Replacing a column of a matrix
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
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I want to replace a column of a matrix. The best I've been able to come up with is
a=RandomReal[9,5,5]
b=ConstantArray[0,5]
replacepos=4
Transpose[ReplacePart[Transpose[a], replacepos -> b]],
which does the trick, but looks awkward to me.
list-manipulation matrix
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
asked Dec 5 at 12:09
Patricio
406
add a comment |
up vote
3
down vote
favorite
I want to replace a column of a matrix. The best I've been able to come up with is
a=RandomReal[9,5,5]
b=ConstantArray[0,5]
replacepos=4
Transpose[ReplacePart[Transpose[a], replacepos -> b]],
which does the trick, but looks awkward to me.
list-manipulation matrix
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
asked Dec 5 at 12:09
Patricio
406
2
Did you search the site before posting? This has been asked a bunch of times...
– Marius Ladegård Meyer
Dec 5 at 12:13
There is the insert function I believe is useful for you.
– Buddha_the_Scientist
Dec 5 at 12:13
a.DiagonalMatrix[1, 1, 1, 0, 1]
– user1066
Dec 5 at 12:53
@MariusLadegårdMeyer, yes I did, but previous questions seemed to look for more complex things. Thanks
– Patricio
Dec 5 at 18:54
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I want to replace a column of a matrix. The best I've been able to come up with is
a=RandomReal[9,5,5]
b=ConstantArray[0,5]
replacepos=4
Transpose[ReplacePart[Transpose[a], replacepos -> b]],
which does the trick, but looks awkward to me.
list-manipulation matrix
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
asked Dec 5 at 12:09
Patricio
406
I want to replace a column of a matrix. The best I've been able to come up with is
a=RandomReal[9,5,5]
b=ConstantArray[0,5]
replacepos=4
Transpose[ReplacePart[Transpose[a], replacepos -> b]],
which does the trick, but looks awkward to me.
list-manipulation matrix
list-manipulation matrix
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
asked Dec 5 at 12:09
Patricio
406
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
asked Dec 5 at 12:09
Patricio
406
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
edited Dec 5 at 12:15
Henrik Schumacher
47.4k466134
47.4k466134
asked Dec 5 at 12:09
Patricio
406
asked Dec 5 at 12:09
Patricio
406
asked Dec 5 at 12:09
Patricio
406
406
2
Did you search the site before posting? This has been asked a bunch of times...
– Marius Ladegård Meyer
Dec 5 at 12:13
There is the insert function I believe is useful for you.
– Buddha_the_Scientist
Dec 5 at 12:13
a.DiagonalMatrix[1, 1, 1, 0, 1]
– user1066
Dec 5 at 12:53
@MariusLadegårdMeyer, yes I did, but previous questions seemed to look for more complex things. Thanks
– Patricio
Dec 5 at 18:54
add a comment |
2
Did you search the site before posting? This has been asked a bunch of times...
– Marius Ladegård Meyer
Dec 5 at 12:13
There is the insert function I believe is useful for you.
– Buddha_the_Scientist
Dec 5 at 12:13
a.DiagonalMatrix[1, 1, 1, 0, 1]
– user1066
Dec 5 at 12:53
@MariusLadegårdMeyer, yes I did, but previous questions seemed to look for more complex things. Thanks
– Patricio
Dec 5 at 18:54
2
2
Did you search the site before posting? This has been asked a bunch of times...
– Marius Ladegård Meyer
Dec 5 at 12:13
Did you search the site before posting? This has been asked a bunch of times...
– Marius Ladegård Meyer
Dec 5 at 12:13
There is the insert function I believe is useful for you.
– Buddha_the_Scientist
Dec 5 at 12:13
There is the insert function I believe is useful for you.
– Buddha_the_Scientist
Dec 5 at 12:13
a.DiagonalMatrix[1, 1, 1, 0, 1]– user1066
Dec 5 at 12:53
a.DiagonalMatrix[1, 1, 1, 0, 1]– user1066
Dec 5 at 12:53
@MariusLadegårdMeyer, yes I did, but previous questions seemed to look for more complex things. Thanks
– Patricio
Dec 5 at 18:54
@MariusLadegårdMeyer, yes I did, but previous questions seemed to look for more complex things. Thanks
– Patricio
Dec 5 at 18:54
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
What about
a[[All, replacepos]] = b
?
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
add a comment |
up vote
2
down vote
Indeed, Transpose has to perform superfluous reorderings which are memory bound. This does the trick and should be fairly efficient:
c = Module[buffer = a,
buffer[[All, replacepos]] = b;
buffer
]
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
add a comment |
up vote
1
down vote
a = RandomReal[9, 5, 5]
(a[[#, 4]] = 0) & /@ Range@5
a
answered Dec 5 at 12:21
J42161217
3,712220
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
What about
a[[All, replacepos]] = b
?
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
add a comment |
up vote
3
down vote
What about
a[[All, replacepos]] = b
?
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
add a comment |
up vote
3
down vote
up vote
3
down vote
What about
a[[All, replacepos]] = b
?
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
What about
a[[All, replacepos]] = b
?
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
answered Dec 5 at 13:15
Ulrich Neumann
6,555515
6,555515
add a comment |
add a comment |
up vote
2
down vote
Indeed, Transpose has to perform superfluous reorderings which are memory bound. This does the trick and should be fairly efficient:
c = Module[buffer = a,
buffer[[All, replacepos]] = b;
buffer
]
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
add a comment |
up vote
2
down vote
Indeed, Transpose has to perform superfluous reorderings which are memory bound. This does the trick and should be fairly efficient:
c = Module[buffer = a,
buffer[[All, replacepos]] = b;
buffer
]
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
add a comment |
up vote
2
down vote
up vote
2
down vote
Indeed, Transpose has to perform superfluous reorderings which are memory bound. This does the trick and should be fairly efficient:
c = Module[buffer = a,
buffer[[All, replacepos]] = b;
buffer
]
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
Indeed, Transpose has to perform superfluous reorderings which are memory bound. This does the trick and should be fairly efficient:
c = Module[buffer = a,
buffer[[All, replacepos]] = b;
buffer
]
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
edited Dec 5 at 14:42
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
answered Dec 5 at 12:13
Henrik Schumacher
47.4k466134
47.4k466134
add a comment |
add a comment |
up vote
1
down vote
a = RandomReal[9, 5, 5]
(a[[#, 4]] = 0) & /@ Range@5
a
answered Dec 5 at 12:21
J42161217
3,712220
add a comment |
up vote
1
down vote
a = RandomReal[9, 5, 5]
(a[[#, 4]] = 0) & /@ Range@5
a
answered Dec 5 at 12:21
J42161217
3,712220
add a comment |
up vote
1
down vote
up vote
1
down vote
a = RandomReal[9, 5, 5]
(a[[#, 4]] = 0) & /@ Range@5
a
answered Dec 5 at 12:21
J42161217
3,712220
a = RandomReal[9, 5, 5]
(a[[#, 4]] = 0) & /@ Range@5
a
answered Dec 5 at 12:21
J42161217
3,712220
answered Dec 5 at 12:21
J42161217
3,712220
answered Dec 5 at 12:21
J42161217
3,712220
answered Dec 5 at 12:21
J42161217
3,712220
3,712220
add a comment |
add a comment |
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2
Did you search the site before posting? This has been asked a bunch of times...
– Marius Ladegård Meyer
Dec 5 at 12:13
There is the insert function I believe is useful for you.
– Buddha_the_Scientist
Dec 5 at 12:13
a.DiagonalMatrix[1, 1, 1, 0, 1]– user1066
Dec 5 at 12:53
@MariusLadegårdMeyer, yes I did, but previous questions seemed to look for more complex things. Thanks
– Patricio
Dec 5 at 18:54