Excel Transpose 1 Dimensional Array into a Multidimensional Array
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-1
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I need to adjust my data so it aligns with the rest,
This is my data, and the trailing off numbers need to be put in a (12 by x) matrix
I have been doing this by hand, so the next line looks like this:
And I want to vertically transpose the rest of these numbers that are currently horizontal by groups of 12 so I do not have to do it by hand.
These are the 'trailing off' numbers that need to be vertically transposed by groups of 12, so into a 12 by ?? matrix:
excel transpose
edited Nov 20 at 12:59
Luuklag
3,40482742
asked Nov 19 at 21:08
John Gustafson
11
add a comment |
up vote
-1
down vote
favorite
I need to adjust my data so it aligns with the rest,
This is my data, and the trailing off numbers need to be put in a (12 by x) matrix
I have been doing this by hand, so the next line looks like this:
And I want to vertically transpose the rest of these numbers that are currently horizontal by groups of 12 so I do not have to do it by hand.
These are the 'trailing off' numbers that need to be vertically transposed by groups of 12, so into a 12 by ?? matrix:
excel transpose
edited Nov 20 at 12:59
Luuklag
3,40482742
asked Nov 19 at 21:08
John Gustafson
11
I'm not following. You say "Matrix" and "Multidimensional Array" but those aren't the same thing. I don't know what "Trailing off numbers" means in regards to your picture. Can you share more info without the technical jargon (unless you literally mean a multidimensional array or a matrix).
– JNevill
Nov 19 at 21:15
Apologies, the trailing off numbers refer to Row 5 which continues on for +100 cells past column Q. So it currently exists as a 1 by 100-somthing matrix, and i need it to be a 12 by something matrix
– John Gustafson
Nov 19 at 21:41
You need to provide step by step explanation of what is your input, what is your desired output and what do you want to do. As it stands at the moment, your question does not make much sense.
– Michal Rosa
Nov 20 at 3:04
Does this help? I edited the original post.
– John Gustafson
Nov 20 at 12:42
Your question is too broad as it stand, but this is a typical job for VBA. You should look into the concepts ofLastColumnandLooping. A fairly simple script should then do the job.
– Luuklag
Nov 20 at 13:00
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I need to adjust my data so it aligns with the rest,
This is my data, and the trailing off numbers need to be put in a (12 by x) matrix
I have been doing this by hand, so the next line looks like this:
And I want to vertically transpose the rest of these numbers that are currently horizontal by groups of 12 so I do not have to do it by hand.
These are the 'trailing off' numbers that need to be vertically transposed by groups of 12, so into a 12 by ?? matrix:
excel transpose
edited Nov 20 at 12:59
Luuklag
3,40482742
asked Nov 19 at 21:08
John Gustafson
11
I need to adjust my data so it aligns with the rest,
This is my data, and the trailing off numbers need to be put in a (12 by x) matrix
I have been doing this by hand, so the next line looks like this:
And I want to vertically transpose the rest of these numbers that are currently horizontal by groups of 12 so I do not have to do it by hand.
These are the 'trailing off' numbers that need to be vertically transposed by groups of 12, so into a 12 by ?? matrix:
excel transpose
excel transpose
edited Nov 20 at 12:59
Luuklag
3,40482742
asked Nov 19 at 21:08
John Gustafson
11
edited Nov 20 at 12:59
Luuklag
3,40482742
asked Nov 19 at 21:08
John Gustafson
11
edited Nov 20 at 12:59
Luuklag
3,40482742
edited Nov 20 at 12:59
Luuklag
3,40482742
edited Nov 20 at 12:59
Luuklag
3,40482742
3,40482742
asked Nov 19 at 21:08
John Gustafson
11
asked Nov 19 at 21:08
John Gustafson
11
asked Nov 19 at 21:08
John Gustafson
11
11
I'm not following. You say "Matrix" and "Multidimensional Array" but those aren't the same thing. I don't know what "Trailing off numbers" means in regards to your picture. Can you share more info without the technical jargon (unless you literally mean a multidimensional array or a matrix).
– JNevill
Nov 19 at 21:15
Apologies, the trailing off numbers refer to Row 5 which continues on for +100 cells past column Q. So it currently exists as a 1 by 100-somthing matrix, and i need it to be a 12 by something matrix
– John Gustafson
Nov 19 at 21:41
You need to provide step by step explanation of what is your input, what is your desired output and what do you want to do. As it stands at the moment, your question does not make much sense.
– Michal Rosa
Nov 20 at 3:04
Does this help? I edited the original post.
– John Gustafson
Nov 20 at 12:42
Your question is too broad as it stand, but this is a typical job for VBA. You should look into the concepts ofLastColumnandLooping. A fairly simple script should then do the job.
– Luuklag
Nov 20 at 13:00
add a comment |
I'm not following. You say "Matrix" and "Multidimensional Array" but those aren't the same thing. I don't know what "Trailing off numbers" means in regards to your picture. Can you share more info without the technical jargon (unless you literally mean a multidimensional array or a matrix).
– JNevill
Nov 19 at 21:15
Apologies, the trailing off numbers refer to Row 5 which continues on for +100 cells past column Q. So it currently exists as a 1 by 100-somthing matrix, and i need it to be a 12 by something matrix
– John Gustafson
Nov 19 at 21:41
You need to provide step by step explanation of what is your input, what is your desired output and what do you want to do. As it stands at the moment, your question does not make much sense.
– Michal Rosa
Nov 20 at 3:04
Does this help? I edited the original post.
– John Gustafson
Nov 20 at 12:42
Your question is too broad as it stand, but this is a typical job for VBA. You should look into the concepts ofLastColumnandLooping. A fairly simple script should then do the job.
– Luuklag
Nov 20 at 13:00
I'm not following. You say "Matrix" and "Multidimensional Array" but those aren't the same thing. I don't know what "Trailing off numbers" means in regards to your picture. Can you share more info without the technical jargon (unless you literally mean a multidimensional array or a matrix).
– JNevill
Nov 19 at 21:15
I'm not following. You say "Matrix" and "Multidimensional Array" but those aren't the same thing. I don't know what "Trailing off numbers" means in regards to your picture. Can you share more info without the technical jargon (unless you literally mean a multidimensional array or a matrix).
– JNevill
Nov 19 at 21:15
Apologies, the trailing off numbers refer to Row 5 which continues on for +100 cells past column Q. So it currently exists as a 1 by 100-somthing matrix, and i need it to be a 12 by something matrix
– John Gustafson
Nov 19 at 21:41
Apologies, the trailing off numbers refer to Row 5 which continues on for +100 cells past column Q. So it currently exists as a 1 by 100-somthing matrix, and i need it to be a 12 by something matrix
– John Gustafson
Nov 19 at 21:41
You need to provide step by step explanation of what is your input, what is your desired output and what do you want to do. As it stands at the moment, your question does not make much sense.
– Michal Rosa
Nov 20 at 3:04
You need to provide step by step explanation of what is your input, what is your desired output and what do you want to do. As it stands at the moment, your question does not make much sense.
– Michal Rosa
Nov 20 at 3:04
Does this help? I edited the original post.
– John Gustafson
Nov 20 at 12:42
Does this help? I edited the original post.
– John Gustafson
Nov 20 at 12:42
Your question is too broad as it stand, but this is a typical job for VBA. You should look into the concepts of
LastColumn and Looping. A fairly simple script should then do the job.– Luuklag
Nov 20 at 13:00
Your question is too broad as it stand, but this is a typical job for VBA. You should look into the concepts of
LastColumn and Looping. A fairly simple script should then do the job.– Luuklag
Nov 20 at 13:00
add a comment |
1 Answer
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You can use the formula as below
I have made the groups of 6 because to show the image properly but you can change that to 12
- A3 to F3 is index you can increase it to 12
- A1 to AT1 is the series to evaluate, you will have your own
- in cell A4 the formula is
=INDEX($A$1:$AT$1,1,6*(ROW()-4)+A$3)
- In your case
6*(ROW()-4)will be 12*(row() - (row number where the formula is to be inserted)
answered Nov 20 at 16:14
usmanhaq
1,093128
add a comment |
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oldest
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up vote
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down vote
You can use the formula as below
I have made the groups of 6 because to show the image properly but you can change that to 12
- A3 to F3 is index you can increase it to 12
- A1 to AT1 is the series to evaluate, you will have your own
- in cell A4 the formula is
=INDEX($A$1:$AT$1,1,6*(ROW()-4)+A$3)
- In your case
6*(ROW()-4)will be 12*(row() - (row number where the formula is to be inserted)
answered Nov 20 at 16:14
usmanhaq
1,093128
add a comment |
up vote
0
down vote
You can use the formula as below
I have made the groups of 6 because to show the image properly but you can change that to 12
- A3 to F3 is index you can increase it to 12
- A1 to AT1 is the series to evaluate, you will have your own
- in cell A4 the formula is
=INDEX($A$1:$AT$1,1,6*(ROW()-4)+A$3)
- In your case
6*(ROW()-4)will be 12*(row() - (row number where the formula is to be inserted)
answered Nov 20 at 16:14
usmanhaq
1,093128
add a comment |
up vote
0
down vote
up vote
0
down vote
You can use the formula as below
I have made the groups of 6 because to show the image properly but you can change that to 12
- A3 to F3 is index you can increase it to 12
- A1 to AT1 is the series to evaluate, you will have your own
- in cell A4 the formula is
=INDEX($A$1:$AT$1,1,6*(ROW()-4)+A$3)
- In your case
6*(ROW()-4)will be 12*(row() - (row number where the formula is to be inserted)
answered Nov 20 at 16:14
usmanhaq
1,093128
You can use the formula as below
I have made the groups of 6 because to show the image properly but you can change that to 12
- A3 to F3 is index you can increase it to 12
- A1 to AT1 is the series to evaluate, you will have your own
- in cell A4 the formula is
=INDEX($A$1:$AT$1,1,6*(ROW()-4)+A$3)
- In your case
6*(ROW()-4)will be 12*(row() - (row number where the formula is to be inserted)
answered Nov 20 at 16:14
usmanhaq
1,093128
answered Nov 20 at 16:14
usmanhaq
1,093128
answered Nov 20 at 16:14
usmanhaq
1,093128
answered Nov 20 at 16:14
usmanhaq
1,093128
1,093128
add a comment |
add a comment |
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I'm not following. You say "Matrix" and "Multidimensional Array" but those aren't the same thing. I don't know what "Trailing off numbers" means in regards to your picture. Can you share more info without the technical jargon (unless you literally mean a multidimensional array or a matrix).
– JNevill
Nov 19 at 21:15
Apologies, the trailing off numbers refer to Row 5 which continues on for +100 cells past column Q. So it currently exists as a 1 by 100-somthing matrix, and i need it to be a 12 by something matrix
– John Gustafson
Nov 19 at 21:41
You need to provide step by step explanation of what is your input, what is your desired output and what do you want to do. As it stands at the moment, your question does not make much sense.
– Michal Rosa
Nov 20 at 3:04
Does this help? I edited the original post.
– John Gustafson
Nov 20 at 12:42
Your question is too broad as it stand, but this is a typical job for VBA. You should look into the concepts of
LastColumnandLooping. A fairly simple script should then do the job.– Luuklag
Nov 20 at 13:00