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I have a list of the form:

toy = {{a, {b, c}}, {d, {e, f}}, {g, {h, j}}, {k, {m, n}}};

and would like to create the list 

{{a, c}, {d, f}, {g, j}, {k, n}}

from toy.

The following code does it but I find this esthetically unsatisfactory because I don't like creating the two independent lists toy[[;; , 1]] and toy[[;; , 2, 2]] and combining them.

Transpose[{toy[[;; , 1]], toy[[;; , 2, 2]]}]

Is there a cleaner more direct way to do this?

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    $\begingroup$ For ex.{#[[1]], #[[2, 2]]} & /@ toy or {#1, #2[[2]]} & @@@ toy $\endgroup$
    – Dr. belisarius
    Commented Jan 27, 2015 at 4:49
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    $\begingroup$ toy /. {x_, {y_, z_}} :> {x, z} $\endgroup$
    – BoLe
    Commented Jan 27, 2015 at 4:50
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    $\begingroup$ Just for something different: toy[[All, 2]] = toy[[All, 2, 2]]; toy $\endgroup$
    – Mike Honeychurch
    Commented Jan 27, 2015 at 4:56
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    $\begingroup$ Without destroying the toy Transpose@{toy[[All,1]],toy[[All,2,2]]} $\endgroup$
    – tchronis
    Commented Jan 27, 2015 at 7:21
  • $\begingroup$ @tchronis That method is already in the question as "unsatisfactory" $\endgroup$
    – Mr.Wizard
    Commented Jan 27, 2015 at 7:25

8 Answers 8

Reset to default
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Is this what you want:

Map[{#[[1]], #[[2, 2]]} &, toy]
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4
$\begingroup$

One should ban all those comments spammers one day... :)

To not duplicate them:

toy[[All , 2, 0]] = #2 &;
toy

{{a, c}, {d, f}, {g, j}, {k, n}}

Keep in mind that this modifies toy.

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6
  • $\begingroup$ +1 nice one :) How come you (and others) use ` ;; ` instead of All? $\endgroup$
    – Mike Honeychurch
    Commented Jan 27, 2015 at 7:24
  • $\begingroup$ @MikeHoneychurch you mean why? 2 is less than 3 characters :) But may be obscuring sometimes and doesn't work for <V9 in every form. Still, I like it :) $\endgroup$
    – Kuba
    Commented Jan 27, 2015 at 7:27
  • $\begingroup$ @Mike Probably to save a character, but even as a terse fanatic I prefer All. First because it is descriptive and ledible, and second because it is backward-compatible. (;; does not work in v7.) It's only one more character so I vote for All. I've even edited it into a few answers here and there. $\endgroup$
    – Mr.Wizard
    Commented Jan 27, 2015 at 7:27
  • $\begingroup$ @Mr.Wizard ok ok... :) $\endgroup$
    – Kuba
    Commented Jan 27, 2015 at 7:27
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    $\begingroup$ @Mr.Wizard I don't! My notebooks are full of ;; :) $\endgroup$
    – Kuba
    Commented Jan 27, 2015 at 7:29
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Maybe:

Extract[{{1}, {2, 2}}] /@ toy
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3
$\begingroup$ 
(Flatten /@ toy)[[;; , {1, 3}]]
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3
$\begingroup$ 
☺ = {#, #2 & @@ #2} & @@@ # &;
☺ @ toy

{{a, c}, {d, f}, {g, j}, {k, n}}

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    $\begingroup$ Or, {#, Last@#2} & @@@ toy $\endgroup$
    – bbgodfrey
    Commented Jan 20, 2018 at 1:02
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    $\begingroup$ @bbgodfrey, yes but then we can't say "look ma, no letters!" $\endgroup$
    – kglr
    Commented Jan 20, 2018 at 2:33
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list = {{a, {b, c}}, {d, {e, f}}, {g, {h, j}}, {k, {m, n}}};

Using ReplaceAt (new in 13.1)

ReplaceAt[list, {_, a_} :> a, {;; , 2}]

{{a, c}, {d, f}, {g, j}, {k, n}}

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1
$\begingroup$ 
toy = {{a, {b, c}}, {d, {e, f}}, {g, {h, j}}, {k, {m, n}}};

Cases[toy, {a_, {b_, c_}} :> {a, c}]
{First@#, Last@Last@#} & /@ toy
Flatten@*ReplacePart[{{2, 1}} -> Nothing] /@ toy
Flatten@*Delete[{2, 1}] /@ toy
#[[{1, -1}]] &@*MapAt[Splice, 2] /@ toy
{First@#, Last@#} & /@ MapAt[Splice, toy, {All, 2}]
Through[{First, Last}[#]] & /@ MapAt[Splice, toy, {All, 2}]

Result:

{{a, c}, {d, f}, {g, j}, {k, n}}

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1
$\begingroup$

Using ReplaceList:

list = {{a, {b, c}}, {d, {e, f}}, {g, {h, j}}, {k, {m, n}}};

ReplaceList[#, {s1_, {s2__}} :> Splice@{s1, First@s2}] & /@ list

(*{{a, c}, {d, f}, {g, j}, {k, n}}*)

Or using Cases:

Cases[#, {s1_, {s2__}} :> {s1, First@s2}] &@list

(*{{a, c}, {d, f}, {g, j}, {k, n}}*)

Or using ReplaceAll:

# /. {s2_, s3 : {__}} :> {s2, Last@s3} &@list

(*{{a, c}, {d, f}, {g, j}, {k, n}}*)
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