Suppose I have this nestlist t={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} How to calculate c=t(i)+t(i+1)
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A pretty general way of attacking problems like this that involve filtering data is to use ListConvolve. For this specific problem:
t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
ListConvolve[{1, 1}, t]
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
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t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
Most[t] + Rest[t]
(* or *)
t[[;;-2]] + t[[2;;]]
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
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It can be done very nicely by the function almost everybody forgets about.
t = Range[0, 10];
FoldPairList[{#1 + #2, #2}&, t]
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
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t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
seq = Total /@ Partition[t, 2, 1]
(* {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} *)
To find the general term of the sequence
c[n_] = FindSequenceFunction[seq, n]
(* -1 + 2 n *)
Verifying,
c /@ Range[Length[seq]] == seq
(* True *)
EDIT: Alternatively,
Clear[c, t]
eqns = {c[n] == t[n] + t[n + 1], t[n] == t[n - 1] + 1, t[1] == 0, t[2] == 1};
sol = RSolve[eqns, {c, t}, n][[1]]
(* {c -> Function[{n}, -1 + 2 n], t -> Function[{n}, -1 + n]} *)
Verifying the solutions,
eqns /. sol
(* {True, True, True, True} *)
c[n] /. sol
(* -1 + 2 n *)
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MovingMap[Total, Range[0, 10], 1]
(* Out: {1, 3, 5, 7, 9, 11, 13, 15, 17, 19} *)
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t[[1 ;; Length@t - 1]] + t[[2 ;; Length@t]]
Or you can use Drop instead.
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$\begingroup$ Actually what I first thought about was
sum(i+j for i, j in zip(t,t[1:])), and I wrote it in mma code. $\endgroup$ – wuyudi Dec 2 '20 at 6:15
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This?
t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
Plus @@@ Partition[t, 2, 1]
Or
t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
MovingMap[Total, t, 1]
{1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
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t = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; Table[t[[i]] + t[[i + 1]], {i, 1, Length@t - 1}]maybe suitable for beginner. $\endgroup$ – cvgmt Dec 2 '20 at 2:25
ListConvolve,Part+Span,Table+Part+Lenghtmight give you an idea. reference.wolfram.com/language/guide/ListManipulation.html could also be a good place to start looking for a possible solution. $\endgroup$ – N0va Dec 2 '20 at 1:57