I've tried 3 methods but all failed to do that.
1st Method
Apply[Flatten, {1, {2, {3, 4}, 5}, 6}, {2}]
2nd Method
Map[Flatten, {1, {2, {3, 4}, 5}, 6}, {2}]
3rd Method
Flatten[{1, {2, {3, 4}, 5}, 6}, {2}]
I wanna get {1, {2, 3, 4, 5}, 6}
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7 Answers
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lst = {1, {2, {3, 4}, 5}, 6};
FlattenAt[lst, {2, 2}]
{1, {2, 3, 4, 5}, 6}
Also
Map[## & @@ # &, lst, {2}]
{1, {2, 3, 4, 5}, 6}
Replace[lst, List -> Sequence, {3}, Heads -> True]
{1, {2, 3, 4, 5}, 6}
And
☺ = ## & @@@ # & /@ # &;
☺ @ lst
{1, {2, 3, 4, 5}, 6}
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4$\begingroup$ Because this is an accepted answer, I suppose it is important to note a principal difference between
FlattenAt[lst, {2, 2}]and the rest of the examples:FlattenAtflattens position, while, for example,Replace[lst, List -> Sequence, {3}, Heads -> True]flattens level. $\endgroup$ Commented Jan 29, 2020 at 13:16
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We've got a few answers already, but here's my 2 cents:
Replace[l_List :> Flatten[l]] /@ {1, {2, {3, 4}, 5}, 6}
{1, {2, 3, 4, 5}, 6}
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$\begingroup$ Is it possible to do this with
Apply? $\endgroup$– kileCommented Jan 28, 2020 at 12:47 -
3$\begingroup$ Using
Applydoesn't make much sense withFlatten, sinceFlattenis a single-argument function (It does have an infrequently-used but useful second argument, but still).Applyis more useful for multi-argument functions likeJoin,List, andPlus. Technically,Flatten[{##}] & @@@ {1, {2, {3, 4}, 5}, 6}works, but only because 1 and 6 are atomic. It wouldn't work with, e.g.,{Sin[1], {2, {3, 4}, 5}, Sin[6]}, so it's not a very good way to do what you want. $\endgroup$ Commented Jan 28, 2020 at 13:01
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This will work:
l = {1, {2, {3, 4}, 5}, 6}
MapAt[Flatten, l, 2]
{1, {2, 3, 4, 5}, 6}
also:
# /. x_ /; Length[x] > 1 :> Flatten@x & /@ l
{1, {2, 3, 4, 5}, 6}
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list = {1, {2, {3, 4}, 5}, 6};
If we want do modify list directly there is ApplyTo (since V 12.2)
list[[2]] //= Flatten;
list
{1, {2, 3, 4, 5}, 6}
A more traditional and explicit approach is
list /. {a_, b_, c_} :> {a, Flatten[b], c}
{1, {2, 3, 4, 5}, 6}
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A couple more:
list = {1, {2, {3, 4}, 5}, 6};
Apply[## &, list, {2}]
Flatten[{##}] & @@@ list
Also if you are looking for specific level control consider levelspec in Replace. For example with:
rl = {a_, x__, z_} :> {a, {x} /. rl, z};
deep = Range[14] /. rl
{1, {2, {3, {4, {5, {6, {7, 8}, 9}, 10}, 11}, 12}, 13}, 14}
Then:
Replace[deep, {x__} :> x, {3, 5}]
{1, {2, {3, 4, 5, 6, {7, 8}, 9, 10, 11, 12}, 13}, 14}
Related:
answered Jan 29, 2020 at 12:53
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lst[[2]]=Flatten@lst[[2]];lst
{1, {2, 3, 4, 5}, 6}
Also:
Delete[lst, {2,2,0}]
{1, {2, 3, 4, 5}, 6}
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1$\begingroup$ The
Delete-solution is so different, nice and funny, that I would have posted it as another answer. $\endgroup$– eldoCommented Oct 11, 2023 at 19:08
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Using Table:
Table[If[ListQ[#[[i]]], Flatten[#[[i]]], #[[i]]], {i, Length@#}] &@list
(*{1, {2, 3, 4, 5}, 6}*)
Also, using Cases:
Cases[list, x_ :> If[ListQ[x], Flatten@x, x]]
(*{1, {2, 3, 4, 5}, 6}*)
answered Oct 11, 2023 at 20:20
{a, {b, c}, {d e}, f} -> {a,b,c,d,e,f}. You might be asking "assuming things are nested x -> (A x B), flatten the deepest" (not the case since presumably you want this to also work on{1,{2,3,4,5},6}). You might be asking "flatten the second element". You might be asking "flatten the second element by one level". You need to either specify the problem unambiguously, and/or provide like 10 all-encompassing edge-case examples (e.g. the empty list, etc.). $\endgroup$