32
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Write a snippet to calculate the mode (most common number) of a list of positive integers.

For example, the mode of

d = [4,3,1,0,6,1,6,4,4,0,3,1,7,7,3,4,1,1,2,8]

is 1, because it occurs the maximum of 5 times.

You may assume that the list is stored in a variable such as d and has a unique mode.

e.g.: Python, 49

max(((i,d.count(i))for i in set(d)), key=lambda x:x[1])

This is , so the shortest solution in bytes wins.

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0

46 Answers 46

29
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Python 2 - 18

max(d,key=d.count)

Since your python answer doesn't seem to print, I expect this is what you want.

Add 6 bytes for print normally.

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3
  • \$\begingroup\$ perfect, think @globby needs to see future :) \$\endgroup\$
    – garg10may
    Dec 19, 2014 at 16:25
  • 14
    \$\begingroup\$ The great thing about this one is that it's not even golfy, it's just Pythonic. The only thing that's been golfed is a space between d, and key=. \$\endgroup\$
    – wchargin
    Dec 19, 2014 at 17:28
  • 8
    \$\begingroup\$ @WChargin: Eh, Pythonic would be to avoid the quadratic runtime by using defaultdict(int) or Counter. Something like Counter(d).most_common()[0]. \$\endgroup\$ Dec 21, 2014 at 3:01
27
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Matlab/Octave, 7 5 bytes

Unsurprisingly there's a built-in function for finding modes. As an anonymous function:

@mode

This returns the most commonly occuring element in the input vector with ties going to the smaller value.

Saved 2 bytes thanks to Dennis!

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4
  • 3
    \$\begingroup\$ +1, clearly the right tool for the job. As it's a builtin, what happens if there's more than one number of highest frequency? \$\endgroup\$ Dec 20, 2014 at 10:55
  • 2
    \$\begingroup\$ @steveverrill According to the documentation (type help mode): "If two, or more, values have the same frequency 'mode` returns the smallest." \$\endgroup\$
    – wchargin
    Dec 21, 2014 at 3:05
  • 1
    \$\begingroup\$ Unnamed functions seem to be allowed (the accepted answer is one), so you could shorten this to @mode. \$\endgroup\$
    – Dennis
    Jun 2, 2016 at 23:41
  • \$\begingroup\$ @Dennis Thanks! Though I admit it's a strange feeling to edit my first answer on the site. \$\endgroup\$
    – Alex A.
    Jun 3, 2016 at 5:18
16
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Pyth - 6

eo/QNQ

Try it online.

Expects input on stdin like [4,3,1,0,6,1,6,4,4,0,3,1,7,7,3,4,1,1,2,8]. Ties are resolved by last occurrence because Python performs stable sorts.

Sorts the list by count the value in the list, then prints the last number of the list.

Q could be replaced with d if you initialized d to contain the value before e.g. =d[4 3 1 0 6 4 4 0 1 7 7 3 4 1 1 2 8)

Python-esque pseudo-code:

Q=eval(input());print(sorted(Q,key=Q.count)[-1])

Full Explanation:

            : Q=eval(input()) (implicit)
e           : ... [-1]
 o   Q      : orderby(lambda N: ...,Q)
  /QN       : count(Q,N)

Pyth's orderby runs exactly like Python's sorted with orderby's first argument being the key argument.

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0
12
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Mathematica, 25 bytes

Last@SortBy[d,d~Count~#&]

or

#&@@SortBy[d,-d~Count~#&]

As in the challenge, this expects the list to be stored in d.

or... 15 bytes

Of course, Mathematica wouldn't be Mathematica if it didn't have a built-in:

#&@@Commonest@d

Commonest returns a list of all most common elements (in case of a tie), and #&@@ is a golfed First@.

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1
  • \$\begingroup\$ another case for mthmca \$\endgroup\$ Apr 13, 2016 at 19:26
9
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Ruby, 22 bytes

d.max_by{|i|d.count i}

Basically a port of my Mathematica answer, except Ruby has a direct max_by so I don't need to sort first.

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1
  • 1
    \$\begingroup\$ I was about to suggest d.max_by d.method:count but that's about a million (aka not even two) bytes longer. Still, it's worth noting that it's possible. \$\endgroup\$
    – anon
    May 10, 2016 at 18:39
9
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R, 33 25 bytes

Thanks @Hugh for the help shortening:

names(sort(-table(d))[1])

The original:

v=table(d);names(v[which.max(v)])

This calculates the frequency of each element in the vector d, then returns the name of the column containing the largest value. The value returned is actually a character string containing the number. It didn't say anywhere that that wasn't okay, so...

Any suggestions to shorten this are welcome!

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1
  • 2
    \$\begingroup\$ names(sort(-table(d))[1]) \$\endgroup\$
    – Hugh
    Dec 20, 2014 at 8:21
9
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CJam, 11 10 bytes

A{A\-,}$0=

Assumes the array in a variable called A. This is basically sorting the array based on the occurrence of each number in the array and then picks the last element of the array.

Example usage

[1 2 3 4 4 2 6 6 6 6]:A;A{aA\/,}$W=

Output

6

1 byte saved thanks to Dennis!

Try it online here

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4
  • \$\begingroup\$ A{A\-,}$0= is one byte shorter. \$\endgroup\$
    – Dennis
    Dec 31, 2014 at 2:47
  • 1
    \$\begingroup\$ As of 0.6.5 it's doable in 8 bytes: Ae`$e_W= \$\endgroup\$ Jun 3, 2015 at 0:09
  • \$\begingroup\$ @MartinEnder Umm... nope. I knew you need to sort first. \$\endgroup\$ Dec 12, 2016 at 18:27
  • \$\begingroup\$ @ErikGolferエリックゴルファー whoops, you're right, needs 9 bytes: $e`$e_W= \$\endgroup\$ Dec 12, 2016 at 18:30
8
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K5, 6 bytes

*>#:'=

The first (*) of the descending elements (>) of the count of each (#:') of the group (=). Step by step:

  i
4 3 1 0 6 1 6 4 4 0 3 1 7 7 3 4 1 1 2 8

  =i
4 3 1 0 6 7 2 8!(0 7 8 15
 1 10 14
 2 5 11 16 17
 3 9
 4 6
 12 13
 ,18
 ,19)

  #:'=i
4 3 1 0 6 7 2 8!4 3 5 2 2 2 1 1

  >#:'=i
1 4 3 7 6 0 8 2

  *>#:'=i
1

try it in your browser!

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8
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Powershell 19

($d|group)[0].Count

(this asumes the array is already on $d)

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8
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J - 12 char

Anonymous function. Sorts list from most to least common, taking first item.

(0{~.\:#/.~)
  • 0{ First of
  • ~. Unique items
  • \: Downsorted by
  • #/.~ Frequencies

Try it for yourself.

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1
  • 1
    \$\begingroup\$ This is really 10 bytes--the function can be assigned without the parens. \$\endgroup\$ May 12, 2016 at 21:20
6
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JavaScript (ES6) 51

Just a single line expression using the preloaded variable d. Sort the array by frequency then get the first element.
Nasty side effect, the original array is altered

d.sort((a,b)=>d.map(w=>t+=(w==b)-(w==a),t=0)&&t)[0]

As usual, using .map instead of .reduce because it's 1 char shorter overall. With .reduce it' almost a clean, non-golfed solution.

d.sort((a,b)=>d.reduce((t,w)=>t+(w==b)-(w==a),0))[0]

At last, a solution using a function, not changing the original array and without globals (62 bytes):

F=d=>[...d].sort((a,b)=>d.reduce((t,w)=>t+(w==b)-(w==a),0))[0]

Test In FireFox/FireBug console

d=[4,3,1,0,6,1,6,4,4,0,3,1,7,7,3,4,1,1,2,8]
d.sort((a,b)=>x.map(w=>t+=(w==b)-(w==a),t=0)&&t)[0]

Output 1

The d array becomes:

[1, 1, 1, 1, 1, 4, 4, 4, 4, 3, 3, 3, 0, 6, 6, 0, 7, 7, 2, 8]
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1
  • \$\begingroup\$ &&t => |t.. \$\endgroup\$
    – l4m2
    Jul 6, 2023 at 8:35
6
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Perl 6, 21 bytes

.Bag.invert.max.value

Example:

$_ = < 4 3 1 0 6 1 6 4 4 0 3 1 7 7 3 4 1 1 2 8 >».Int;

say .Bag.invert.max.value; # implicitly calls $_.Bag…

If there is a tie it will print the larger of the ones that tied.


The .Bag method on a List or an Array creates a quantified hash that associates the total count of how many times a given value was seen with that value.

bag(4(4), 3(3), 1(5), 0(2), 6(2), 7(2), 2, 8)

The .invert method creates a List of the pairs in the bag with the key and the value swapped. ( The reason we call this is for the next method to do what we want )

4 => 4,  3 => 3,  5 => 1,  2 => 0,  2 => 6,  2 => 7,  1 => 2,  1 => 8

The .max method on a List of Pairs returns the biggest Pair comparing the keys first and in the case of a tie comparing the values.
( This is because that is how multi infix:<cmp>(Pair:D \a, Pair:D \b) determines which is larger )

5 => 1

The .value method returns the value from the Pair. ( It would have been the key we were after if it wasn't for the .invert call earlier )

1

If you want to return all of the values that tied in the case of a tie:

say @list.Bag.classify(*.value).max.value».key

The .classify method returns a list of pairs where the keys are from calling the Whatever lambda *.value with each of the Pairs.

1 => [2 => 1, 8 => 1],
2 => [0 => 2, 6 => 2, 7 => 2],
3 => [3 => 3],
4 => [4 => 4],
5 => [1 => 5]

Then we call .max to get the largest Pair.

"5" => [1 => 5]

A call to .value gets us the original Pairs from the Bag ( just one in this case )

1 => 5

Then we use >>.key to call the .key method on every Pair in the list, so that we end up with a list of the values that were seen the most.

1
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5
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Python - 32

max((x.count(i),i)for i in x)[1]

Don't see an 18 character solution anywhere in the future to be honest.

EDIT: I stand corrected, and impressed.

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5
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05AB1E, 3 bytes

.MJ

Explanation:

.M  # Gets the most frequent element in the [implicit] input
  J # Converts to a string, needed as the program would output "[1]" instead of "1" without this.

If you want to store the array in a variable instead of using input, just push the array to the stack at the start of the program.

Try it online!

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4
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JavaScript, ES6, 71 bytes

A bit long, can be golfed a lot.

f=a=>(c=b=[],a.map(x=>b[x]?b[x]++:b[x]=1),b.map((x,i)=>c[x]=i),c.pop())

This creates a function f which can be called like f([1,1,1,2,1,2,3,4,1,5]) and will return 1.

Try it on your latest Firefox's Console.

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3
  • \$\begingroup\$ Off-topic, but I just realized how relevant your username is to PCG.SE. :P \$\endgroup\$
    – user344
    Dec 20, 2014 at 11:16
  • \$\begingroup\$ @nyuszika7h heh. Although I have had this username long before I even know PPCG existed. \$\endgroup\$
    – Optimizer
    Dec 20, 2014 at 11:20
  • \$\begingroup\$ f=a=>(c=b=[],a.map(x=>b[x]++-1?0:b[x]=1),b.map((x,i)=>c[x]=i),c.pop()) is 1 byte shorter. \$\endgroup\$
    – Bálint
    May 11, 2016 at 5:59
4
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Dyalog APL, 12 characters

d[⊃⍒+/∘.=⍨d]

∘.=⍨d is the same as d∘.=d, reflexive outer product of =. It creates a boolean matrix comparing every pair of elements in d.

+/ sums that matrix along one of the axes and produces a vector.

grades the vector, i.e. sorts it by indices. (As the glyphs suggest, grades in descending order and would grade in ascending order.)

takes the first index from the grading—the index of the largest element of d.

d[...] returns that element.

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1
  • \$\begingroup\$ +/∘.=⍨d counts for each element of d. ⊢∘≢⌸d counts for each element of ∪d, so the indices don't correspond to those of d. Counterexample: d←1 1 2 2 2. To make it work: (∪d)[⊃⍒⊢∘≢⌸d] or (⊃⍒⊢∘≢⌸d)⊃∪d. \$\endgroup\$
    – ngn
    Dec 31, 2014 at 3:15
4
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Jelly, 2 bytes

Æṃ

Try it online!

Jelly finally got a mode builtin! A non-builtin version however is 5 bytes:

ċ@Þ`Ṫ

Try it online!

How they work

Æṃ - Main link. Takes a list d on the left
Æṃ - Return the mode of d

Well that was unimpressive. How about the 5 byte, non-builtin version?

ċ@Þ`Ṫ - Main link. Takes d on the left
   `  - Use d as both the left and right argument to:
  Þ   -   Sort d by:
ċ@    -     Count occurrences in d
    Ṫ - Take the last element (i.e. the element with a maximal count in d)
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3
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C# - 49

Can't really compete using C# but oh well:

Assuming d is the array

d.GroupBy(i=>i).OrderBy(a=>a.Count()).Last().Key;

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3
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bash - 29 27 characters

sort|uniq -c|sort -nr|sed q

Using it:

sort|uniq -c|sort -nr|sed q
4
3
1
0
6
1
6
4
4
0
3
1
7
7
3
4
1
1
2
8
[ctrl-D]
5 1

i.e. "1" is the mode, and it appears five times.

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3
  • \$\begingroup\$ sort|uniq -c|sort -nr|sed q saves a couple of characters \$\endgroup\$ Dec 19, 2014 at 20:06
  • \$\begingroup\$ I posted the same answer, but you were faster :) \$\endgroup\$
    – pgy
    Dec 19, 2014 at 20:10
  • \$\begingroup\$ @pgy - thank you - have updated! \$\endgroup\$
    – user15259
    Dec 19, 2014 at 20:25
3
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GolfScript, 10 bytes

a{a\-,}$0=

From this answer I wrote to Tips for golfing in GolfScript. Expects the input in an array named a, returns result on stack. (To read input from an array on the stack, prepend : for 11 bytes; to read input from stdin (in the format [1 2 1 3 7]), also prepend ~ for 12 bytes.)

This code works by iterating over the input array, subtracting each element from the original array, and counting the number of elements left. This is then used as a key to sort the original array by, and the first element of the sorted array is returned.

Online demo.

Ps. Thanks to Peter Taylor for pointing out this challenge to me.

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3
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Java 8, 83 Bytes

d.stream().max((x,y)->Collections.frequency(d,x)-Collections.frequency(d,y)).get();

d must be a Collection<Integer>.


If Collections can be statically imported:
59 Bytes

d.stream().max((x,y)->frequency(d,x)-frequency(d,y)).get();
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3
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Haskell, 42 39 bytes

f s=snd$maximum[([1|y<-s,y==x],x)|x<-s]

Try it online!

Edit: Thans to Zgarb for -3 bytes

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2
  • 1
    \$\begingroup\$ I think sum is unnecessary here. \$\endgroup\$
    – Zgarb
    Nov 29, 2017 at 11:44
  • \$\begingroup\$ @Zgarb Right, I actually used exactly the same trick in a previous answer. Thanks for reminding me! \$\endgroup\$
    – Laikoni
    Nov 29, 2017 at 15:03
3
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Jelly, 7 bytes

ṢŒrṪÞṪṪ

Try it online!

Longer than K :(

Assumes the first argument has the array. There are no variables in Jelly.

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3
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JavaScript (Node.js), 45 bytes

x=>x.map(a=t=>a[t]=0+a[t]||[1+t]).sort()[0]-1

Try it online!

42 without 0 support

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2
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Java 8 : 184 bytes

Stream.of(A).collect(Collectors.groupingBy(i -> i, Collectors.counting())).entrySet().stream().sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())).findFirst().get().getKey();

Input A must be of type Integer[]. Note java.util.* and java.util.stream.* need to be imported, however in the spirit oneliner they are left out.

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2
  • \$\begingroup\$ downvoting because of ... ? \$\endgroup\$ Jan 4, 2015 at 20:56
  • 1
    \$\begingroup\$ I know it's been more than two years, but you can remove the spaces at (i->i,Collectors.counting()). \$\endgroup\$ Feb 1, 2017 at 12:18
2
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Bash + unix tools, 62 bytes

Expects the array in the STDIN. The input format does not count, as long as the numbers are non-negative integers.

grep -o [0-9]\*|sort|uniq -c|sort -n|awk 'END{print $2}'

Edited: escaped wildcard in grep argument. Now it can be run safely in non-empty directories. Thanks to manatwork.

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2
  • 1
    \$\begingroup\$ Best if run in an empty directory. Otherwise [0-9]* may get expanded to matching file names. \$\endgroup\$
    – manatwork
    Nov 29, 2015 at 13:26
  • \$\begingroup\$ Alternatively, put ' around the argument to grep. \$\endgroup\$ Nov 29, 2015 at 19:41
2
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Perl, 27 bytes

$Q[$a{$_}++]=$_ for@F;pop@Q

Returns the last most common value in case of a tie.

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2
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jq, 29 characters

group_by(.)|max_by(length)[0]

Sample run:

bash-4.3$ jq 'group_by(.)|max_by(length)[0]' <<< '[4,3,1,0,6,1,6,4,4,0,3,1,7,7,3,4,1,1,2,8]'
1

On-line test

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2
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PHP, 53 50 bytes

<?=array_flip($c=array_count_values($d))[max($c)];

Run like this:

echo '<?php $d=$argv;?><?=array_flip($c=array_count_values($d))[max($c)]; echo"\n";' | php -- 4 3 1 0 6 1 6 4 4 0 3 1 7 7 3 4 1 1 2 8

Tweaks

  • Saved 3 bytes by making use of the freedom to assume the input is assigned to a variable d
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2
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Haskell 78

import Data.List
import Data.Ord
g=head.maximumBy(comparing length).group.sort

If the imports are ignored, it's 45.

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2
  • 1
    \$\begingroup\$ You can save 4 bytes by using point-free style and 2 bytes by using maximumBy instead of last.sortBy. The new code would become g=head.maximumBy(comparing length).group.sort. \$\endgroup\$
    – Hjulle
    Nov 28, 2017 at 15:54
  • \$\begingroup\$ 1.) Anonymous functions are allowed, so you can drop the g=. 2.) You can replace maximumBy(comparing length) by snd.maximum.map((,)=<<length) which doesn't need to import Ord, for a total of 62 bytes: Try it online! \$\endgroup\$
    – Laikoni
    Nov 29, 2017 at 11:08

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