I want to make a table with following data,
$a_{ij} = b_{ij}$ for $1\le i <j\le n-2$ and $|i-j| \ge 2$.
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2 Answers
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7
Try
n = 7;
Table[If[Abs[i - j] > 2, a[i, j] == b[i, j], Nothing], {i, 1,n - 2}, {j, i+1, n - 2}] // Flatten
(*{a[1, 4] == b[1, 4], a[1, 5] == b[1, 5], a[2, 5] == b[2, 5],a[4, 1] == b[4, 1], a[5, 1] == b[5, 1], a[5, 2] == b[5, 2]}*)
answered Jul 25, 2019 at 13:07
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$\begingroup$ Thanks! How to get only non-zero entries of the table? Otherwise it looks a bit clumsy. $\endgroup$ Commented Jul 25, 2019 at 13:18
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$\begingroup$ Without knowing
a,bit's difficult to distinguish non-zero entries! $\endgroup$ Commented Jul 25, 2019 at 13:22 -
$\begingroup$ I just meant the trivial
0orNullentries that are appearing without referring toaorb.. $\endgroup$ Commented Jul 25, 2019 at 13:33 -
1$\begingroup$
Nothinginside If is what you're looking for. My answer is edited. $\endgroup$ Commented Jul 25, 2019 at 13:39 -
$\begingroup$ Thanks! This is what I was looking for and I understand it too. $\endgroup$ Commented Jul 25, 2019 at 13:42
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3
Try
n = 7;
t=Table[a[i, j] == b[i, j], {j, 1,n - 2}, {i, 1,j - 2}]/.{}->Sequence[]
giving
{{a[1,3]==b[1,3]},
{a[1,4]==b[1,4],a[2,4]==b[2,4]},
{a[1,5]==b[1,5],a[2,5]==b[2,5],a[3,5]==b[3,5]}}
And you can use Flatten on that if you don't want the extra {}
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$\begingroup$ This is great! But I have no clue how it's working! Can you explain that
{}->Sequence[]part? Also can I generalize this to different conditions, say,|i-j| >= 5or something.. $\endgroup$ Commented Jul 25, 2019 at 13:31 -
1$\begingroup$ If you try that without the
/.{}->Sequence[]you will see that it leaves in "empty" rows where none of the items match both your conditions, so I used that "trick" to discard empty rows. For your generalized conditions tryn = 12; t=Table[ a[i, j] == b[i, j], {j, 1,n - 2}, {i, 1,j - 5}]/.{}->Sequence[]$\endgroup$– BillCommented Jul 25, 2019 at 13:36 -
$\begingroup$ Yes, that's perfect. That should be imposed. $\endgroup$ Commented Jul 25, 2019 at 13:40