Dimension 1 and dimension 2 in octave
$begingroup$
I have this in octave:
octave:139> r
r =
3 1
7 2
octave:140> r(2,:)
ans =
7 2
octave:141> r(1,1)
ans = 3
octave:142> r(1,2)
ans = 1
This kind of indexing indicates to me that the dimension 1 is row and dimension 2 is column.
But down below when I specify dimension 1, I get max along the columns (now rows)
octave:143> max(r,,1)
ans =
7 2
And here when I specify dimension 2, I get max along the rows (now colums)
octave:144> max(r,,2)
ans =
3
7
Why so ?
octave
asked Feb 28 '17 at 15:23
abcabc
1165
$endgroup$
add a comment |
$begingroup$
I have this in octave:
octave:139> r
r =
3 1
7 2
octave:140> r(2,:)
ans =
7 2
octave:141> r(1,1)
ans = 3
octave:142> r(1,2)
ans = 1
This kind of indexing indicates to me that the dimension 1 is row and dimension 2 is column.
But down below when I specify dimension 1, I get max along the columns (now rows)
octave:143> max(r,,1)
ans =
7 2
And here when I specify dimension 2, I get max along the rows (now colums)
octave:144> max(r,,2)
ans =
3
7
Why so ?
octave
asked Feb 28 '17 at 15:23
abcabc
1165
$endgroup$
add a comment |
$begingroup$
I have this in octave:
octave:139> r
r =
3 1
7 2
octave:140> r(2,:)
ans =
7 2
octave:141> r(1,1)
ans = 3
octave:142> r(1,2)
ans = 1
This kind of indexing indicates to me that the dimension 1 is row and dimension 2 is column.
But down below when I specify dimension 1, I get max along the columns (now rows)
octave:143> max(r,,1)
ans =
7 2
And here when I specify dimension 2, I get max along the rows (now colums)
octave:144> max(r,,2)
ans =
3
7
Why so ?
octave
asked Feb 28 '17 at 15:23
abcabc
1165
$endgroup$
I have this in octave:
octave:139> r
r =
3 1
7 2
octave:140> r(2,:)
ans =
7 2
octave:141> r(1,1)
ans = 3
octave:142> r(1,2)
ans = 1
This kind of indexing indicates to me that the dimension 1 is row and dimension 2 is column.
But down below when I specify dimension 1, I get max along the columns (now rows)
octave:143> max(r,,1)
ans =
7 2
And here when I specify dimension 2, I get max along the rows (now colums)
octave:144> max(r,,2)
ans =
3
7
Why so ?
octave
octave
asked Feb 28 '17 at 15:23
abcabc
1165
asked Feb 28 '17 at 15:23
abcabc
1165
edited Feb 28 '17 at 18:04
abc
asked Feb 28 '17 at 15:23
abcabc
1165
asked Feb 28 '17 at 15:23
abcabc
1165
asked Feb 28 '17 at 15:23
abcabc
1165
1165
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yeah, that can be a bit confusing. Try to think about it this way:
1 = which row
2 = which column
max(r,,1) = "get the biggest row" (which of course might be a mixture of different rows).
max(r,,2) = get the biggest column
Hope this helps to see the reason for this. For matrices it would be possible to use a different convention which might look more intuitve for you. But for higher rank objects (i.e. more than two indices), this convention is the only sensible.
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
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add a comment |
$begingroup$
I think a more sensible answer is that the DIM input tells Octave which dimension to take as the independent variable. That is, for example in sum(X, DIM) with DIM =1, at each step the summation is done along the vertical direction (first dimension), which corresponds to the rows as you have observed in your first examples, and the result is a function of the horizontal direction, much like in a double integral.
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Yeah, that can be a bit confusing. Try to think about it this way:
1 = which row
2 = which column
max(r,,1) = "get the biggest row" (which of course might be a mixture of different rows).
max(r,,2) = get the biggest column
Hope this helps to see the reason for this. For matrices it would be possible to use a different convention which might look more intuitve for you. But for higher rank objects (i.e. more than two indices), this convention is the only sensible.
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
$endgroup$
add a comment |
$begingroup$
Yeah, that can be a bit confusing. Try to think about it this way:
1 = which row
2 = which column
max(r,,1) = "get the biggest row" (which of course might be a mixture of different rows).
max(r,,2) = get the biggest column
Hope this helps to see the reason for this. For matrices it would be possible to use a different convention which might look more intuitve for you. But for higher rank objects (i.e. more than two indices), this convention is the only sensible.
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
$endgroup$
add a comment |
$begingroup$
Yeah, that can be a bit confusing. Try to think about it this way:
1 = which row
2 = which column
max(r,,1) = "get the biggest row" (which of course might be a mixture of different rows).
max(r,,2) = get the biggest column
Hope this helps to see the reason for this. For matrices it would be possible to use a different convention which might look more intuitve for you. But for higher rank objects (i.e. more than two indices), this convention is the only sensible.
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
$endgroup$
Yeah, that can be a bit confusing. Try to think about it this way:
1 = which row
2 = which column
max(r,,1) = "get the biggest row" (which of course might be a mixture of different rows).
max(r,,2) = get the biggest column
Hope this helps to see the reason for this. For matrices it would be possible to use a different convention which might look more intuitve for you. But for higher rank objects (i.e. more than two indices), this convention is the only sensible.
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
answered Feb 28 '17 at 16:10
SimonSimon
2,285216
2,285216
add a comment |
add a comment |
$begingroup$
I think a more sensible answer is that the DIM input tells Octave which dimension to take as the independent variable. That is, for example in sum(X, DIM) with DIM =1, at each step the summation is done along the vertical direction (first dimension), which corresponds to the rows as you have observed in your first examples, and the result is a function of the horizontal direction, much like in a double integral.
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
$endgroup$
add a comment |
$begingroup$
I think a more sensible answer is that the DIM input tells Octave which dimension to take as the independent variable. That is, for example in sum(X, DIM) with DIM =1, at each step the summation is done along the vertical direction (first dimension), which corresponds to the rows as you have observed in your first examples, and the result is a function of the horizontal direction, much like in a double integral.
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
$endgroup$
add a comment |
$begingroup$
I think a more sensible answer is that the DIM input tells Octave which dimension to take as the independent variable. That is, for example in sum(X, DIM) with DIM =1, at each step the summation is done along the vertical direction (first dimension), which corresponds to the rows as you have observed in your first examples, and the result is a function of the horizontal direction, much like in a double integral.
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
$endgroup$
I think a more sensible answer is that the DIM input tells Octave which dimension to take as the independent variable. That is, for example in sum(X, DIM) with DIM =1, at each step the summation is done along the vertical direction (first dimension), which corresponds to the rows as you have observed in your first examples, and the result is a function of the horizontal direction, much like in a double integral.
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
answered Dec 12 '18 at 18:06
arsaKasraarsaKasra
1011
1011
add a comment |
add a comment |
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