8
Since you want to test the last element of each {x, y, z, a1[x]}, I'd modify C.E.'s suggestion in the comments to make sure that cases where Pi/2 appears at any other locations don't get thrown away:
DeleteCases[t, {_, _, _, Pi/2}, {3}] /. {} -> Nothing
This pattern first throws away the structure you don't want, and then the replacement drops the ...
4
I would suggest not to flatten the table, since the relevant information is contained in its formatting and can be extracted by Part and Total.
T = ParallelTable[{x, y, z, k, fff[x, y, z, k]}, {x, 0, 1, 0.1}, {y, 0.5, 5.5, 1}, {z, 0.2, 2, 0.2}, {k, 0.2, 1, 0.2}];
fvalues = T[[All, All, All, All, 5]];
xy = T[[All, All, 1, 1, 1 ;; 2]];
averages = Total[fvalues,...
3
You may use Query with GroupBy.
assoc =
Query[
GroupBy[#[[;; 2]] &]
, Mean@#[[All, -1]] &
]@tabb;
Short@assoc
<|{0., 0.5} -> 2.074*10^12, <<64>> , {1., 5.5} -> 0.00023595|>
This gives the result as an Association with the keys the x-y pairs.
assoc[{0., 0.5}]
2.074*10^12
However, this can be easily converted ...
3
Lot's of ways to achieve this. A few follow:
list1 = {"list1", 1/x, -1 + x, -(x/(-2 + x)), -1 + 1/x, -1 + 2 x};
list2 = {"list2", {{a, b}}, {{c, d}}, {{a, c}, {b, d}}, {{e, f}, {g, h}}, {{k, t}}};
list3 = {"No.", 1, 2, 3, 4, 5};
Grid[Transpose[{list3, list2, list1}], Frame -> All]
This gets a bit closer:
list1 = {"...
2
As mentioned by @BobHanlon in the comments, the solution is to use RuleDelayed in the following way:
tr[a^2]^5 /. tr[a^k_]^l_ :> t[Sequence @@ Table[k, {i, 1, l}]]
(* t[2,2,2,2,2] *)
This produces the desired output.
2
Table already puts it's element in a row. To build your Listplot, it looks lie you just need to Flatten.
tst1 = {0, 0, 0, 8, 15, 18, 19, 20, 25, 30, 28, 20, 18, 16, 14, 12, 11, 10, 8, 6, 4, 3, 2, 1, 0.5};
tst2 = {0, 0, 0, Table[tst1[[i]], {i, 4, 25}]} // Flatten
ListPlot[Table[{tst1[[i]], tst2[[i]]}, {i, 25}]]
1
Try inspecting the plots of {s1,s2,f1[...]-1} and {s1,s2,f2[...]-1} where ... is your parameter set. (I've replaced your functions with f1 and f2 using NIntegrate to speed things up):
f1[S1_, S2_, γ_, β1_, β2_, δ1_, δ2_,α11_, α12_] := (γ*β1)/δ1*NIntegrate[Exp[-γ*a]*(1 - Exp[-δ1*a])*Exp[-S1*(α11*β1)/δ1 (1 - Exp[-δ1*a]) - S2*(α12*β2)/δ2 (1 - Exp[-δ2*a])], {a,0,...
1
Since both of them are an arithmetic sequence (a sequence with equal step size) you can use Differences:
h = Table[i, {i, 0, 5, 1}];
p = Table[j, {j, 0, 255, 51}];
If you apply once, you'll get step size:
Differences[h]
(* Out: {1,1,1,1,1} *)
Differences[p]
(* Out: {51,51,51,51,51} *)
Now you can use various ways to equate them, here we'll use another ...
1
Thanks to the @cvgmt's feedback, alternatives to Transpose:
Code 1
Thread[{list1, list2}]
% == Transpose[{list1, list2}]
(* Out: True *)
Code 2
Can only be used on two lists, each of them should be 1-dimensional:
Inner[List, list1, list2, List]
% == Transpose[{list1, list2}]
(* Out: True *)
Speed Comparison
lg[n_] := ConstantArray[RandomReal[{-1, 1}, 10^...
Only top voted, non community-wiki answers of a minimum length are eligible
