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I am trying to perform a calculation on a list containing sublists. More exact: The x-axis values are time data, progressing from zero, the y-axis data are data points, which I want to normalize. I want the y data to have their minimum at zero. Therefore, I look for the smallest y value in each sublist and add or subtract (depending if it is positive or negative) from all other y values of that sublist. I want to perform the calculation for all sublists. I think I haven't quite understood what the Slot (#) exactly does?!

I post here just an example list illustrating my problem, since the real list is very long.

My approach:

list={{{0,-0.1},{1,0.5},{2,0.7},{3,1.2},{4,0.6},{5,0.5},{6,1.3}},{{0,0.1},{1,0.5},{2,0.3},{3,0.7},{4,0.8},{5,1.1},{6,1.2}}}

list//{#[[1]],(#[[2]]-Min[#[[All,2]]])}&/@#&

This approach gives the following errors:

Part::partw: Part 1 of {} does not exist. >>

Part::partw: Part 2 of {} does not exist. >>

Part::partw: Part 1 of {} does not exist. >>

General::stop: Further output of Part::partw will be suppressed during this calculation. >>

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  • $\begingroup$ You might be interested in Rescale[]. $\endgroup$
    – J. M.'s eventual burnout ♦
    Aug 27, 2015 at 11:27
  • $\begingroup$ Cool! I didn't know that command. But just out of interest, is there a way to solve my problem with something similar I tried? $\endgroup$
    – Niki
    Aug 27, 2015 at 11:31

2 Answers 2

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First of all, your code does not return an error on my machine.

Second, using "something similar to what you tried", you might want to do

Transpose[{#[[All, 1]], #[[All, 2]] - Min[#[[All, 2]]]}] & /@ list
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Just based on your text:

f[u_] := Module[{a, b}, {a, b} = Transpose@u; 
  Transpose[{a, b - Min@b}]]

f/@list yields:

{{{0, 0.}, {1, 0.6}, {2, 0.8}, {3, 1.3}, {4, 0.7}, {5, 0.6}, {6, 1.4}}, {{0, 0.}, {1, 0.4}, {2, 0.2}, {3, 0.6}, {4, 0.7}, {5, 1.}, {6, 1.1}}}

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  • $\begingroup$ Puh, thanks a lot for the answer, it's very elegant. I have to wrap my head round it! I'm not very familiar with such constructs. $\endgroup$
    – Niki
    Aug 27, 2015 at 12:25
  • $\begingroup$ @Niki just play around...essentially transposing rectangular array allows you deal with columns as you like then re: transposing restores initial configuration with transformed rows...play and enjoy:) $\endgroup$
    – ubpdqn
    Aug 27, 2015 at 23:50

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