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I would like to generate a table using a function of two variables. I tried the following:

Table[{a, b, f[a, b]}, {a, 0, 5}, {b, 0, 5}] 

which does work as I had hoped. I want the table to produce a, b, f[a, b] in columns using Grid. Thanks for suggestions!

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  • $\begingroup$ Table[{a, b, f[a, b]}, {a, 0, 5}, {b, 0, 5}] // Grid ? Table[{a, b, f[a, b]}, {a, 0, 5}, {b, 0, 5}] // TableForm ? $\endgroup$ – murray Dec 10 '16 at 22:43
  • $\begingroup$ Thank you, these work, but table iterates a1 b1, F[a1,b1] then a1,b2,F[a1,b2] etc I would like a1,b1,F[a1,b1] then a2,b2,F[a2,b2]. Can I do this? $\endgroup$ – CMoller Dec 11 '16 at 0:02
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It is not clear how many columns that you want. Some of the possibilities

data = Table[{a, b, f[a, b]}, {a, 0, 5}, {b, 0, 5}];

data // Flatten[#, 1] & // Grid

enter image description here

Row[Riffle[Grid /@ Partition[data // Flatten[#, 1] &, 18], Spacer[25]]]

enter image description here

Row[Riffle[Grid /@ Partition[data // Flatten[#, 1] &, 12], 
  Spacer[25]]]

enter image description here

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  • $\begingroup$ Sorry I'm not used to this site and did not realize that Bob Hanion has given me a good answer. Thank you! $\endgroup$ – CMoller Dec 11 '16 at 0:09
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    $\begingroup$ @user3429864: I'm confused - you accepted an answer which gives a result different from what you said you wanted in your comment responding to mine on your original post. By contrast, corey979's answer does give what you said in that comment. $\endgroup$ – murray Dec 12 '16 at 1:44
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Judging from your comment, you want this:

MapThread[{#1, #2, f[#1, #2]} &, {{1, 3, 5, 7, 9}, {2, 4, 6, 8, 10}}] // Grid

enter image description here


Or if the a's and b's take the same corresponding values:

Table[{a, a, f[a, a]}, {a, 1, 5}] // Grid

enter image description here

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