26
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Task

Given (by any means) a sorted floating point dataset, return (by any means and within 1‰ of the correct value) the interquartile mean.

One possible algorithm

  1. Discard the lowest and highest quarters of the data points.
  2. Calculate the average (sum divided by count) of the remaining data points.

Note: If the dataset size is not evenly split-able in four, you will have to weigh the datapoints that are shared by sub-sets. See Example evaluation 2 below.

Example evaluation 1

Given {1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38}

  1. Data count is 12, so we remove lowest and highest 3 datapoints:
    {1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38}
  2. Average of the remaining 6 datapoints:
    (5 + 6 + 6 + 7 + 7 + 8) / 6 = 6.5

Example evaluation 2

Given {1, 3, 5, 7, 9, 11, 13, 15, 17}

  1. Count is 9, so each quarter has 2¼ datapoints:
    {1, 2, (0.25×5), (0.75×5), 7, 9, 11, (0.75×13), (0.25×13), 15, 17}
  2. Average of the remaining 4.5 datapoints:
    (0.75×5 + 7 + 9 + 11 + 0.75×13) / 4.5 = 9
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22 Answers 22

5
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Scilab, 8 bytes

trimmean

See the documentation. By default, discard=50, so the IQM is computed.

EDIT: You know, this is a trivial built-in answer, so I’m marking it as CW.

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  • \$\begingroup\$ I guess this will be the winner. Well done. \$\endgroup\$ – Adám Jul 26 '16 at 19:30
8
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Pyth, 11 10 bytes

.O><lQS*4Ql
.OsPtc4S*4

Test suite.

How it works

It quadruplicates the input list to ensure that the data count is divisible by 4.

It still needs sorting, because *4 applies to the whole list instead of to each individual element.

Then, it splits the list into four equal parts, then takes away the first and the last part.

The remaining list is flattened and the average is taken.

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8
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MATL, 12 11 bytes

4Y"G"6L)]Ym

Input is a horizontal vector, with the format

[1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38]

or

[1 3 4 5 6 6 7 7 8 8 9 38]

Try it online!

Explanation

4Y"    % Input horizontal vector implicitly. Repeat each element 4 times (run-length
       % decoding). The resulting array is still sorted.
G"     % Push input, for each: repeat as many times as the input size
  6L)  %   Remove first and last elements, by applying the index "2:end-1"
]      % End for each
Ym     % Compute mean. Display implicitly
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  • \$\begingroup\$ I don't get it. How does 6L) remove the first and last elements? When I do it, it pushes a bunch of complex numbers. \$\endgroup\$ – DJMcMayhem Jul 24 '16 at 17:11
  • 5
    \$\begingroup\$ @DrGreenEggsandIronMan Complex numbers can be used for that in MATL. The imaginary unit stands for the end of the array, and if there are two of three numbers they define a range. So [2, -1+i] when used as an index means 2:end-1 \$\endgroup\$ – Luis Mendo Jul 24 '16 at 17:19
7
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Snowman, 66 bytes

}vg","aS:10sB;aM4aRAsOal`,4nD,`aG0AaGal`NdE`AaL1AfL:nA;alaF,nDtSsP

Try it online!

Uses the same algorithm as @LeakyNun's answers.

}         enable variables b, e, and g
vg        read a line of input into b
","aS     split on commas (in-place)
:10sB;aM  convert each element in resulting array to number ("frombase(10)-map")
4aR       repeat the array 4 times
AsO       sort the array
al        take the length and put it in e without consuming b (the array)
`,        swap b and e, then move e to g; now b=length g=array
4nD       divide b by 4 (4 was stored in e, which is why the array was moved)
,`        move the array and length/4 back to their original positions
aG        split the array into groups of length (length/4)
0AaG      take all elements with index >0 (i.e. remove the first element)
al        store the length of the new array in e again
`NdE`     bring it up to b, decrement, and put it back
AaL       take all elements with index <length-1 (i.e. remove last)
1AfL      flatten the array 1 level deep
:nA;      push a block that adds two numbers (to e)
al        store the length of this new array in g
aF        fold b over e (sum the numbers)
,         move g (the length) into e
nD        divide the sum by the length, resulting in the average
tSsP      to-string and print
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  • 2
    \$\begingroup\$ This language looks awful. I love it. \$\endgroup\$ – Mego Jul 25 '16 at 9:11
6
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Python 3, 50 bytes

lambda n:sum(sorted(n*4)[len(n):-len(n)])/len(n)/2

Ideone it!

How it works

It is a translation of my answer in Pyth.

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5
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Jelly, 14 13 12 bytes

x4ṫL‘$ḣLN$S÷LH
x4ṫLḊḣLN$S÷LH
x4œs4ḊṖFS÷LH

Try it online!

Test suite.

How it works

It is a translation of my answer in Pyth.

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  • \$\begingroup\$ I'm pretty sure this can be shortened, as I can do it 15 in APL. \$\endgroup\$ – Adám Jul 24 '16 at 14:15
  • \$\begingroup\$ @Adám Please post your solution (so that I may copy haha) \$\endgroup\$ – Leaky Nun Jul 24 '16 at 14:16
  • \$\begingroup\$ I want to give Marinus a chance... \$\endgroup\$ – Adám Jul 24 '16 at 14:20
  • \$\begingroup\$ Here you go! \$\endgroup\$ – Adám May 18 '17 at 15:10
  • \$\begingroup\$ Enough of a chance after more than 9 months, certainly \$\endgroup\$ – Luis Mendo May 18 '17 at 15:14
4
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Pyke, 16 13 bytes

4*S4ftOsDsRl/

Try it here!

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  • \$\begingroup\$ You broke my streak... \$\endgroup\$ – Leaky Nun Jul 24 '16 at 14:43
  • \$\begingroup\$ I'm so sorry :(​​​​ \$\endgroup\$ – Blue Jul 24 '16 at 14:45
4
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Brachylog, 21 bytes

:3jo@4brbcLl/N,L+:N*.

Try it online! or verify multiple test cases

Explanation

This is basically @LeakyNun's Pyth answer algorithm.

:3j      Append 3 copies of the input to itself
o@4      Sort and split in 4 lists of equal length
brb      Remove the head and the tail of the list of lists
cL       Concatenate the 2 sublists into a list L
l/N,     N is the inverse of the length of L
L+:N*.   Output is the product of N and the sum of the elements of L

The only small trick there is in multiplying by the inverse of the length instead of dividing by the length, because division between 2 integers is integer division.

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3
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Octave, 44 bytes

@(x)mean(reshape(~~(1:4)'*x,[],4)(:,2:3)(:))

This defines an anonymous function.

The input is a horizontal vector.

Try it on ideone.

Explanation

The input horizontal vector is first matrix-multiplied (*) by a column vector of four ones (built with ~~(1:4)'). The result is a four-column matrix where each row is a copy of the input vector. This is then reshaped, while maintaining the linear order of the elements, into a 4-column matrix (reshape(...,[],4)). The center two columns are kept ((:,2:3)) and linearized into a single column ((:)), of which the mean is computed (mean(...)).

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  • \$\begingroup\$ You can save 1 byte with the more readable [x;x;x;x] instead of ~~(1:4)'*x \$\endgroup\$ – Tom Carpenter Jul 26 '16 at 2:08
  • \$\begingroup\$ @(x)mean([x;x;x;x](:)((b=numel(x))+1:3*b)) is also 2 bytes less. That was why I had come up with, but it's basically the same as your approach. \$\endgroup\$ – Tom Carpenter Jul 26 '16 at 2:12
  • \$\begingroup\$ @TomCarpenter I don't think it's that similar. I think you should post it as a separate answer \$\endgroup\$ – Luis Mendo Jul 26 '16 at 9:14
3
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J, 20 18 bytes

2 bytes thanks to @miles

#-:@%~-@#+/@}.#}.4#]
-@#(+/%#)@}.#}.4#]

Try it online! (Online interpreter)

Usage

>> f =: -@#(+/%#)@}.#}.4#]
>> f 1 3 5 7 9 11 13 15 17
<< 9

How it works

It is a translation of my answer in Pyth.

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  • \$\begingroup\$ Online J interpreter. \$\endgroup\$ – Adám Jul 24 '16 at 14:31
  • \$\begingroup\$ @Adám Thanks, added. \$\endgroup\$ – Leaky Nun Jul 24 '16 at 14:32
  • 2
    \$\begingroup\$ You can just directly take the average of the middle portion -@#(+/%#)@}.#}.4#] for 18 bytes. \$\endgroup\$ – miles Jul 24 '16 at 16:23
2
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Actually, 20 15 13 bytes

;l╗;+;+S╜@t╜τ@HΣ╜τ@/
;l;τ;a;+;+StHΣ/
;l;τ;aττStHΣ/

Try it online!

How it works

It is a translation of my answer in Pyth.

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  • \$\begingroup\$ For once, an Actually answer which is readable (in Greek). \$\endgroup\$ – Adám Jul 24 '16 at 15:02
  • \$\begingroup\$ @Adám Pyth uses ASCII. \$\endgroup\$ – Leaky Nun Jul 24 '16 at 15:03
2
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Octave, 42bytes

Another anonymous function for Octave.

@(x)mean([x;x;x;x](:)((b=numel(x))+1:3*b))

You can try it online. Simply enter that command, and then do ans([1 2 4 5 6 9]) or whatever numbers are required.

This one starts by creating from the input array one with 4 of each input element by first concatenating four copies vertically, and then flattening it vertically. This maintains the sort order.

Then is extracts the range of elements from the length of the input array plus 1 up to three times the length of the input array. Because the new array is four times longer, this chops off the upper and lower quartiles.

Finally the mean of the new array is returned.

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2
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05AB1E, 15 bytes

€D€D¹gô¦¨˜DOsg/

Explanation

€D€D             # quadruple each element in list
    ¹gô          # split into pieces the size of input
       ¦¨˜       # remove the first and last and flatten the middle 2
          DOsg/  # sum and divide by length

Try it online

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2
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APL (Dyalog), 15 bytes

IQM←(+/÷≢)≢↓-∘≢↓4∘/

Try it online!

4∘/ quadruplicate each element

-∘≢↓ drop as many trailing elements as there are elements in the arguments

≢↓ drop as many leading elements as there are element in the argument

() apply the following tacit function:

+/ the sum

÷ divided by

 the tally

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1
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JavaScript (ES6), 75 bytes

a=>a.concat(a,a,a).sort(g=(x,y)=>x-y).slice(l=a.length,-l).reduce(g,0)/l/-2

Uses the obvious quadruplicate-and-sort approach, and I get to use reduce, which is nice. The only trickery here is to save 4 bytes by reusing the sort comparator to subtract all the array elements from zero, which gives me -2l times the answer I want.

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1
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Golfscript, 28 29 bytes

~.4*$\,.@/1>2<{+}*{+}*'/'@2*
~.4*$\,.@/1>2<{+}*{+}*\2*-1?*

Try it online!

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1
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Actually, 12 bytes

4α;l¼≈;±(Htæ

Try it online! (currently doesn't work because TIO is a few versions behind)

Explanation:

4α;l¼≈;±(Htæ
4α            repeat each element 4 times
  ;l¼≈        length divided by 4, as integer
      ;±      copy, unary negate
        (Ht   remove first and last quartiles
           æ  mean
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1
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Mathematica, 51 bytes

Mean@#[[(l=1+Length@#/4);;-l]]&@Sort@Join[#,#,#,#]&

Sorts four copies of the list (to prevent issues with list length not multiples of four), takes part "1 quarter the length of resulting list plus 1" through to the "1/4 length list + 1 from the end", takes their Mean.

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1
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Java 146 126 Bytes

Such java much verbose!

float m(float[]n){float r=0;int l=n.length,i=l/4;r-=(n[i])*(l%4)/4;r+=n[i*3]*(4-(l%4))/4;for(;i<l*3/4;r+=n[i],i++);return r/l*2;}

Older Ungolfed partially readable with test cases

/**
 *
 * @author rohan
 */
public Golf{

float m(float[]n){
//declarations 
float r=0;
int x,i=0,l=n.length;
//sum the array 
for(float m:n){r+=m;}
//remove the excess
for(;i<l/4;r-=n[i]+n[l-i-1],i++);
//weight the quartiles
r-=(n[l/4]+n[l*3/4])*(l%4)/4;
//return the sum/length but multiply by two since only half of the set is averaged
return r/l*2;
    }
static void interQuartileMean(float... set){
    System.out.println(new Golf().m(set));
}
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
    //test cases pass with flying colours
        interQuartileMean(1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38);
        interQuartileMean(1, 3, 5, 7, 9, 11, 13, 15, 17);   
    }

}
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1
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Clojure, 82 81 bytes

Edit: 1 byte less by re-writing "didvide by 2 n" part.

#(let[n(count %)](*(/ n)0.5(apply +(subvec(vec(for[i % j(range 4)]i))n(* 3 n)))))

Previous:

#(let[n(count %)](/(apply +(subvec(vec(for[i % j(range 4)]i))n(* 3 n)))(* 2.0 n)))

Uses for to generate 4 repeated values, using float 2.0 not to have fractional results, the rest is just standard.

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1
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R, 17 11 bytes

mean(n,0.25)

Assuming n is the input vector in the standard R form n=c(1, 2, 3, ...).

This is in no way surprising since R can be considered “THE language for statistical computing” and has many statistical built-ins.

UPDATE. Saved 6 bytes thanks to rturnbull because trim is the first optional argument by default!

Test cases:

a <- c(1, 3, 4, 5, 6, 6, 7, 7, 8, 8, 9, 38)
b <- c(1, 3, 5, 7, 9, 11, 13, 15, 17)
mean(a,trim=0.25) # Returns 6.5
mean(b,trim=0.25) # Returns 9
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  • \$\begingroup\$ Since trim is the default second argument, you don't need to name it; 0.25 can be shortened to .25 or 1/4. This saves you six bytes. \$\endgroup\$ – rturnbull Dec 26 '16 at 11:48
0
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Excel, 17 bytes

=TRIMMEAN(A:A,.5)

Relaxed input format make this easy. Input one per row in Column A.

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