Mixing
Raise a table of pre-existing data to the power of of one of the variables
Clash Royale CLAN TAG#URR8PPP
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NOTE: I have edited this question to give a table whose first elements don't count up from 0 to 5.
I'm using a sample table here, given by Table[a/b + a^(1/2), b, 1, 6, a, 1, 6] - simply because it's small and easy. This gives me the following data:
testtable1 = 2, 2 + Sqrt[2], 3 + Sqrt[3], 6, 5 + Sqrt[5],
6 + Sqrt[6], 3/2, 1 + Sqrt[2], 3/2 + Sqrt[3], 4, 5/2 + Sqrt[5],
3 + Sqrt[6], 4/3, 2/3 + Sqrt[2], 1 + Sqrt[3], 10/3,
5/3 + Sqrt[5], 2 + Sqrt[6], 5/4, 1/2 + Sqrt[2], 3/4 + Sqrt[3], 3,
5/4 + Sqrt[5], 3/2 + Sqrt[6], 6/5, 2/5 + Sqrt[2], 3/5 + Sqrt[3],
14/5, 1 + Sqrt[5], 6/5 + Sqrt[6], 7/6, 1/3 + Sqrt[2],
1/2 + Sqrt[3], 8/3, 5/6 + Sqrt[5], 1 + Sqrt[6]
But now I want to raise that table to the power of b - without going back to create a new table. Of course, it's easy to say Table[(a/b + a^(1/2))^b, b, 1, 6, a, 1, 6] - but that's only because this is a simple table. I'm looking for an operation that I can apply generically to any testtable1, because the calculations involved in getting to my actual testtable1 are very big and very slow, and hit the error limits of Mathematica - i.e., they become inaccurate. So, I want to crunch the data I have already generated rather than modify the original calculation.
Is this possible? Maybe I need to turn testtable1 into a dataset? Pointers on how to tackle this would be much appreciated.
table data
asked Aug 29 at 8:31
Richard Burke-Ward
1827
add a comment |Â
up vote
3
down vote
favorite
NOTE: I have edited this question to give a table whose first elements don't count up from 0 to 5.
I'm using a sample table here, given by Table[a/b + a^(1/2), b, 1, 6, a, 1, 6] - simply because it's small and easy. This gives me the following data:
testtable1 = 2, 2 + Sqrt[2], 3 + Sqrt[3], 6, 5 + Sqrt[5],
6 + Sqrt[6], 3/2, 1 + Sqrt[2], 3/2 + Sqrt[3], 4, 5/2 + Sqrt[5],
3 + Sqrt[6], 4/3, 2/3 + Sqrt[2], 1 + Sqrt[3], 10/3,
5/3 + Sqrt[5], 2 + Sqrt[6], 5/4, 1/2 + Sqrt[2], 3/4 + Sqrt[3], 3,
5/4 + Sqrt[5], 3/2 + Sqrt[6], 6/5, 2/5 + Sqrt[2], 3/5 + Sqrt[3],
14/5, 1 + Sqrt[5], 6/5 + Sqrt[6], 7/6, 1/3 + Sqrt[2],
1/2 + Sqrt[3], 8/3, 5/6 + Sqrt[5], 1 + Sqrt[6]
But now I want to raise that table to the power of b - without going back to create a new table. Of course, it's easy to say Table[(a/b + a^(1/2))^b, b, 1, 6, a, 1, 6] - but that's only because this is a simple table. I'm looking for an operation that I can apply generically to any testtable1, because the calculations involved in getting to my actual testtable1 are very big and very slow, and hit the error limits of Mathematica - i.e., they become inaccurate. So, I want to crunch the data I have already generated rather than modify the original calculation.
Is this possible? Maybe I need to turn testtable1 into a dataset? Pointers on how to tackle this would be much appreciated.
table data
asked Aug 29 at 8:31
Richard Burke-Ward
1827
Do you prefer exact computations or is it okay to use (inexact) floating point numbers? The latter would speed up your calculations tremendously.
– Henrik Schumacher
Aug 29 at 9:07
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
NOTE: I have edited this question to give a table whose first elements don't count up from 0 to 5.
I'm using a sample table here, given by Table[a/b + a^(1/2), b, 1, 6, a, 1, 6] - simply because it's small and easy. This gives me the following data:
testtable1 = 2, 2 + Sqrt[2], 3 + Sqrt[3], 6, 5 + Sqrt[5],
6 + Sqrt[6], 3/2, 1 + Sqrt[2], 3/2 + Sqrt[3], 4, 5/2 + Sqrt[5],
3 + Sqrt[6], 4/3, 2/3 + Sqrt[2], 1 + Sqrt[3], 10/3,
5/3 + Sqrt[5], 2 + Sqrt[6], 5/4, 1/2 + Sqrt[2], 3/4 + Sqrt[3], 3,
5/4 + Sqrt[5], 3/2 + Sqrt[6], 6/5, 2/5 + Sqrt[2], 3/5 + Sqrt[3],
14/5, 1 + Sqrt[5], 6/5 + Sqrt[6], 7/6, 1/3 + Sqrt[2],
1/2 + Sqrt[3], 8/3, 5/6 + Sqrt[5], 1 + Sqrt[6]
But now I want to raise that table to the power of b - without going back to create a new table. Of course, it's easy to say Table[(a/b + a^(1/2))^b, b, 1, 6, a, 1, 6] - but that's only because this is a simple table. I'm looking for an operation that I can apply generically to any testtable1, because the calculations involved in getting to my actual testtable1 are very big and very slow, and hit the error limits of Mathematica - i.e., they become inaccurate. So, I want to crunch the data I have already generated rather than modify the original calculation.
Is this possible? Maybe I need to turn testtable1 into a dataset? Pointers on how to tackle this would be much appreciated.
table data
asked Aug 29 at 8:31
Richard Burke-Ward
1827
NOTE: I have edited this question to give a table whose first elements don't count up from 0 to 5.
I'm using a sample table here, given by Table[a/b + a^(1/2), b, 1, 6, a, 1, 6] - simply because it's small and easy. This gives me the following data:
testtable1 = 2, 2 + Sqrt[2], 3 + Sqrt[3], 6, 5 + Sqrt[5],
6 + Sqrt[6], 3/2, 1 + Sqrt[2], 3/2 + Sqrt[3], 4, 5/2 + Sqrt[5],
3 + Sqrt[6], 4/3, 2/3 + Sqrt[2], 1 + Sqrt[3], 10/3,
5/3 + Sqrt[5], 2 + Sqrt[6], 5/4, 1/2 + Sqrt[2], 3/4 + Sqrt[3], 3,
5/4 + Sqrt[5], 3/2 + Sqrt[6], 6/5, 2/5 + Sqrt[2], 3/5 + Sqrt[3],
14/5, 1 + Sqrt[5], 6/5 + Sqrt[6], 7/6, 1/3 + Sqrt[2],
1/2 + Sqrt[3], 8/3, 5/6 + Sqrt[5], 1 + Sqrt[6]
But now I want to raise that table to the power of b - without going back to create a new table. Of course, it's easy to say Table[(a/b + a^(1/2))^b, b, 1, 6, a, 1, 6] - but that's only because this is a simple table. I'm looking for an operation that I can apply generically to any testtable1, because the calculations involved in getting to my actual testtable1 are very big and very slow, and hit the error limits of Mathematica - i.e., they become inaccurate. So, I want to crunch the data I have already generated rather than modify the original calculation.
Is this possible? Maybe I need to turn testtable1 into a dataset? Pointers on how to tackle this would be much appreciated.
table data
asked Aug 29 at 8:31
Richard Burke-Ward
1827
edited Aug 29 at 9:03
asked Aug 29 at 8:31
Richard Burke-Ward
1827
asked Aug 29 at 8:31
Richard Burke-Ward
1827
asked Aug 29 at 8:31
Richard Burke-Ward
1827
1827
Do you prefer exact computations or is it okay to use (inexact) floating point numbers? The latter would speed up your calculations tremendously.
– Henrik Schumacher
Aug 29 at 9:07
add a comment |Â
Do you prefer exact computations or is it okay to use (inexact) floating point numbers? The latter would speed up your calculations tremendously.
– Henrik Schumacher
Aug 29 at 9:07
Do you prefer exact computations or is it okay to use (inexact) floating point numbers? The latter would speed up your calculations tremendously.
– Henrik Schumacher
Aug 29 at 9:07
Do you prefer exact computations or is it okay to use (inexact) floating point numbers? The latter would speed up your calculations tremendously.
– Henrik Schumacher
Aug 29 at 9:07
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Suppose you generate you table equivalently by
alist = Range[1, 6];
blist = Range[1, 6];
testtable1 = Table[a/b + a^(1/2), b, blist, a, alist];
Then
Table[(a/b + a^(1/2))^b, b, blist, a, alist]
can be obtained also by
testtable1^blist
and
Table[(a/b + a^(1/2))^a, b, blist, a, alist]
can be obtained by
testtable1^ConstantArray[alist, Length[blist]]
The key observation is that ^ (a.k.a. Power) has the attribute Listable.
For a preexisting table
testtable2 = RandomReal[-1, 1, 1000, 2000];
the following should raise each row to the power of its row count:
poweredbyrow = testtable2^Range[1, Length[testtable2]];
The same for powering by column number:
poweredbycol =
testtable2^ConstantArray[
Range[1, Dimensions[testtable2][[2]]],
Length[testtable2]
];
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from0to5- in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.
– Richard Burke-Ward
Aug 29 at 8:58
Well, if you generate your table bya = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can usetesttable1^b;. Btw.: Using the listaas powers can be done as follows:testtable1^ConstantArray[a, Length[b]];
– Henrik Schumacher
Aug 29 at 9:01
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
 |Â
show 5 more comments
up vote
3
down vote
row $k$ raised to the power $k$:
MapIndexed[#^#2[[1]] &, testtable1, 1] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & 2+sqrt2 & 3+sqrt3 & 6 & 5+sqrt5 & 6+sqrt6 \
frac94 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^2 & 16 & left(frac52+sqrt5right)^2 &
left(3+sqrt6right)^2 \
frac6427 & left(frac23+sqrt2right)^3 & left(1+sqrt3right)^3 & frac100027 &
left(frac53+sqrt5right)^3 & left(2+sqrt6right)^3 \
frac625256 & left(frac12+sqrt2right)^4 & left(frac34+sqrt3right)^4 & 81 &
left(frac54+sqrt5right)^4 & left(frac32+sqrt6right)^4 \
frac77763125 & left(frac25+sqrt2right)^5 & left(frac35+sqrt3right)^5 & frac5378243125 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^5 \
frac11764946656 & left(frac13+sqrt2right)^6 & left(frac12+sqrt3right)^6 & frac262144729 &
left(frac56+sqrt5right)^6 & left(1+sqrt6right)^6 \
endarray
right)$
column $k$ raised to the power $k$:
MapIndexed[#^#2[[2]] &, testtable1, 2] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & left(2+sqrt2right)^2 & left(3+sqrt3right)^3 & 1296 & left(5+sqrt5right)^5 & left(6+sqrt6right)^6
\
frac32 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^3 & 256 & left(frac52+sqrt5right)^5
& left(3+sqrt6right)^6 \
frac43 & left(frac23+sqrt2right)^2 & left(1+sqrt3right)^3 & frac1000081 &
left(frac53+sqrt5right)^5 & left(2+sqrt6right)^6 \
frac54 & left(frac12+sqrt2right)^2 & left(frac34+sqrt3right)^3 & 81 &
left(frac54+sqrt5right)^5 & left(frac32+sqrt6right)^6 \
frac65 & left(frac25+sqrt2right)^2 & left(frac35+sqrt3right)^3 & frac38416625 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^6 \
frac76 & left(frac13+sqrt2right)^2 & left(frac12+sqrt3right)^3 & frac409681 &
left(frac56+sqrt5right)^5 & left(1+sqrt6right)^6 \
endarray
right) $
Also:
Transpose @ MapIndexed[#^#2[[1]] &, Transpose @ testtable1, 1]
same result
answered Aug 29 at 8:42
kglr
158k8183382
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Suppose you generate you table equivalently by
alist = Range[1, 6];
blist = Range[1, 6];
testtable1 = Table[a/b + a^(1/2), b, blist, a, alist];
Then
Table[(a/b + a^(1/2))^b, b, blist, a, alist]
can be obtained also by
testtable1^blist
and
Table[(a/b + a^(1/2))^a, b, blist, a, alist]
can be obtained by
testtable1^ConstantArray[alist, Length[blist]]
The key observation is that ^ (a.k.a. Power) has the attribute Listable.
For a preexisting table
testtable2 = RandomReal[-1, 1, 1000, 2000];
the following should raise each row to the power of its row count:
poweredbyrow = testtable2^Range[1, Length[testtable2]];
The same for powering by column number:
poweredbycol =
testtable2^ConstantArray[
Range[1, Dimensions[testtable2][[2]]],
Length[testtable2]
];
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from0to5- in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.
– Richard Burke-Ward
Aug 29 at 8:58
Well, if you generate your table bya = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can usetesttable1^b;. Btw.: Using the listaas powers can be done as follows:testtable1^ConstantArray[a, Length[b]];
– Henrik Schumacher
Aug 29 at 9:01
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
 |Â
show 5 more comments
up vote
2
down vote
accepted
Suppose you generate you table equivalently by
alist = Range[1, 6];
blist = Range[1, 6];
testtable1 = Table[a/b + a^(1/2), b, blist, a, alist];
Then
Table[(a/b + a^(1/2))^b, b, blist, a, alist]
can be obtained also by
testtable1^blist
and
Table[(a/b + a^(1/2))^a, b, blist, a, alist]
can be obtained by
testtable1^ConstantArray[alist, Length[blist]]
The key observation is that ^ (a.k.a. Power) has the attribute Listable.
For a preexisting table
testtable2 = RandomReal[-1, 1, 1000, 2000];
the following should raise each row to the power of its row count:
poweredbyrow = testtable2^Range[1, Length[testtable2]];
The same for powering by column number:
poweredbycol =
testtable2^ConstantArray[
Range[1, Dimensions[testtable2][[2]]],
Length[testtable2]
];
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from0to5- in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.
– Richard Burke-Ward
Aug 29 at 8:58
Well, if you generate your table bya = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can usetesttable1^b;. Btw.: Using the listaas powers can be done as follows:testtable1^ConstantArray[a, Length[b]];
– Henrik Schumacher
Aug 29 at 9:01
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
 |Â
show 5 more comments
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Suppose you generate you table equivalently by
alist = Range[1, 6];
blist = Range[1, 6];
testtable1 = Table[a/b + a^(1/2), b, blist, a, alist];
Then
Table[(a/b + a^(1/2))^b, b, blist, a, alist]
can be obtained also by
testtable1^blist
and
Table[(a/b + a^(1/2))^a, b, blist, a, alist]
can be obtained by
testtable1^ConstantArray[alist, Length[blist]]
The key observation is that ^ (a.k.a. Power) has the attribute Listable.
For a preexisting table
testtable2 = RandomReal[-1, 1, 1000, 2000];
the following should raise each row to the power of its row count:
poweredbyrow = testtable2^Range[1, Length[testtable2]];
The same for powering by column number:
poweredbycol =
testtable2^ConstantArray[
Range[1, Dimensions[testtable2][[2]]],
Length[testtable2]
];
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
Suppose you generate you table equivalently by
alist = Range[1, 6];
blist = Range[1, 6];
testtable1 = Table[a/b + a^(1/2), b, blist, a, alist];
Then
Table[(a/b + a^(1/2))^b, b, blist, a, alist]
can be obtained also by
testtable1^blist
and
Table[(a/b + a^(1/2))^a, b, blist, a, alist]
can be obtained by
testtable1^ConstantArray[alist, Length[blist]]
The key observation is that ^ (a.k.a. Power) has the attribute Listable.
For a preexisting table
testtable2 = RandomReal[-1, 1, 1000, 2000];
the following should raise each row to the power of its row count:
poweredbyrow = testtable2^Range[1, Length[testtable2]];
The same for powering by column number:
poweredbycol =
testtable2^ConstantArray[
Range[1, Dimensions[testtable2][[2]]],
Length[testtable2]
];
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
edited Aug 29 at 9:10
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
answered Aug 29 at 8:43
Henrik Schumacher
36.7k249103
36.7k249103
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from0to5- in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.
– Richard Burke-Ward
Aug 29 at 8:58
Well, if you generate your table bya = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can usetesttable1^b;. Btw.: Using the listaas powers can be done as follows:testtable1^ConstantArray[a, Length[b]];
– Henrik Schumacher
Aug 29 at 9:01
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
 |Â
show 5 more comments
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from0to5- in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.
– Richard Burke-Ward
Aug 29 at 8:58
Well, if you generate your table bya = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can usetesttable1^b;. Btw.: Using the listaas powers can be done as follows:testtable1^ConstantArray[a, Length[b]];
– Henrik Schumacher
Aug 29 at 9:01
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from
0 to 5 - in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.– Richard Burke-Ward
Aug 29 at 8:58
Correct me if I'm wrong, but this solution essentially relies on the fact that the first item in each entry in the table counts up from
0 to 5 - in which case, I gave a poor example, because that's not necessarily the case... I'll edit the original post.– Richard Burke-Ward
Aug 29 at 8:58
Well, if you generate your table by
a = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can use testtable1^b;. Btw.: Using the list a as powers can be done as follows: testtable1^ConstantArray[a, Length[b]];– Henrik Schumacher
Aug 29 at 9:01
Well, if you generate your table by
a = RandomReal[-1, 1, 1000]; b = RandomReal[-1, 1, 2000]; testtable1 = Outer[Plus, b, a];, then you can use testtable1^b;. Btw.: Using the list a as powers can be done as follows: testtable1^ConstantArray[a, Length[b]];– Henrik Schumacher
Aug 29 at 9:01
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Hi Henrik (again!). Thanks for helping. This is really useful, but again it's dependent on the specific nature of the table. My situation is that I have a very big pre-existing array/table of data, defined only by its name. I want to raise each row to the power of the number of that row. Row 5, raise to the power of 5; row 2, raise to the power of 2...
– Richard Burke-Ward
Aug 29 at 9:09
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Dear Richard, always at your service ;) I exanded my answer. Does it help you?
– Henrik Schumacher
Aug 29 at 9:15
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
Thanks again, Henrik. It's going to take me a bit of time to wade through that. I'm sure it's exactly right, but I need to go all the way back to my original calculations, and figure out each step of your cunning plan :-). I'll mark as answered after that, but would you mind checking back in, say, half an hour just to see if I have any further questions? Really appreciate your help.
– Richard Burke-Ward
Aug 29 at 9:18
 |Â
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up vote
3
down vote
row $k$ raised to the power $k$:
MapIndexed[#^#2[[1]] &, testtable1, 1] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & 2+sqrt2 & 3+sqrt3 & 6 & 5+sqrt5 & 6+sqrt6 \
frac94 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^2 & 16 & left(frac52+sqrt5right)^2 &
left(3+sqrt6right)^2 \
frac6427 & left(frac23+sqrt2right)^3 & left(1+sqrt3right)^3 & frac100027 &
left(frac53+sqrt5right)^3 & left(2+sqrt6right)^3 \
frac625256 & left(frac12+sqrt2right)^4 & left(frac34+sqrt3right)^4 & 81 &
left(frac54+sqrt5right)^4 & left(frac32+sqrt6right)^4 \
frac77763125 & left(frac25+sqrt2right)^5 & left(frac35+sqrt3right)^5 & frac5378243125 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^5 \
frac11764946656 & left(frac13+sqrt2right)^6 & left(frac12+sqrt3right)^6 & frac262144729 &
left(frac56+sqrt5right)^6 & left(1+sqrt6right)^6 \
endarray
right)$
column $k$ raised to the power $k$:
MapIndexed[#^#2[[2]] &, testtable1, 2] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & left(2+sqrt2right)^2 & left(3+sqrt3right)^3 & 1296 & left(5+sqrt5right)^5 & left(6+sqrt6right)^6
\
frac32 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^3 & 256 & left(frac52+sqrt5right)^5
& left(3+sqrt6right)^6 \
frac43 & left(frac23+sqrt2right)^2 & left(1+sqrt3right)^3 & frac1000081 &
left(frac53+sqrt5right)^5 & left(2+sqrt6right)^6 \
frac54 & left(frac12+sqrt2right)^2 & left(frac34+sqrt3right)^3 & 81 &
left(frac54+sqrt5right)^5 & left(frac32+sqrt6right)^6 \
frac65 & left(frac25+sqrt2right)^2 & left(frac35+sqrt3right)^3 & frac38416625 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^6 \
frac76 & left(frac13+sqrt2right)^2 & left(frac12+sqrt3right)^3 & frac409681 &
left(frac56+sqrt5right)^5 & left(1+sqrt6right)^6 \
endarray
right) $
Also:
Transpose @ MapIndexed[#^#2[[1]] &, Transpose @ testtable1, 1]
same result
answered Aug 29 at 8:42
kglr
158k8183382
add a comment |Â
up vote
3
down vote
row $k$ raised to the power $k$:
MapIndexed[#^#2[[1]] &, testtable1, 1] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & 2+sqrt2 & 3+sqrt3 & 6 & 5+sqrt5 & 6+sqrt6 \
frac94 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^2 & 16 & left(frac52+sqrt5right)^2 &
left(3+sqrt6right)^2 \
frac6427 & left(frac23+sqrt2right)^3 & left(1+sqrt3right)^3 & frac100027 &
left(frac53+sqrt5right)^3 & left(2+sqrt6right)^3 \
frac625256 & left(frac12+sqrt2right)^4 & left(frac34+sqrt3right)^4 & 81 &
left(frac54+sqrt5right)^4 & left(frac32+sqrt6right)^4 \
frac77763125 & left(frac25+sqrt2right)^5 & left(frac35+sqrt3right)^5 & frac5378243125 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^5 \
frac11764946656 & left(frac13+sqrt2right)^6 & left(frac12+sqrt3right)^6 & frac262144729 &
left(frac56+sqrt5right)^6 & left(1+sqrt6right)^6 \
endarray
right)$
column $k$ raised to the power $k$:
MapIndexed[#^#2[[2]] &, testtable1, 2] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & left(2+sqrt2right)^2 & left(3+sqrt3right)^3 & 1296 & left(5+sqrt5right)^5 & left(6+sqrt6right)^6
\
frac32 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^3 & 256 & left(frac52+sqrt5right)^5
& left(3+sqrt6right)^6 \
frac43 & left(frac23+sqrt2right)^2 & left(1+sqrt3right)^3 & frac1000081 &
left(frac53+sqrt5right)^5 & left(2+sqrt6right)^6 \
frac54 & left(frac12+sqrt2right)^2 & left(frac34+sqrt3right)^3 & 81 &
left(frac54+sqrt5right)^5 & left(frac32+sqrt6right)^6 \
frac65 & left(frac25+sqrt2right)^2 & left(frac35+sqrt3right)^3 & frac38416625 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^6 \
frac76 & left(frac13+sqrt2right)^2 & left(frac12+sqrt3right)^3 & frac409681 &
left(frac56+sqrt5right)^5 & left(1+sqrt6right)^6 \
endarray
right) $
Also:
Transpose @ MapIndexed[#^#2[[1]] &, Transpose @ testtable1, 1]
same result
answered Aug 29 at 8:42
kglr
158k8183382
add a comment |Â
up vote
3
down vote
up vote
3
down vote
row $k$ raised to the power $k$:
MapIndexed[#^#2[[1]] &, testtable1, 1] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & 2+sqrt2 & 3+sqrt3 & 6 & 5+sqrt5 & 6+sqrt6 \
frac94 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^2 & 16 & left(frac52+sqrt5right)^2 &
left(3+sqrt6right)^2 \
frac6427 & left(frac23+sqrt2right)^3 & left(1+sqrt3right)^3 & frac100027 &
left(frac53+sqrt5right)^3 & left(2+sqrt6right)^3 \
frac625256 & left(frac12+sqrt2right)^4 & left(frac34+sqrt3right)^4 & 81 &
left(frac54+sqrt5right)^4 & left(frac32+sqrt6right)^4 \
frac77763125 & left(frac25+sqrt2right)^5 & left(frac35+sqrt3right)^5 & frac5378243125 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^5 \
frac11764946656 & left(frac13+sqrt2right)^6 & left(frac12+sqrt3right)^6 & frac262144729 &
left(frac56+sqrt5right)^6 & left(1+sqrt6right)^6 \
endarray
right)$
column $k$ raised to the power $k$:
MapIndexed[#^#2[[2]] &, testtable1, 2] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & left(2+sqrt2right)^2 & left(3+sqrt3right)^3 & 1296 & left(5+sqrt5right)^5 & left(6+sqrt6right)^6
\
frac32 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^3 & 256 & left(frac52+sqrt5right)^5
& left(3+sqrt6right)^6 \
frac43 & left(frac23+sqrt2right)^2 & left(1+sqrt3right)^3 & frac1000081 &
left(frac53+sqrt5right)^5 & left(2+sqrt6right)^6 \
frac54 & left(frac12+sqrt2right)^2 & left(frac34+sqrt3right)^3 & 81 &
left(frac54+sqrt5right)^5 & left(frac32+sqrt6right)^6 \
frac65 & left(frac25+sqrt2right)^2 & left(frac35+sqrt3right)^3 & frac38416625 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^6 \
frac76 & left(frac13+sqrt2right)^2 & left(frac12+sqrt3right)^3 & frac409681 &
left(frac56+sqrt5right)^5 & left(1+sqrt6right)^6 \
endarray
right) $
Also:
Transpose @ MapIndexed[#^#2[[1]] &, Transpose @ testtable1, 1]
same result
answered Aug 29 at 8:42
kglr
158k8183382
row $k$ raised to the power $k$:
MapIndexed[#^#2[[1]] &, testtable1, 1] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & 2+sqrt2 & 3+sqrt3 & 6 & 5+sqrt5 & 6+sqrt6 \
frac94 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^2 & 16 & left(frac52+sqrt5right)^2 &
left(3+sqrt6right)^2 \
frac6427 & left(frac23+sqrt2right)^3 & left(1+sqrt3right)^3 & frac100027 &
left(frac53+sqrt5right)^3 & left(2+sqrt6right)^3 \
frac625256 & left(frac12+sqrt2right)^4 & left(frac34+sqrt3right)^4 & 81 &
left(frac54+sqrt5right)^4 & left(frac32+sqrt6right)^4 \
frac77763125 & left(frac25+sqrt2right)^5 & left(frac35+sqrt3right)^5 & frac5378243125 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^5 \
frac11764946656 & left(frac13+sqrt2right)^6 & left(frac12+sqrt3right)^6 & frac262144729 &
left(frac56+sqrt5right)^6 & left(1+sqrt6right)^6 \
endarray
right)$
column $k$ raised to the power $k$:
MapIndexed[#^#2[[2]] &, testtable1, 2] // MatrixForm // TeXForm
$left(
beginarraycccccc
2 & left(2+sqrt2right)^2 & left(3+sqrt3right)^3 & 1296 & left(5+sqrt5right)^5 & left(6+sqrt6right)^6
\
frac32 & left(1+sqrt2right)^2 & left(frac32+sqrt3right)^3 & 256 & left(frac52+sqrt5right)^5
& left(3+sqrt6right)^6 \
frac43 & left(frac23+sqrt2right)^2 & left(1+sqrt3right)^3 & frac1000081 &
left(frac53+sqrt5right)^5 & left(2+sqrt6right)^6 \
frac54 & left(frac12+sqrt2right)^2 & left(frac34+sqrt3right)^3 & 81 &
left(frac54+sqrt5right)^5 & left(frac32+sqrt6right)^6 \
frac65 & left(frac25+sqrt2right)^2 & left(frac35+sqrt3right)^3 & frac38416625 &
left(1+sqrt5right)^5 & left(frac65+sqrt6right)^6 \
frac76 & left(frac13+sqrt2right)^2 & left(frac12+sqrt3right)^3 & frac409681 &
left(frac56+sqrt5right)^5 & left(1+sqrt6right)^6 \
endarray
right) $
Also:
Transpose @ MapIndexed[#^#2[[1]] &, Transpose @ testtable1, 1]
same result
answered Aug 29 at 8:42
kglr
158k8183382
edited Aug 29 at 9:30
answered Aug 29 at 8:42
kglr
158k8183382
answered Aug 29 at 8:42
kglr
158k8183382
answered Aug 29 at 8:42
kglr
158k8183382
158k8183382
add a comment |Â
add a comment |Â
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Do you prefer exact computations or is it okay to use (inexact) floating point numbers? The latter would speed up your calculations tremendously.
– Henrik Schumacher
Aug 29 at 9:07