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I have a table given by

example1=Table[N[x^(x/r), 5],{r, 1, 11},{x, 1, 10}];
example2=Accumulate[example1];
example2

I would like to use example2 to create a new table that uses example2's data, but only shows every nth row of the example2 table - in this case (for example), every 2nd, 3rd or 4th row.

How do I do this?

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  • 1
    $\begingroup$ Don't assign something in TableForm to a variable. TableForm is meant for display only, and it cannot be used for computing (without removing it first). Thus, use example2 = Accumulate[...];, then use example2 // TableForm for display or example2[[;; ;; 3]] to take every third row. Look up these in the docs: Part, Span, TableForm. $\endgroup$
    – Szabolcs
    Commented Aug 26, 2018 at 12:30
  • $\begingroup$ OK, so let's lose the TableForm. Point taken. Question now edited. But I'm new enough to all this that, whilst I can see that in principle Span and Part could give me what I want, I'm not entirely sure how to put it into practice... $\endgroup$ Commented Aug 26, 2018 at 12:39
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    $\begingroup$ But I showed you the exact code you need in my first comment—[[ is Part and ;; is Span. Here it is again: example2[[;; ;; 3]] to take every third row. $\endgroup$
    – Szabolcs
    Commented Aug 26, 2018 at 12:43

1 Answer 1

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example2[[;;;;2]] 

{{1.0000, 4.0000, 27.000, 256.00, 3125.0, 46656., 8.2354*10^5, 1.6777*10^7, 3.8742*10^8, 1.0000*10^10},
{3.0000, 7.5874, 35.196, 278.35, 3195.5, 46908., 8.2454*10^5, 1.6782*10^7, 3.8744*10^8, 1.0000*10^10},
{5.0000, 10.321, 39.409, 285.38, 3208.0, 46931., 8.2459*10^5, 1.6782*10^7, 3.8744*10^8, 1.0000*10^10},
{7.0000, 12.800, 42.742, 290.11, 3215.0, 46942., 8.2461*10^5, 1.6782*10^7, 3.8744*10^8, 1.0000*10^10},
{9.0000, 15.156, 45.694, 293.96, 3220.2, 46949., 8.2462*10^5, 1.6782*10^7, 3.8744*10^8, 1.0000*10^10}}

Alternatively, you can Drop every other row starting with the second row:

Drop[example2, {2, -1, 2}]

same result

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  • $\begingroup$ OK, got it now! Much appreciated :-) $\endgroup$ Commented Aug 26, 2018 at 12:43
  • $\begingroup$ @Richard, my pleasure. $\endgroup$
    – kglr
    Commented Aug 26, 2018 at 12:43

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