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Jun 16, 2020 at 9:23 history edited CommunityBot
Commonmark migration
Jun 24, 2015 at 14:42 vote accept anderstood
Jun 24, 2015 at 7:04 answer added Mr.Wizard timeline score: 6
Jun 24, 2015 at 6:43 history edited Mr.Wizard CC BY-SA 3.0
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Jun 6, 2015 at 11:44 answer added Albert Retey timeline score: 5
Jun 6, 2015 at 5:11 comment added Kattern Yes, quite strange. Maybe you can substitute the variable like ReleaseHold[Hold[ParallelTable[a[1], {a[1], 0, 10}]] /. a[1] -> i] first. Let wait to see whether there is a better solution.
Jun 6, 2015 at 5:05 comment added anderstood @kattern: Well... changing Table to ParallelTable works in this case. I still don't get it, why does ParallelTable[a[1], {a[1], 0, 10}]not work?
Jun 6, 2015 at 5:02 comment added anderstood @kattern Try this standalone code (which works): With[{iter = Sequence @@ Table[{gag[j], -1, 1}, {j, 1, 3}]}, Table[f[Table[gag[i], {i, 1, 3}]], iter]]!. How can I make it work with ParallelTableinstead of Table?
Jun 6, 2015 at 5:01 comment added Kattern Still do not understand, can you please write an executable Table version of what you what?
Jun 6, 2015 at 4:56 comment added anderstood @kattern: Yes, that's my question: why does it work with Table but not with ParallelTable. With[{iter = Sequence @@ Table[{a[j],-1, n}, {j, 1, n}]}, Table[f[Table[a[i], {i, 1, n}]], iter]] if you prefer...
Jun 6, 2015 at 4:54 comment added Kattern It appears ParallelTable[a[1], {a[1], 0, 10}] not working with variable like a[1], since ParallelTable[f[i], {i, 0, 10}] does work. `ParallelTable[f[Table[a[i],{i,1,n}]],Table[a[j],{j,1,n}]] does not make sense.
Jun 6, 2015 at 2:01 history edited anderstood CC BY-SA 3.0
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Jun 6, 2015 at 1:56 history asked anderstood CC BY-SA 3.0