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I have the following matrix:

data = {{0.795718, 0.737, 0.813, 1.27301}, {0.846782, 0.767, 0.86,1.2859}, 
  {0.742211, 0.691, 0.767, 1.27714}, {0.754719, 0.706, 0.783,1.27319},  
  {0.743044, 0.691, 0.783, 1.26768}, {0.745167, 0.706,0.783, 1.28189},  
  {0.815733, 0.752, 0.86,1.25376}, {0.551127, 0.553,0.553, 1.2084}};

each row is made up of a list of 4 elements: {x1,x2,x3,y}. I would like to apply a rule to the matrix that allows me to select, row by row, only one of the first 3 elements, the one that is closest to a certain value, let us say xbest (eg 0.8 in the example with the data above), let us call this element xwin. Eventually, I would like have a new matrix with rows of 2 elements only: {xwin,y}. Your help is appreciated.

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This seems like a good case for NearestTo:

NearestTo[x]
is an operator form that yields Nearest[elems, x] when applied to a list elems.

nTo = NearestTo[xbest, 1];
Join[nTo /@ data[[All, ;; 3]], data[[All, {4}]], 2]

{{0.795718, 1.27301}, {0.767, 1.2859}, {0.767, 1.27714}, {0.783, 1.27319}, {0.783, 1.26768}, {0.783, 1.28189}, {0.815733, 1.25376}, {0.553, 1.2084}}

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  • $\begingroup$ Nice! I didn't know of this new function! Do you know how does it compare with Nearest in terms of efficiency? $\endgroup$ – Fraccalo Jun 29 '18 at 8:23
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    $\begingroup$ @Fraccalo, it seems that for long lists they are similar, but shorter lists, to my surprise, Nearest is much faster: dt=RandomInteger[100000,s]; NearestTo[99]@dt// RepeatedTiming versus Nearest[dt, 99]// RepeatedTiming for s=100 and s = 10^6. $\endgroup$ – kglr Jun 29 '18 at 8:54
  • $\begingroup$ I am using v10.0 and NearestTo is not defined there. How to circumvent this? and how to get the element (from the first 3) that is the furthest away from xBest? $\endgroup$ – Luigi Jun 29 '18 at 9:23
  • $\begingroup$ @Luigi, NearestTo is new in v11.3. I would go with Fraccalo's answer for finding nearest and farthest elements. $\endgroup$ – kglr Jun 29 '18 at 9:31
  • $\begingroup$ @kglr Thanks for the benchmark! And yes, the discrepancy in the speed depending on the list length is quite weird $\endgroup$ – Fraccalo Jun 29 '18 at 10:12
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What about:

xBest = 0.8;
{First@Nearest[#[[1 ;; 3]], xBest], #[[4]]} & /@ data

And for the "Farthest" function:

xBest = 0.8;
{First@Nearest[#[[1 ;; 3]], xBest, 
     DistanceFunction -> (-Abs[#1 - #2] &)], #[[4]]} & /@ data
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    $\begingroup$ could work! however, if you apply it to data, you see that the last row has duplicated value (3 elements rather than 2). With xBest=0.8058 you get {{0.813, 1.27301}, {0.767, 1.2859}, {0.767, 1.27714}, {0.783, 1.27319}, {0.783, 1.26768}, {0.783, 1.28189}, {0.815733, 1.25376}, {0.553, 0.553, 1.2084}} as result. $\endgroup$ – Luigi Jun 28 '18 at 20:40
  • $\begingroup$ Check the last edit :D $\endgroup$ – Fraccalo Jun 28 '18 at 20:43
  • $\begingroup$ Now, how to select the element that is furthest away from xBest. In other words, what is the opposite of Nearest? thanks! $\endgroup$ – Luigi Jun 29 '18 at 6:43
  • $\begingroup$ See new edit to my answer $\endgroup$ – Fraccalo Jun 29 '18 at 8:06

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