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I have a list of lists that are not of the same length (for example, {{6}, {14}, {6, 26, 30}}), but I have to compare every element in the $n^{\text{th}}$ sublist with the corresponding $n^{\text{th}}$ odd number. So, 6 in the first sublist with 1 (first odd number), 14 in the second sublist with 3 (second odd number), 6 with 5 (third odd number), 26 with 5, 30 with 5, etc.

I tried to make another list with odd numbers, but I really don't know how to program the case when the comparison is with only one element in the sublist (e.g. sublists 1 and 2) and with multi-element sublists (e.g., 3rd sublist).

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    $\begingroup$ Your last few questions have made me curious: what is your actual problem that has you dealing with lists of lists with different lengths? $\endgroup$
    – J. M.'s lack of A.I. ♦
    Sep 5, 2012 at 9:14

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MapIndexed is your friend :) The part specification for each of the sublists is supplied as a second argument to the function being mapped. With this, you can construct your $n^{\text{th}}$ odd number to compare against. For example:

list = {{6}, {14}, {6, 26, 3}, {5, 10}};
MapIndexed[Thread[# > First@(2 #2 - 1)] &, list]
(* {{True}, {True}, {True, True, False}, {False, True}} *)

Replace > (Greater) with < (Less) or == (Equal) or any other comparison function as per your needs.

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Something like :

data = {{6}, {14}, {6, 2, 30}};

odds = 2 # + 1 & /@ Range[0, Length[data] - 1];

MapThread[Thread[Greater[#1, #2]] &, {data, odds}]

(* {{True}, {True}, {True, False, True}} *)

If the condition is :

cond[x_, y_] := If[x > y, Pi y, 0]

then one needs to wrap it with Unevaluated :

MapThread[Thread[Unevaluated[cond[#1, #2]]] &, {data, odds}]

(* {{\[Pi]}, {3 \[Pi]}, {5 \[Pi], 0, 5 \[Pi]}} *)
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  • $\begingroup$ I'd have done odds = Range[1, 2 Length[data] - 1, 2] myself... $\endgroup$
    – J. M.'s lack of A.I. ♦
    Sep 5, 2012 at 8:24
  • $\begingroup$ And if I don't want Mathematica to write {True, True} but to calculate something with numbers that satisfy conditions? For example: If True, Print[\pi*(2k+1)]? $\endgroup$
    – WayneGacy
    Sep 5, 2012 at 8:27
  • $\begingroup$ @WayneGacy Just change the function Greater to what you need. $\endgroup$
    – b.gates.you.know.what
    Sep 5, 2012 at 8:29
  • $\begingroup$ MapThread[Thread[If[#1 == 3.14*#2, Print[#1]]] &, {data, odds}] {If[{6} == 3.14, Print[{6}]], If[{14} == 9.42, Print[{14}]], If[{6, 2, 30} == 15.7, Print[{6, 2, 30}]]} If I put that line in Mathematica, the output is like line above. I don't want to write that, only that sublist whose elements satisfy condition. THANK you! $\endgroup$
    – WayneGacy
    Sep 5, 2012 at 8:40
  • $\begingroup$ @WayneGacy Please see edit. $\endgroup$
    – b.gates.you.know.what
    Sep 5, 2012 at 9:08

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