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I have the following sample list (my actual list is, of course, much longer):

A={{{"15", "CG"}, {"391", "CG"}, {"412", "CC3"}}, {{"3", "CG"}, 
    {"16", "CG"}, {"392", "CG"}}};

I would like to map an arbitrary function f onto the string representation of numbers, like this:

{{{f["15"], "CG"}, {f["391"], "CG"}, {f["412"], "CC3"}}, {{f["3"], "CG"}, 
  {f["16"], "CG"}, {f["392"], "CG"}}}

Is there a straightforward, succinct way of doing this using Map, MapAt, or something else? Unlike Map, it seems that MapAt does not have an option with levelspec.

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11 Answers 11

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22
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A combination of Map and MapAt perhaps?

Map[MapAt[f, #, 1] &, A, {2}]

(* ===>  {{{f["15"], "CG"}, {f["391"], "CG"}, {f["412"], "CC3"}}, 
         {{f["3"], "CG"}, {f["16"], "CG"}, {f["392"], "CG"}}} 
*)
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To map f to number strings appearing at any position or depth:

Map[If[StringMatchQ[#, NumberString], f[#], #] &, A, {-1}]
(* => {{{f["15"], "CG"}, {f["391"], "CG"}, {f["412"], "CC3"}}, {{f["3"], "CG"}, {f["16"], "CG"}, {f["392"], "CG"}}} *)

another example:

b = {{{"15", "CG"}, {{"391", "CG"}, "230"}, {"412", "CC3"}}, {{"3", 
 "CG"}, {"392", {"CG", {"CG", "345"}}}}};
Map[If[StringMatchQ[#, NumberString], f[#], #] &, b, {-1}]
(* => {{{f["15"], "CG"}, {{f["391"], "CG"}, f["230"]}, {f["412"], "CC3"}}, {{f["3"], "CG"}, {f["392"], {"CG", {"CG", f["345"]}}}}} *)
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    $\begingroup$ This one deserves more credit, +1 $\endgroup$
    – Rojo
    May 18, 2012 at 20:31
15
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More options:

Apply[{f[#1], ##2} &, A, {2}]

{f[#1], ##2} & @@@ # & /@ A

ReplacePart[A, {i_, j_, 1} :> f @ A[[i, j, 1]] ]

ReplacePart[A, p:{_, _, 1} :> f @ Extract[A, p] ]
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  • $\begingroup$ @David I'm not sure how to read that; is the code hard to read? (I think ReplacePart is quite readable myself.) $\endgroup$
    – Mr.Wizard
    May 19, 2012 at 13:21
  • $\begingroup$ The second example would be difficult to read for the newly initiated in mma. My +1 more accurately reflects my amazement at your skill in crafting so many different solutions. $\endgroup$
    – DavidC
    May 19, 2012 at 13:43
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A[[All, All, 1]] = Map[f, A[[All, All, 1]], {2}]

I think the neatest given the question is probably a mix between David Carraher and kguler's

A /. x_String /; StringMatchQ[x, NumberString] :> f[x]

or a little bit more efficiently, searching only the lowest levels

Replace[A, x_String /; StringMatchQ[x, NumberString] :> f[x], {-1}]
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6
  • $\begingroup$ This solution intrigues me, but I don't think it all got posted $\endgroup$
    – Jagra
    May 18, 2012 at 20:16
  • $\begingroup$ It makes the changes in place. Why does it intrigue you @Jagra? $\endgroup$
    – Rojo
    May 18, 2012 at 20:19
  • $\begingroup$ A construct (A[[All,All,1]] =), that I just never thought of using before. Except I can't get it work properly ;-( $\endgroup$
    – Jagra
    May 18, 2012 at 20:23
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    $\begingroup$ @Jagra, since the changes are made in place, you can use it only once. If you run it twice, you get f[f["23"]], etc $\endgroup$
    – Rojo
    May 18, 2012 at 20:25
  • 1
    $\begingroup$ Also, don't look at the output of that line. Look at the value of A after running the line $\endgroup$
    – Rojo
    May 18, 2012 at 20:26
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Try a very simple idea: 

A = {{{"15", "CG"}, {"391", "CG"}, {"412", "CC3"}}, {{"3", 
     "CG"}, {"16", "CG"}, {"392", "CG"}}};

g[{x_, y_}] := {f[x], y};

Map[g, A, {2}]

{{{f["15"], "CG"}, {f["391"], "CG"}, {f["412"], "CC3"}}, {{f["3"], 
   "CG"}, {f["16"], "CG"}, {f["392"], "CG"}}}
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11
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I like to use @@ (or @@@ on lists) since #1 is easier to read than #[[1]] in the middle (muddle) of more brackets:

Map[{f[#1], #2} & @@ # &, A, {2}]

{{{f["15"], "CG"}, {f["391"], "CG"}, {f["412"], "CC3"}}, {{f["3"], 
   "CG"}, {f["16"], "CG"}, {f["392"], "CG"}}}
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10
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Perhaps

 g = # /. {a_String, b_String} :>  {f[a], b} &;
 g@A

yielding

{{{f["15"], "CG"}, {f["391"], "CG"}, {f["412"], "CC3"}}, {{f["3"], 
"CG"}, {f["16"], "CG"}, {f["392"], "CG"}}}
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0
9
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Another possibility, which I don't think has been given (but is similar to other answers): 

Map[{f@First@#, Last@#} &, A, {2}]
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9
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As @Kuba shows here, now in version 9 you can do:

MapAt[f,A,{All,All,1}]
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    $\begingroup$ It might be nice to mention that Kuba brought this to your attention. $\endgroup$
    – Mr.Wizard
    Aug 27, 2013 at 4:25
  • $\begingroup$ I agree. Done :) $\endgroup$
    – Murta
    Aug 27, 2013 at 11:24
  • $\begingroup$ even MapAt[f,A,{;;,;;,1}] $\endgroup$
    – garej
    Jan 29, 2016 at 9:22
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Map on map does the job nicely:

{f[#[[1]]], #[[2]]} & /@ # & /@ A

{{{f[15],CG},{f[391],CG},{f[412],CC3}},{{f[3],CG},{f[16],CG},{f[392],CG}}}
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7
$\begingroup$ 
A /. a_?(StringMatchQ[#, DigitCharacter ..] &) :> f[a] // Quiet
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