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I have a bunch of datasets which look like:

3 7 1 6 5
8 2 4 1 2
8 1 5 2 5
...

and I need to make a list of pairs such that the first element makes a pair with other elements in the same line.

3 7 
3 1
3 6
...

I usually use Table because it is fast enough for $10^3$ elements, but as my dataset gets bigger, it's no longer efficient. For example:

data = RandomInteger[{1, 10}, {10^7, 5}];
Flatten[Table[{data[[i, 1]], data[[i, j]]}, {i, 1, Length@data}, {j, 
 2, 5}], 1]//AbsoluteTiming//First
(* 39.828744 *)

What is the most efficient way to find this pair list?

Timing

Finally I had a chance to do a timing on my real machine:

(* Kuba *)
0.726869
(* ciao *)
1.297926
(* nested table*)
2.146111
(* Shutao Tang *)
11.290944
(* Mr.Wizard♦ *)
57.232387

Although it seems results can be (completely) different if I do a timing on my old laptop!

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2
  • $\begingroup$ Are the elements always a machine size Integer? Or always a machine-precision Real? Or mixture of different types of data? $\endgroup$
    – Michael E2
    Commented Apr 11, 2015 at 21:38
  • 2
    $\begingroup$ @MichaelE2: They're always Integer. $\endgroup$
    – Mahdi
    Commented Apr 11, 2015 at 21:49

6 Answers 6

Reset to default
16
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About 4x faster:

Partition[Flatten @ data[[All, {1, 2, 1, 3, 1, 4, 1, 5}]], 2]
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5
  • 1
    $\begingroup$ Shouldn't you make this more general? $\endgroup$
    – Mr.Wizard
    Commented Apr 11, 2015 at 22:32
  • $\begingroup$ It depends. If it is slow, probably doesn't matter. If it is fast, ok. But atm I'm confused because I'm testing your and rasher's method with result that your is 20x slower than OP's. So as soon as I figure it out I will tell you :P $\endgroup$
    – Kuba
    Commented Apr 11, 2015 at 22:35
  • $\begingroup$ LOL -- looks like I wrote a function for the wrong shape of data. Okay, +1, and back to the drawing board. $\endgroup$
    – Mr.Wizard
    Commented Apr 11, 2015 at 22:43
  • $\begingroup$ @Mr.Wizard Ok :) I have to go but I will make it more general tomorrow, if it is worth it :). Feel free to delete my non relevant comments when you change something. Good night. :) $\endgroup$
    – Kuba
    Commented Apr 11, 2015 at 22:46
  • $\begingroup$ Ah, clean. I feel herp-a-derp for not thinking of that... +1 $\endgroup$
    – ciao
    Commented Apr 12, 2015 at 1:54
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I think you'll find this faster:

pairem[data_] := 
 Module[{c = ConstantArray[0, 8*(Length@data)], 
   p = Flatten@data[[All, 2 ;;]], p2 = data[[All, 1]]},
  c[[2 ;; ;; 2]] = p;
  c[[1 ;; ;; 2]] = Flatten@Transpose[ConstantArray[p2, 4]];
  Partition[c, 2]]
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1
  • $\begingroup$ @Pickett I don't know, I still get only 2.5 speed up here, maybe it is hardware related. $\endgroup$
    – Kuba
    Commented Apr 11, 2015 at 22:27
5
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Edit: this method is optimized for long sublists which is exactly the opposite of your example. Sorry. I'll post if I find anything applicable.

This is fairly clean and on larger data at least as fast as rasher's code:

fn = ArrayFlatten[{{First@#, Rest@# ~Partition~ 1 }}] &;

Test:

fn /@ {{3, 7, 1, 6, 5}, {8, 2, 4, 1, 2}, {8, 1, 5, 2, 5}}

{{{3, 7}, {3, 1}, {3, 6}, {3, 5}},
 {{8, 2}, {8, 4}, {8, 1}, {8, 2}},
 {{8, 1}, {8, 5}, {8, 2}, {8, 5}}}

x = RandomInteger[99, {5000, 5000}];

fn /@ x   // AbsoluteTiming // First
pairem[x] // AbsoluteTiming // First

0.334019

0.383022

A related question: Prepend 0 to sublists

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  • 3
    $\begingroup$ Try with x = RandomInteger[99, {10^6, 5}] which is really the case. $\endgroup$
    – Kuba
    Commented Apr 11, 2015 at 22:39
  • $\begingroup$ @Kuba, x = RandomInteger[99, {10^6, 5}];pairem[x]; // AbsoluteTiming fn /@ x; // AbsoluteTiming ===> {0.2158204, Null},{13.1357421, Null} $\endgroup$
    – xyz
    Commented Apr 12, 2015 at 5:12
4
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Here is a refinement of @Kuba's clever Part syntax:

data = RandomInteger[{1,10},{10^7,5}];

r1 = Partition[
    Flatten @ data[[All,{1,2,1,3,1,4,1,5}]],
    2
]; //AbsoluteTiming

r2 = ArrayReshape[
    data[[All, {1,2,1,3,1,4,1,5}]],
    {4 Length[data], 2}
]; //AbsoluteTiming

r1===r2

{1.55516, Null}

{0.568824, Null}

True

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3
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My trail:

data = RandomInteger[{1, 10}, {8, 5}]

{{1, 1, 9, 10, 6}, {7, 2, 8, 5, 4}, {2, 1, 10, 1, 7}, {9, 1, 4, 5, 2},
   {6, 10, 6, 5, 10}, {2, 10, 1, 7, 4}, {3, 5, 3, 2, 2}, {1, 2, 9, 6, 2}}

Thread@{#1, {##2}} & @@@ data

{
   {{1, 1}, {1, 9}, {1, 10}, {1, 6}}, {{7, 2}, {7, 8}, {7, 5}, {7, 4}},
   {{2, 1}, {2, 10}, {2, 1}, {2, 7}}, {{9, 1}, {9, 4}, {9, 5}, {9, 2}},
   {{6, 10}, {6, 6}, {6, 5}, {6, 10}}, {{2, 10}, {2, 1}, {2, 7}, {2, 4}},  
   {{3, 5}, {3, 3}, {3, 2}, {3, 2}}, {{1, 2}, {1,9}, {1, 6}, {1, 2}}
}

Performance test

x = RandomInteger[99, {10^6, 5}];
Thread@{#1, {##2}} & @@@ x; // AbsoluteTiming

{2.3535156, Null}
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2
  • $\begingroup$ Pretty, so +1, but much slower even than OP code, who is looking for efficiency. $\endgroup$
    – ciao
    Commented Apr 12, 2015 at 5:01
  • $\begingroup$ @rasher, The high-efficiency methods are pairem and @Kuba method currently $\endgroup$
    – xyz
    Commented Apr 12, 2015 at 5:22
2
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Just for fun a solution with Riffle:

riffle[d_] := Module[{data = Transpose[d]},
  Partition[Flatten[Transpose[Rest[Riffle[data, {First[data]}]]]], 2]
  ]
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