How to delete two specific columns in a matrix in mathematica [duplicate]

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This question already has an answer here:

• 
  Correct way to remove matrix columns?
  
  3 answers
  
  

Say I have a matrix of random integers, which is given by

A=RandomInteger[25, 9, 11]

How can specific columns of this matrix (say 4th and 7th columns) be deleted?

list-manipulation

share|improve this question

asked Sep 6 at 11:20

Soumyajit Roy

906

marked as duplicate by corey979, José Antonio Díaz Navas, MikeLimaOscar, Öskå, kirma yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Delete[Transpose@A, 4, 7] // Transpose
  – Okkes Dulgerci
  Sep 6 at 12:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  mA[[All, Complement[Range[11], 4, 7]]]
  – Alan
  Sep 6 at 13:18
  

  
  
  
  

add a comment | 

up vote
7
down vote

favorite

This question already has an answer here:

• 
  Correct way to remove matrix columns?
  
  3 answers
  
  

Say I have a matrix of random integers, which is given by

A=RandomInteger[25, 9, 11]

How can specific columns of this matrix (say 4th and 7th columns) be deleted?

list-manipulation

share|improve this question

asked Sep 6 at 11:20

Soumyajit Roy

906

marked as duplicate by corey979, José Antonio Díaz Navas, MikeLimaOscar, Öskå, kirma yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Delete[Transpose@A, 4, 7] // Transpose
  – Okkes Dulgerci
  Sep 6 at 12:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  mA[[All, Complement[Range[11], 4, 7]]]
  – Alan
  Sep 6 at 13:18
  

  
  
  
  

add a comment | 

up vote
7
down vote

favorite

up vote
7
down vote

favorite

This question already has an answer here:

• 
  Correct way to remove matrix columns?
  
  3 answers
  
  

Say I have a matrix of random integers, which is given by

A=RandomInteger[25, 9, 11]

How can specific columns of this matrix (say 4th and 7th columns) be deleted?

list-manipulation

share|improve this question

asked Sep 6 at 11:20

Soumyajit Roy

906

This question already has an answer here:

• 
  Correct way to remove matrix columns?
  
  3 answers
  
  

Say I have a matrix of random integers, which is given by

A=RandomInteger[25, 9, 11]

How can specific columns of this matrix (say 4th and 7th columns) be deleted?

This question already has an answer here:

• 
  Correct way to remove matrix columns?
  
  3 answers
  
  

list-manipulation

share|improve this question

asked Sep 6 at 11:20

Soumyajit Roy

906

share|improve this question

share|improve this question

asked Sep 6 at 11:20

Soumyajit Roy

906

asked Sep 6 at 11:20

Soumyajit Roy

906

asked Sep 6 at 11:20

Soumyajit Roy

906

906

marked as duplicate by corey979, José Antonio Díaz Navas, MikeLimaOscar, Öskå, kirma yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by corey979, José Antonio Díaz Navas, MikeLimaOscar, Öskå, kirma yesterday

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Delete[Transpose@A, 4, 7] // Transpose
  – Okkes Dulgerci
  Sep 6 at 12:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  mA[[All, Complement[Range[11], 4, 7]]]
  – Alan
  Sep 6 at 13:18
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Delete[Transpose@A, 4, 7] // Transpose
  – Okkes Dulgerci
  Sep 6 at 12:51
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  mA[[All, Complement[Range[11], 4, 7]]]
  – Alan
  Sep 6 at 13:18
  

  
  
  
  

1

1

Delete[Transpose@A, 4, 7] // Transpose
– Okkes Dulgerci
Sep 6 at 12:51

Delete[Transpose@A, 4, 7] // Transpose
– Okkes Dulgerci
Sep 6 at 12:51

1

1

mA[[All, Complement[Range[11], 4, 7]]]
– Alan
Sep 6 at 13:18

mA[[All, Complement[Range[11], 4, 7]]]
– Alan
Sep 6 at 13:18

add a comment | 

7 Answers
7

active

oldest

votes

up vote
7
down vote

I learned this method long time ago from Mr Wizard answer, it works well. But there are many other ways

a = RandomInteger[25, 9, 11];
a // MatrixForm

ReplacePart[a, _, 4, _, 7 :> Sequence];
MatrixForm[%]

share|improve this answer

answered Sep 6 at 11:37

Nasser

56.7k485203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
  – m_goldberg
  Sep 6 at 17:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Again, my obligatory warning: This method unpacks arrays.
  – Henrik Schumacher
  Sep 6 at 19:48
  

  
  
  
  

add a comment | 

up vote
7
down vote

The fasted way to do this is to use Part and construct the indices you need to take using Complement:

dropColumns[mat_?MatrixQ, columns : __Integer] := With[
 columnsToTake = Complement[
 Range[Dimensions[mat][[2]]],
 columns
 ]
 ,
 mat[[All, columnsToTake]]
];
a = RandomInteger[25, 9, 11];
a // MatrixForm
dropColumns[a, 2, 5] // MatrixForm

share|improve this answer

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

add a comment | 

up vote
4
down vote

A = RandomInteger[25, 9, 11];
A // MatrixForm
DeleteCol[Matrix_, indexlist_] := 
 Block[internalMatrix = Matrix, i, k = 0, 
 internallist = SortBy[indexlist, Smaller],
 For[i = 1, i <= Length[internallist], i++, 
 internalMatrix = 
 Delete[Transpose[internalMatrix], internallist[[i]] - k];
 internalMatrix = Transpose[internalMatrix];
 k++;
 ];
 internalMatrix
 ];
DeleteCol[A, 4, 7];
% // MatrixForm
DeleteCol[A, 7, 4];
% // MatrixForm

share|improve this answer

edited Sep 6 at 18:15

answered Sep 6 at 12:08

Diogo

1,686416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can you delete two or more columns?
  – Okkes Dulgerci
  Sep 6 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Okkes Dulgerci now it's possible to delete two or more columns.
  – Diogo
  Sep 6 at 17:28
  

  
  
  
  

add a comment | 

up vote
2
down vote

Here is another way to do it. I assume that columns you want delete is ordered. i.e. 4,7 will work but 7,4 won't.

 SeedRandom@2;
 A = RandomInteger[25, 9, 11];
 MatrixForm@A

$A=left(
beginarrayccccccccccc
23 & 3 & 18 & 11 & 10 & 17 & 8 & 3 & 8 & 0 & 19 \
9 & 23 & 25 & 24 & 14 & 4 & 3 & 4 & 12 & 12 & 8 \
8 & 18 & 6 & 3 & 1 & 14 & 21 & 4 & 14 & 10 & 20 \
22 & 8 & 10 & 20 & 19 & 1 & 9 & 12 & 0 & 19 & 11 \
25 & 10 & 8 & 7 & 18 & 7 & 9 & 23 & 1 & 6 & 15 \
12 & 1 & 0 & 14 & 19 & 12 & 2 & 5 & 3 & 7 & 5 \
23 & 24 & 1 & 14 & 3 & 2 & 22 & 16 & 21 & 4 & 11 \
4 & 2 & 3 & 20 & 24 & 8 & 10 & 3 & 6 & 12 & 19 \
20 & 24 & 6 & 1 & 13 & 10 & 8 & 21 & 5 & 6 & 21 \
endarray
right)$

deleteColumns[mat_?MatrixQ, col_] := 
 With[column = col - Range[0, Length@col - 1], 
 Fold[Delete[#, #2] &, Transpose@A, column] // Transpose]
deleteColumns[A, 4, 7, 10] // MatrixForm

$A=left(
beginarraycccccccc
23 & 3 & 18 & 10 & 17 & 3 & 8 & 19 \
9 & 23 & 25 & 14 & 4 & 4 & 12 & 8 \
8 & 18 & 6 & 1 & 14 & 4 & 14 & 20 \
22 & 8 & 10 & 19 & 1 & 12 & 0 & 11 \
25 & 10 & 8 & 18 & 7 & 23 & 1 & 15 \
12 & 1 & 0 & 19 & 12 & 5 & 3 & 5 \
23 & 24 & 1 & 3 & 2 & 16 & 21 & 11 \
4 & 2 & 3 & 24 & 8 & 3 & 6 & 19 \
20 & 24 & 6 & 13 & 10 & 21 & 5 & 21 \
endarray
right)$

share|improve this answer

edited Sep 6 at 13:30

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

add a comment | 

up vote
2
down vote

I am confused by the complexity of many of these responses. The single command

Drop[A, None, 4,7,3]

is sufficient to remove columns $4$ and $7$, in steps of $3$ (so as not to remove columms $5$ and $6$). If you wanted to remove more than two columns that are not separated by an equal number of columns, then you would need to use Drop more than once, starting from the rightmost column to be removed; e.g., to remove columns $1$, $3$, and $9$, you could just do

Drop[Drop[A, None, 9], None, 1,3,2]

or equivalently,

Drop[Drop[A, None, 3,9,6], None, 1]

If you had an arbitrary sorted list of column indices to remove, which we might call c, then Fold is trivially applied:

Fold[Drop[#1, None, #2]&, A, Reverse[c]]

share|improve this answer

edited Sep 7 at 1:02

answered Sep 7 at 0:57

heropup

1,529712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
  – Sjoerd Smit
  Sep 7 at 8:15
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

MapThread[Delete, A, ConstantArray[4, 7, Length[A]]]

share|improve this answer

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

add a comment | 

up vote
1
down vote

You can also use a combination of Fold and Drop:

ClearAll[dropCols1]
dropCols1 = Fold[Drop[#, None, #2] &, #, List /@ Reverse @ Sort[#2]] &;

Examples:

m = Array[Subscript[a, ##] &, 9, 9];

dropCols1[m, 4, 7] // MatrixForm // TeXForm

$smallleft(
beginarrayccccccc
a_1,1 & a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,1 & a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,1 & a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,1 & a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,1 & a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,1 & a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,1 & a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,1 & a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,1 & a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

dropCols1[m, 4, 7, 1] // MatrixForm // TeXForm

$smallleft(
beginarraycccccc
a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

Alternatively, (1) assign ##& (or Nothing in versions 10+) to the desired columns (dropCols2) or (2) use a combination of MapAt and ##&& (dropCols3):

ClearAll[dropCols2, dropCols3]
dropCols2 = Module[a = #, a[[All, #2]] = ## &; a] &;
dropCols3 = MapAt[## & &, #, All, #2] &;

Equal @@ (#[m, 4, 7] & /@ dropCols1, dropCols2, dropCols3)

True

Equal @@ (#[m, 4, 7, 1] & /@ dropCols1, dropCols2, dropCols3)

True

share|improve this answer

edited Sep 7 at 0:53

answered Sep 6 at 22:50

kglr

159k8183383

add a comment | 

7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
7
down vote

I learned this method long time ago from Mr Wizard answer, it works well. But there are many other ways

a = RandomInteger[25, 9, 11];
a // MatrixForm

ReplacePart[a, _, 4, _, 7 :> Sequence];
MatrixForm[%]

share|improve this answer

answered Sep 6 at 11:37

Nasser

56.7k485203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
  – m_goldberg
  Sep 6 at 17:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Again, my obligatory warning: This method unpacks arrays.
  – Henrik Schumacher
  Sep 6 at 19:48
  

  
  
  
  

add a comment | 

up vote
7
down vote

I learned this method long time ago from Mr Wizard answer, it works well. But there are many other ways

a = RandomInteger[25, 9, 11];
a // MatrixForm

ReplacePart[a, _, 4, _, 7 :> Sequence];
MatrixForm[%]

share|improve this answer

answered Sep 6 at 11:37

Nasser

56.7k485203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
  – m_goldberg
  Sep 6 at 17:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Again, my obligatory warning: This method unpacks arrays.
  – Henrik Schumacher
  Sep 6 at 19:48
  

  
  
  
  

add a comment | 

up vote
7
down vote

up vote
7
down vote

I learned this method long time ago from Mr Wizard answer, it works well. But there are many other ways

a = RandomInteger[25, 9, 11];
a // MatrixForm

ReplacePart[a, _, 4, _, 7 :> Sequence];
MatrixForm[%]

share|improve this answer

answered Sep 6 at 11:37

Nasser

56.7k485203

I learned this method long time ago from Mr Wizard answer, it works well. But there are many other ways

a = RandomInteger[25, 9, 11];
a // MatrixForm

ReplacePart[a, _, 4, _, 7 :> Sequence];
MatrixForm[%]

share|improve this answer

answered Sep 6 at 11:37

Nasser

56.7k485203

share|improve this answer

share|improve this answer

answered Sep 6 at 11:37

Nasser

56.7k485203

answered Sep 6 at 11:37

Nasser

56.7k485203

answered Sep 6 at 11:37

Nasser

56.7k485203

56.7k485203

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
  – m_goldberg
  Sep 6 at 17:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Again, my obligatory warning: This method unpacks arrays.
  – Henrik Schumacher
  Sep 6 at 19:48
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
  – m_goldberg
  Sep 6 at 17:56
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Again, my obligatory warning: This method unpacks arrays.
  – Henrik Schumacher
  Sep 6 at 19:48
  

  
  
  
  

Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
– m_goldberg
Sep 6 at 17:56

Some variations on this theme. Mr.Wizard's (## &) can be substituted for Sequence. In recent versions of Mathematica, substituting Nothing works too.
– m_goldberg
Sep 6 at 17:56

1

1

Again, my obligatory warning: This method unpacks arrays.
– Henrik Schumacher
Sep 6 at 19:48

Again, my obligatory warning: This method unpacks arrays.
– Henrik Schumacher
Sep 6 at 19:48

add a comment | 

up vote
7
down vote

The fasted way to do this is to use Part and construct the indices you need to take using Complement:

dropColumns[mat_?MatrixQ, columns : __Integer] := With[
 columnsToTake = Complement[
 Range[Dimensions[mat][[2]]],
 columns
 ]
 ,
 mat[[All, columnsToTake]]
];
a = RandomInteger[25, 9, 11];
a // MatrixForm
dropColumns[a, 2, 5] // MatrixForm

share|improve this answer

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

add a comment | 

up vote
7
down vote

The fasted way to do this is to use Part and construct the indices you need to take using Complement:

dropColumns[mat_?MatrixQ, columns : __Integer] := With[
 columnsToTake = Complement[
 Range[Dimensions[mat][[2]]],
 columns
 ]
 ,
 mat[[All, columnsToTake]]
];
a = RandomInteger[25, 9, 11];
a // MatrixForm
dropColumns[a, 2, 5] // MatrixForm

share|improve this answer

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

add a comment | 

up vote
7
down vote

up vote
7
down vote

The fasted way to do this is to use Part and construct the indices you need to take using Complement:

dropColumns[mat_?MatrixQ, columns : __Integer] := With[
 columnsToTake = Complement[
 Range[Dimensions[mat][[2]]],
 columns
 ]
 ,
 mat[[All, columnsToTake]]
];
a = RandomInteger[25, 9, 11];
a // MatrixForm
dropColumns[a, 2, 5] // MatrixForm

share|improve this answer

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

The fasted way to do this is to use Part and construct the indices you need to take using Complement:

dropColumns[mat_?MatrixQ, columns : __Integer] := With[
 columnsToTake = Complement[
 Range[Dimensions[mat][[2]]],
 columns
 ]
 ,
 mat[[All, columnsToTake]]
];
a = RandomInteger[25, 9, 11];
a // MatrixForm
dropColumns[a, 2, 5] // MatrixForm

share|improve this answer

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

share|improve this answer

share|improve this answer

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

edited Sep 6 at 12:49

Okkes Dulgerci

3,0641615

3,0641615

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

answered Sep 6 at 12:02

Sjoerd Smit

2,490314

2,490314

add a comment | 

add a comment | 

up vote
4
down vote

A = RandomInteger[25, 9, 11];
A // MatrixForm
DeleteCol[Matrix_, indexlist_] := 
 Block[internalMatrix = Matrix, i, k = 0, 
 internallist = SortBy[indexlist, Smaller],
 For[i = 1, i <= Length[internallist], i++, 
 internalMatrix = 
 Delete[Transpose[internalMatrix], internallist[[i]] - k];
 internalMatrix = Transpose[internalMatrix];
 k++;
 ];
 internalMatrix
 ];
DeleteCol[A, 4, 7];
% // MatrixForm
DeleteCol[A, 7, 4];
% // MatrixForm

share|improve this answer

edited Sep 6 at 18:15

answered Sep 6 at 12:08

Diogo

1,686416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can you delete two or more columns?
  – Okkes Dulgerci
  Sep 6 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Okkes Dulgerci now it's possible to delete two or more columns.
  – Diogo
  Sep 6 at 17:28
  

  
  
  
  

add a comment | 

up vote
4
down vote

A = RandomInteger[25, 9, 11];
A // MatrixForm
DeleteCol[Matrix_, indexlist_] := 
 Block[internalMatrix = Matrix, i, k = 0, 
 internallist = SortBy[indexlist, Smaller],
 For[i = 1, i <= Length[internallist], i++, 
 internalMatrix = 
 Delete[Transpose[internalMatrix], internallist[[i]] - k];
 internalMatrix = Transpose[internalMatrix];
 k++;
 ];
 internalMatrix
 ];
DeleteCol[A, 4, 7];
% // MatrixForm
DeleteCol[A, 7, 4];
% // MatrixForm

share|improve this answer

edited Sep 6 at 18:15

answered Sep 6 at 12:08

Diogo

1,686416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can you delete two or more columns?
  – Okkes Dulgerci
  Sep 6 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Okkes Dulgerci now it's possible to delete two or more columns.
  – Diogo
  Sep 6 at 17:28
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

A = RandomInteger[25, 9, 11];
A // MatrixForm
DeleteCol[Matrix_, indexlist_] := 
 Block[internalMatrix = Matrix, i, k = 0, 
 internallist = SortBy[indexlist, Smaller],
 For[i = 1, i <= Length[internallist], i++, 
 internalMatrix = 
 Delete[Transpose[internalMatrix], internallist[[i]] - k];
 internalMatrix = Transpose[internalMatrix];
 k++;
 ];
 internalMatrix
 ];
DeleteCol[A, 4, 7];
% // MatrixForm
DeleteCol[A, 7, 4];
% // MatrixForm

share|improve this answer

edited Sep 6 at 18:15

answered Sep 6 at 12:08

Diogo

1,686416

A = RandomInteger[25, 9, 11];
A // MatrixForm
DeleteCol[Matrix_, indexlist_] := 
 Block[internalMatrix = Matrix, i, k = 0, 
 internallist = SortBy[indexlist, Smaller],
 For[i = 1, i <= Length[internallist], i++, 
 internalMatrix = 
 Delete[Transpose[internalMatrix], internallist[[i]] - k];
 internalMatrix = Transpose[internalMatrix];
 k++;
 ];
 internalMatrix
 ];
DeleteCol[A, 4, 7];
% // MatrixForm
DeleteCol[A, 7, 4];
% // MatrixForm

share|improve this answer

edited Sep 6 at 18:15

answered Sep 6 at 12:08

Diogo

1,686416

share|improve this answer

share|improve this answer

edited Sep 6 at 18:15

edited Sep 6 at 18:15

edited Sep 6 at 18:15

answered Sep 6 at 12:08

Diogo

1,686416

answered Sep 6 at 12:08

Diogo

1,686416

answered Sep 6 at 12:08

Diogo

1,686416

1,686416

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can you delete two or more columns?
  – Okkes Dulgerci
  Sep 6 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Okkes Dulgerci now it's possible to delete two or more columns.
  – Diogo
  Sep 6 at 17:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  How can you delete two or more columns?
  – Okkes Dulgerci
  Sep 6 at 13:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Okkes Dulgerci now it's possible to delete two or more columns.
  – Diogo
  Sep 6 at 17:28
  

  
  
  
  

How can you delete two or more columns?
– Okkes Dulgerci
Sep 6 at 13:25

How can you delete two or more columns?
– Okkes Dulgerci
Sep 6 at 13:25

@Okkes Dulgerci now it's possible to delete two or more columns.
– Diogo
Sep 6 at 17:28

@Okkes Dulgerci now it's possible to delete two or more columns.
– Diogo
Sep 6 at 17:28

add a comment | 

up vote
2
down vote

Here is another way to do it. I assume that columns you want delete is ordered. i.e. 4,7 will work but 7,4 won't.

 SeedRandom@2;
 A = RandomInteger[25, 9, 11];
 MatrixForm@A

$A=left(
beginarrayccccccccccc
23 & 3 & 18 & 11 & 10 & 17 & 8 & 3 & 8 & 0 & 19 \
9 & 23 & 25 & 24 & 14 & 4 & 3 & 4 & 12 & 12 & 8 \
8 & 18 & 6 & 3 & 1 & 14 & 21 & 4 & 14 & 10 & 20 \
22 & 8 & 10 & 20 & 19 & 1 & 9 & 12 & 0 & 19 & 11 \
25 & 10 & 8 & 7 & 18 & 7 & 9 & 23 & 1 & 6 & 15 \
12 & 1 & 0 & 14 & 19 & 12 & 2 & 5 & 3 & 7 & 5 \
23 & 24 & 1 & 14 & 3 & 2 & 22 & 16 & 21 & 4 & 11 \
4 & 2 & 3 & 20 & 24 & 8 & 10 & 3 & 6 & 12 & 19 \
20 & 24 & 6 & 1 & 13 & 10 & 8 & 21 & 5 & 6 & 21 \
endarray
right)$

deleteColumns[mat_?MatrixQ, col_] := 
 With[column = col - Range[0, Length@col - 1], 
 Fold[Delete[#, #2] &, Transpose@A, column] // Transpose]
deleteColumns[A, 4, 7, 10] // MatrixForm

$A=left(
beginarraycccccccc
23 & 3 & 18 & 10 & 17 & 3 & 8 & 19 \
9 & 23 & 25 & 14 & 4 & 4 & 12 & 8 \
8 & 18 & 6 & 1 & 14 & 4 & 14 & 20 \
22 & 8 & 10 & 19 & 1 & 12 & 0 & 11 \
25 & 10 & 8 & 18 & 7 & 23 & 1 & 15 \
12 & 1 & 0 & 19 & 12 & 5 & 3 & 5 \
23 & 24 & 1 & 3 & 2 & 16 & 21 & 11 \
4 & 2 & 3 & 24 & 8 & 3 & 6 & 19 \
20 & 24 & 6 & 13 & 10 & 21 & 5 & 21 \
endarray
right)$

share|improve this answer

edited Sep 6 at 13:30

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

add a comment | 

up vote
2
down vote

Here is another way to do it. I assume that columns you want delete is ordered. i.e. 4,7 will work but 7,4 won't.

 SeedRandom@2;
 A = RandomInteger[25, 9, 11];
 MatrixForm@A

$A=left(
beginarrayccccccccccc
23 & 3 & 18 & 11 & 10 & 17 & 8 & 3 & 8 & 0 & 19 \
9 & 23 & 25 & 24 & 14 & 4 & 3 & 4 & 12 & 12 & 8 \
8 & 18 & 6 & 3 & 1 & 14 & 21 & 4 & 14 & 10 & 20 \
22 & 8 & 10 & 20 & 19 & 1 & 9 & 12 & 0 & 19 & 11 \
25 & 10 & 8 & 7 & 18 & 7 & 9 & 23 & 1 & 6 & 15 \
12 & 1 & 0 & 14 & 19 & 12 & 2 & 5 & 3 & 7 & 5 \
23 & 24 & 1 & 14 & 3 & 2 & 22 & 16 & 21 & 4 & 11 \
4 & 2 & 3 & 20 & 24 & 8 & 10 & 3 & 6 & 12 & 19 \
20 & 24 & 6 & 1 & 13 & 10 & 8 & 21 & 5 & 6 & 21 \
endarray
right)$

deleteColumns[mat_?MatrixQ, col_] := 
 With[column = col - Range[0, Length@col - 1], 
 Fold[Delete[#, #2] &, Transpose@A, column] // Transpose]
deleteColumns[A, 4, 7, 10] // MatrixForm

$A=left(
beginarraycccccccc
23 & 3 & 18 & 10 & 17 & 3 & 8 & 19 \
9 & 23 & 25 & 14 & 4 & 4 & 12 & 8 \
8 & 18 & 6 & 1 & 14 & 4 & 14 & 20 \
22 & 8 & 10 & 19 & 1 & 12 & 0 & 11 \
25 & 10 & 8 & 18 & 7 & 23 & 1 & 15 \
12 & 1 & 0 & 19 & 12 & 5 & 3 & 5 \
23 & 24 & 1 & 3 & 2 & 16 & 21 & 11 \
4 & 2 & 3 & 24 & 8 & 3 & 6 & 19 \
20 & 24 & 6 & 13 & 10 & 21 & 5 & 21 \
endarray
right)$

share|improve this answer

edited Sep 6 at 13:30

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

add a comment | 

up vote
2
down vote

up vote
2
down vote

Here is another way to do it. I assume that columns you want delete is ordered. i.e. 4,7 will work but 7,4 won't.

 SeedRandom@2;
 A = RandomInteger[25, 9, 11];
 MatrixForm@A

$A=left(
beginarrayccccccccccc
23 & 3 & 18 & 11 & 10 & 17 & 8 & 3 & 8 & 0 & 19 \
9 & 23 & 25 & 24 & 14 & 4 & 3 & 4 & 12 & 12 & 8 \
8 & 18 & 6 & 3 & 1 & 14 & 21 & 4 & 14 & 10 & 20 \
22 & 8 & 10 & 20 & 19 & 1 & 9 & 12 & 0 & 19 & 11 \
25 & 10 & 8 & 7 & 18 & 7 & 9 & 23 & 1 & 6 & 15 \
12 & 1 & 0 & 14 & 19 & 12 & 2 & 5 & 3 & 7 & 5 \
23 & 24 & 1 & 14 & 3 & 2 & 22 & 16 & 21 & 4 & 11 \
4 & 2 & 3 & 20 & 24 & 8 & 10 & 3 & 6 & 12 & 19 \
20 & 24 & 6 & 1 & 13 & 10 & 8 & 21 & 5 & 6 & 21 \
endarray
right)$

deleteColumns[mat_?MatrixQ, col_] := 
 With[column = col - Range[0, Length@col - 1], 
 Fold[Delete[#, #2] &, Transpose@A, column] // Transpose]
deleteColumns[A, 4, 7, 10] // MatrixForm

$A=left(
beginarraycccccccc
23 & 3 & 18 & 10 & 17 & 3 & 8 & 19 \
9 & 23 & 25 & 14 & 4 & 4 & 12 & 8 \
8 & 18 & 6 & 1 & 14 & 4 & 14 & 20 \
22 & 8 & 10 & 19 & 1 & 12 & 0 & 11 \
25 & 10 & 8 & 18 & 7 & 23 & 1 & 15 \
12 & 1 & 0 & 19 & 12 & 5 & 3 & 5 \
23 & 24 & 1 & 3 & 2 & 16 & 21 & 11 \
4 & 2 & 3 & 24 & 8 & 3 & 6 & 19 \
20 & 24 & 6 & 13 & 10 & 21 & 5 & 21 \
endarray
right)$

share|improve this answer

edited Sep 6 at 13:30

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

Here is another way to do it. I assume that columns you want delete is ordered. i.e. 4,7 will work but 7,4 won't.

 SeedRandom@2;
 A = RandomInteger[25, 9, 11];
 MatrixForm@A

$A=left(
beginarrayccccccccccc
23 & 3 & 18 & 11 & 10 & 17 & 8 & 3 & 8 & 0 & 19 \
9 & 23 & 25 & 24 & 14 & 4 & 3 & 4 & 12 & 12 & 8 \
8 & 18 & 6 & 3 & 1 & 14 & 21 & 4 & 14 & 10 & 20 \
22 & 8 & 10 & 20 & 19 & 1 & 9 & 12 & 0 & 19 & 11 \
25 & 10 & 8 & 7 & 18 & 7 & 9 & 23 & 1 & 6 & 15 \
12 & 1 & 0 & 14 & 19 & 12 & 2 & 5 & 3 & 7 & 5 \
23 & 24 & 1 & 14 & 3 & 2 & 22 & 16 & 21 & 4 & 11 \
4 & 2 & 3 & 20 & 24 & 8 & 10 & 3 & 6 & 12 & 19 \
20 & 24 & 6 & 1 & 13 & 10 & 8 & 21 & 5 & 6 & 21 \
endarray
right)$

deleteColumns[mat_?MatrixQ, col_] := 
 With[column = col - Range[0, Length@col - 1], 
 Fold[Delete[#, #2] &, Transpose@A, column] // Transpose]
deleteColumns[A, 4, 7, 10] // MatrixForm

$A=left(
beginarraycccccccc
23 & 3 & 18 & 10 & 17 & 3 & 8 & 19 \
9 & 23 & 25 & 14 & 4 & 4 & 12 & 8 \
8 & 18 & 6 & 1 & 14 & 4 & 14 & 20 \
22 & 8 & 10 & 19 & 1 & 12 & 0 & 11 \
25 & 10 & 8 & 18 & 7 & 23 & 1 & 15 \
12 & 1 & 0 & 19 & 12 & 5 & 3 & 5 \
23 & 24 & 1 & 3 & 2 & 16 & 21 & 11 \
4 & 2 & 3 & 24 & 8 & 3 & 6 & 19 \
20 & 24 & 6 & 13 & 10 & 21 & 5 & 21 \
endarray
right)$

share|improve this answer

edited Sep 6 at 13:30

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

share|improve this answer

share|improve this answer

edited Sep 6 at 13:30

edited Sep 6 at 13:30

edited Sep 6 at 13:30

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

answered Sep 6 at 13:17

Okkes Dulgerci

3,0641615

3,0641615

add a comment | 

add a comment | 

up vote
2
down vote

I am confused by the complexity of many of these responses. The single command

Drop[A, None, 4,7,3]

is sufficient to remove columns $4$ and $7$, in steps of $3$ (so as not to remove columms $5$ and $6$). If you wanted to remove more than two columns that are not separated by an equal number of columns, then you would need to use Drop more than once, starting from the rightmost column to be removed; e.g., to remove columns $1$, $3$, and $9$, you could just do

Drop[Drop[A, None, 9], None, 1,3,2]

or equivalently,

Drop[Drop[A, None, 3,9,6], None, 1]

If you had an arbitrary sorted list of column indices to remove, which we might call c, then Fold is trivially applied:

Fold[Drop[#1, None, #2]&, A, Reverse[c]]

share|improve this answer

edited Sep 7 at 1:02

answered Sep 7 at 0:57

heropup

1,529712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
  – Sjoerd Smit
  Sep 7 at 8:15
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

I am confused by the complexity of many of these responses. The single command

Drop[A, None, 4,7,3]

is sufficient to remove columns $4$ and $7$, in steps of $3$ (so as not to remove columms $5$ and $6$). If you wanted to remove more than two columns that are not separated by an equal number of columns, then you would need to use Drop more than once, starting from the rightmost column to be removed; e.g., to remove columns $1$, $3$, and $9$, you could just do

Drop[Drop[A, None, 9], None, 1,3,2]

or equivalently,

Drop[Drop[A, None, 3,9,6], None, 1]

If you had an arbitrary sorted list of column indices to remove, which we might call c, then Fold is trivially applied:

Fold[Drop[#1, None, #2]&, A, Reverse[c]]

share|improve this answer

edited Sep 7 at 1:02

answered Sep 7 at 0:57

heropup

1,529712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
  – Sjoerd Smit
  Sep 7 at 8:15
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

I am confused by the complexity of many of these responses. The single command

Drop[A, None, 4,7,3]

is sufficient to remove columns $4$ and $7$, in steps of $3$ (so as not to remove columms $5$ and $6$). If you wanted to remove more than two columns that are not separated by an equal number of columns, then you would need to use Drop more than once, starting from the rightmost column to be removed; e.g., to remove columns $1$, $3$, and $9$, you could just do

Drop[Drop[A, None, 9], None, 1,3,2]

or equivalently,

Drop[Drop[A, None, 3,9,6], None, 1]

If you had an arbitrary sorted list of column indices to remove, which we might call c, then Fold is trivially applied:

Fold[Drop[#1, None, #2]&, A, Reverse[c]]

share|improve this answer

edited Sep 7 at 1:02

answered Sep 7 at 0:57

heropup

1,529712

I am confused by the complexity of many of these responses. The single command

Drop[A, None, 4,7,3]

is sufficient to remove columns $4$ and $7$, in steps of $3$ (so as not to remove columms $5$ and $6$). If you wanted to remove more than two columns that are not separated by an equal number of columns, then you would need to use Drop more than once, starting from the rightmost column to be removed; e.g., to remove columns $1$, $3$, and $9$, you could just do

Drop[Drop[A, None, 9], None, 1,3,2]

or equivalently,

Drop[Drop[A, None, 3,9,6], None, 1]

If you had an arbitrary sorted list of column indices to remove, which we might call c, then Fold is trivially applied:

Fold[Drop[#1, None, #2]&, A, Reverse[c]]

share|improve this answer

edited Sep 7 at 1:02

answered Sep 7 at 0:57

heropup

1,529712

share|improve this answer

share|improve this answer

edited Sep 7 at 1:02

edited Sep 7 at 1:02

edited Sep 7 at 1:02

answered Sep 7 at 0:57

heropup

1,529712

answered Sep 7 at 0:57

heropup

1,529712

answered Sep 7 at 0:57

heropup

1,529712

1,529712

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
  – Sjoerd Smit
  Sep 7 at 8:15
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
  – Sjoerd Smit
  Sep 7 at 8:15
  
  

  
  
  
  

I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
– Sjoerd Smit
Sep 7 at 8:15

I considered Drop, but because you cannot give give it an arbitrary list of column indices, I decided that Part is more suited for the job. Especially because repeated calls to Drop is going to be slow for large matrices.
– Sjoerd Smit
Sep 7 at 8:15

add a comment | 

up vote
1
down vote

MapThread[Delete, A, ConstantArray[4, 7, Length[A]]]

share|improve this answer

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

add a comment | 

up vote
1
down vote

MapThread[Delete, A, ConstantArray[4, 7, Length[A]]]

share|improve this answer

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

add a comment | 

up vote
1
down vote

up vote
1
down vote

MapThread[Delete, A, ConstantArray[4, 7, Length[A]]]

share|improve this answer

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

MapThread[Delete, A, ConstantArray[4, 7, Length[A]]]

share|improve this answer

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

share|improve this answer

share|improve this answer

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

answered Sep 6 at 14:02

Chris Degnen

21.4k23281

21.4k23281

add a comment | 

add a comment | 

up vote
1
down vote

You can also use a combination of Fold and Drop:

ClearAll[dropCols1]
dropCols1 = Fold[Drop[#, None, #2] &, #, List /@ Reverse @ Sort[#2]] &;

Examples:

m = Array[Subscript[a, ##] &, 9, 9];

dropCols1[m, 4, 7] // MatrixForm // TeXForm

$smallleft(
beginarrayccccccc
a_1,1 & a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,1 & a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,1 & a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,1 & a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,1 & a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,1 & a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,1 & a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,1 & a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,1 & a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

dropCols1[m, 4, 7, 1] // MatrixForm // TeXForm

$smallleft(
beginarraycccccc
a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

Alternatively, (1) assign ##& (or Nothing in versions 10+) to the desired columns (dropCols2) or (2) use a combination of MapAt and ##&& (dropCols3):

ClearAll[dropCols2, dropCols3]
dropCols2 = Module[a = #, a[[All, #2]] = ## &; a] &;
dropCols3 = MapAt[## & &, #, All, #2] &;

Equal @@ (#[m, 4, 7] & /@ dropCols1, dropCols2, dropCols3)

True

Equal @@ (#[m, 4, 7, 1] & /@ dropCols1, dropCols2, dropCols3)

True

share|improve this answer

edited Sep 7 at 0:53

answered Sep 6 at 22:50

kglr

159k8183383

add a comment | 

up vote
1
down vote

You can also use a combination of Fold and Drop:

ClearAll[dropCols1]
dropCols1 = Fold[Drop[#, None, #2] &, #, List /@ Reverse @ Sort[#2]] &;

Examples:

m = Array[Subscript[a, ##] &, 9, 9];

dropCols1[m, 4, 7] // MatrixForm // TeXForm

$smallleft(
beginarrayccccccc
a_1,1 & a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,1 & a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,1 & a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,1 & a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,1 & a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,1 & a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,1 & a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,1 & a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,1 & a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

dropCols1[m, 4, 7, 1] // MatrixForm // TeXForm

$smallleft(
beginarraycccccc
a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

Alternatively, (1) assign ##& (or Nothing in versions 10+) to the desired columns (dropCols2) or (2) use a combination of MapAt and ##&& (dropCols3):

ClearAll[dropCols2, dropCols3]
dropCols2 = Module[a = #, a[[All, #2]] = ## &; a] &;
dropCols3 = MapAt[## & &, #, All, #2] &;

Equal @@ (#[m, 4, 7] & /@ dropCols1, dropCols2, dropCols3)

True

Equal @@ (#[m, 4, 7, 1] & /@ dropCols1, dropCols2, dropCols3)

True

share|improve this answer

edited Sep 7 at 0:53

answered Sep 6 at 22:50

kglr

159k8183383

add a comment | 

up vote
1
down vote

up vote
1
down vote

You can also use a combination of Fold and Drop:

ClearAll[dropCols1]
dropCols1 = Fold[Drop[#, None, #2] &, #, List /@ Reverse @ Sort[#2]] &;

Examples:

m = Array[Subscript[a, ##] &, 9, 9];

dropCols1[m, 4, 7] // MatrixForm // TeXForm

$smallleft(
beginarrayccccccc
a_1,1 & a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,1 & a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,1 & a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,1 & a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,1 & a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,1 & a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,1 & a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,1 & a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,1 & a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

dropCols1[m, 4, 7, 1] // MatrixForm // TeXForm

$smallleft(
beginarraycccccc
a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

Alternatively, (1) assign ##& (or Nothing in versions 10+) to the desired columns (dropCols2) or (2) use a combination of MapAt and ##&& (dropCols3):

ClearAll[dropCols2, dropCols3]
dropCols2 = Module[a = #, a[[All, #2]] = ## &; a] &;
dropCols3 = MapAt[## & &, #, All, #2] &;

Equal @@ (#[m, 4, 7] & /@ dropCols1, dropCols2, dropCols3)

True

Equal @@ (#[m, 4, 7, 1] & /@ dropCols1, dropCols2, dropCols3)

True

share|improve this answer

edited Sep 7 at 0:53

answered Sep 6 at 22:50

kglr

159k8183383

You can also use a combination of Fold and Drop:

ClearAll[dropCols1]
dropCols1 = Fold[Drop[#, None, #2] &, #, List /@ Reverse @ Sort[#2]] &;

Examples:

m = Array[Subscript[a, ##] &, 9, 9];

dropCols1[m, 4, 7] // MatrixForm // TeXForm

$smallleft(
beginarrayccccccc
a_1,1 & a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,1 & a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,1 & a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,1 & a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,1 & a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,1 & a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,1 & a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,1 & a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,1 & a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

dropCols1[m, 4, 7, 1] // MatrixForm // TeXForm

$smallleft(
beginarraycccccc
a_1,2 & a_1,3 & a_1,5 & a_1,6 & a_1,8 & a_1,9 \
a_2,2 & a_2,3 & a_2,5 & a_2,6 & a_2,8 & a_2,9 \
a_3,2 & a_3,3 & a_3,5 & a_3,6 & a_3,8 & a_3,9 \
a_4,2 & a_4,3 & a_4,5 & a_4,6 & a_4,8 & a_4,9 \
a_5,2 & a_5,3 & a_5,5 & a_5,6 & a_5,8 & a_5,9 \
a_6,2 & a_6,3 & a_6,5 & a_6,6 & a_6,8 & a_6,9 \
a_7,2 & a_7,3 & a_7,5 & a_7,6 & a_7,8 & a_7,9 \
a_8,2 & a_8,3 & a_8,5 & a_8,6 & a_8,8 & a_8,9 \
a_9,2 & a_9,3 & a_9,5 & a_9,6 & a_9,8 & a_9,9 \
endarray
right)$

Alternatively, (1) assign ##& (or Nothing in versions 10+) to the desired columns (dropCols2) or (2) use a combination of MapAt and ##&& (dropCols3):

ClearAll[dropCols2, dropCols3]
dropCols2 = Module[a = #, a[[All, #2]] = ## &; a] &;
dropCols3 = MapAt[## & &, #, All, #2] &;

Equal @@ (#[m, 4, 7] & /@ dropCols1, dropCols2, dropCols3)

True

Equal @@ (#[m, 4, 7, 1] & /@ dropCols1, dropCols2, dropCols3)

True

share|improve this answer

edited Sep 7 at 0:53

answered Sep 6 at 22:50

kglr

159k8183383

share|improve this answer

share|improve this answer

edited Sep 7 at 0:53

edited Sep 7 at 0:53

edited Sep 7 at 0:53

answered Sep 6 at 22:50

kglr

159k8183383

answered Sep 6 at 22:50

kglr

159k8183383

answered Sep 6 at 22:50

kglr

159k8183383

159k8183383

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