Elements Replacement
Clash Royale CLAN TAG#URR8PPP
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Please I need help on how to replace elements in a matrix with another elements. For example, I have a (5,5) matrix, and I like to replace elements from All rows and column 3 to 5, with zero. such that (All, 3;;5) replace with zero.
m = MatrixForm[Partition[Range[25]^2, 5]]
I like to replace all the elements in the specified area with zero:
list-manipulation matrix replacement
$endgroup$
add a comment |
$begingroup$
Please I need help on how to replace elements in a matrix with another elements. For example, I have a (5,5) matrix, and I like to replace elements from All rows and column 3 to 5, with zero. such that (All, 3;;5) replace with zero.
m = MatrixForm[Partition[Range[25]^2, 5]]
I like to replace all the elements in the specified area with zero:
list-manipulation matrix replacement
$endgroup$
add a comment |
$begingroup$
Please I need help on how to replace elements in a matrix with another elements. For example, I have a (5,5) matrix, and I like to replace elements from All rows and column 3 to 5, with zero. such that (All, 3;;5) replace with zero.
m = MatrixForm[Partition[Range[25]^2, 5]]
I like to replace all the elements in the specified area with zero:
list-manipulation matrix replacement
$endgroup$
Please I need help on how to replace elements in a matrix with another elements. For example, I have a (5,5) matrix, and I like to replace elements from All rows and column 3 to 5, with zero. such that (All, 3;;5) replace with zero.
m = MatrixForm[Partition[Range[25]^2, 5]]
I like to replace all the elements in the specified area with zero:
list-manipulation matrix replacement
list-manipulation matrix replacement
asked Feb 16 at 10:24
GeopGeop
473
473
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add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Don't use MatrixForm
when defining the matrix. It's only a display wrapper, as explained here.
m = Partition[Range[25]^2, 5]
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
Replacement:
m[[All, 3 ;; 5]] = 0;
m
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289, 0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
add a comment |
$begingroup$
m = Partition[Range[25]^2, 5]
A few more alternatives:
SparseArray[m[[All, ;; 2]], Dimensions[m]]
PadRight[m[[All, ;; 2]], Dimensions@m]
MapAt[0 &, m , All, 3 ;; ]
all give
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289,
0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Don't use MatrixForm
when defining the matrix. It's only a display wrapper, as explained here.
m = Partition[Range[25]^2, 5]
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
Replacement:
m[[All, 3 ;; 5]] = 0;
m
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289, 0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
add a comment |
$begingroup$
Don't use MatrixForm
when defining the matrix. It's only a display wrapper, as explained here.
m = Partition[Range[25]^2, 5]
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
Replacement:
m[[All, 3 ;; 5]] = 0;
m
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289, 0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
add a comment |
$begingroup$
Don't use MatrixForm
when defining the matrix. It's only a display wrapper, as explained here.
m = Partition[Range[25]^2, 5]
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
Replacement:
m[[All, 3 ;; 5]] = 0;
m
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289, 0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
Don't use MatrixForm
when defining the matrix. It's only a display wrapper, as explained here.
m = Partition[Range[25]^2, 5]
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625
Replacement:
m[[All, 3 ;; 5]] = 0;
m
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289, 0, 0, 0, 441, 484, 0, 0, 0
edited Feb 17 at 4:36
answered Feb 16 at 10:33
RomanRoman
3,030718
3,030718
add a comment |
add a comment |
$begingroup$
m = Partition[Range[25]^2, 5]
A few more alternatives:
SparseArray[m[[All, ;; 2]], Dimensions[m]]
PadRight[m[[All, ;; 2]], Dimensions@m]
MapAt[0 &, m , All, 3 ;; ]
all give
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289,
0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
add a comment |
$begingroup$
m = Partition[Range[25]^2, 5]
A few more alternatives:
SparseArray[m[[All, ;; 2]], Dimensions[m]]
PadRight[m[[All, ;; 2]], Dimensions@m]
MapAt[0 &, m , All, 3 ;; ]
all give
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289,
0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
add a comment |
$begingroup$
m = Partition[Range[25]^2, 5]
A few more alternatives:
SparseArray[m[[All, ;; 2]], Dimensions[m]]
PadRight[m[[All, ;; 2]], Dimensions@m]
MapAt[0 &, m , All, 3 ;; ]
all give
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289,
0, 0, 0, 441, 484, 0, 0, 0
$endgroup$
m = Partition[Range[25]^2, 5]
A few more alternatives:
SparseArray[m[[All, ;; 2]], Dimensions[m]]
PadRight[m[[All, ;; 2]], Dimensions@m]
MapAt[0 &, m , All, 3 ;; ]
all give
1, 4, 0, 0, 0, 36, 49, 0, 0, 0, 121, 144, 0, 0, 0, 256, 289,
0, 0, 0, 441, 484, 0, 0, 0
answered Feb 16 at 11:42
kglrkglr
188k10204422
188k10204422
add a comment |
add a comment |
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