},{j1,...,jq}];
DropOperator [A，{i1，..，ip}，{j1，...，jq}] =(A [[ik]] [[jl]])(1 Table[A[[ik']][[jl']],{k',p'},{l','q}]
DropOperator[A,{i1,..,ip},{j1,...,jq}]=(A[[ik]][[jl]]) (1<=k'<=p', 1<=l'<=q') =Table[A[[ik']][[jl']],{k',p'},{l','q}]
尽管如上所述的Table可以解决问题，但以这种方式使用Table的效率非常低.
While Table as described above does the trick, it is highly inefficient to use Table in that manner.
我只想举个例子，我以@ belisarius为例:
Just to give an idea, I took @ belisarius example:
In: First@Timing[a = RandomInteger[1000, {5000, 5000}];]
Out:0.218
In:Clear[b,c]
In:First@Timing[
b = Table[
If[i < 100, If[j < 100, a[[i]][[j]], a[[i]][[j + 1]]],
If[j < 100, a[[i + 1]][[j]], a[[i + 1]][[j + 1]]]], {i,
4999}, {j, 4999}]]
Out:140.807
In:First@Timing[c = Drop[a, {100}, {100}]]
Out:0.093
In:c===b
Out:True
注意:关于早期文章中Drop的使用，我也考虑过使用它，但是当我检查文档时，没有建议以@belisarius和@daniel的方式完成它.如果可以在将来的版本中朝那个方向更新文档，那将是有帮助的.
Note: With respect to the use of Drop in the earlier post, I thought about using it as well, but when I checked the documentation, there was no suggestion of getting it done the way @belisarius and @daniel suggested. If the documentation could be updated in that direction in future releases, it would be helpful.
推荐答案
Part在切片数组时直接支持索引列表.以下定义利用了这一点:
Part directly supports lists of indices when slicing arrays. The following definitions exploit that:
takeOperator[a_?MatrixQ, rows_List, cols_List] :=
a[[rows, cols]]
dropOperator[a_?MatrixQ, rows_List, cols_List] :=
a[[##]]& @@ complementaryIndices[a, rows, cols]
complementaryIndices[a_?MatrixQ, rows_List, cols_List] :=
Complement @@@ Transpose @ {Range /@ Dimensions @ a, {rows, cols}}
示例用法:
a = RandomInteger[1000, {5000, 5000}];
First @ Timing @ takeOperator[a, Range[1, 5000, 2], Range[1, 5000, 2]]
(* 0.016 *)
First @ Timing @ dropOperator[a, Range[1, 5000, 2], Range[1, 5000, 2]]
(* 0.015 *)
这篇关于在Mathematica中选择/删除矩阵中行/列的列表的有效方法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！