Finding median of list in Python

Question

How do you find the median of a list in Python  The list can be of any size and the numbers are not guaranteed to be in any particular order   If the list contains an even number of elements  the function should return the average of the middle two   Here are some examples  sorted for display purposes    median  1      1 median  1  1      1 median  1  1  2  4      1 5 median  0  2  5  6  8  9  9      6 median  0  0  0  0  4  4  6  8      2

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def median array       if len array   lt  1          return None      if len array    2    0          median    array len array   2-1  len array   2 1           return sum median    len median      else          return array len array   2

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Do yourself  def median numbers        quot  quot  quot      Calculate median of a list numbers       param numbers  the numbers to be calculated       return  median value of numbers        gt  gt  gt  median  1  3  3  6  7  8  9       6      gt  gt  gt  median  1  2  3  4  5  6  8  9       4 5      gt  gt  gt  import statistics      gt  gt  gt  import random      gt  gt  gt  numbers   random sample range -50  50   k 100       gt  gt  gt  statistics median numbers     median numbers      True      quot  quot  quot      numbers   sorted numbers      mid index   len numbers     2     return            numbers mid index    numbers mid index - 1     2 if mid index   2    0         else numbers mid index          if   name       quot   main   quot       from doctest import testmod      testmod    source from

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median Function  def median midlist       midlist sort       lens   len midlist      if lens   2    0           midl    lens   2          res   midlist midl      else          odd    lens   2  -1         ev    lens   2           res   float midlist odd    midlist ev     float 2      return res

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Of course you can use build in functions  but if you would like to create your own you can do something like this  The trick here is to use   operator that flip positive number to negative  For instance  2 -  -3 and using negative in for list in Python will count items from the end  So if you have mid    2 then it will take third element from beginning and third item from the end   def median data       data sort       mid   len data     2     return  data mid    data  mid     2

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fuction median  def median d       d np sort d      n2 int len d  2      r n2 2     if  r  0           med d n2       else          med  d n2    data m 1     2     return med

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Here s a cleaner solution   def median lst       quotient  remainder   divmod len lst   2      if remainder          return sorted lst  quotient      return sum sorted lst  quotient - 1 quotient   1     2    Note  Answer changed to incorporate suggestion in comments

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def median array          Calculate median of the given list                TODO  use statistics median in Python 3     array   sorted array      half  odd   divmod len array   2      if odd          return array half      return  array half - 1    array half     2 0

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Here what I came up with during this exercise in Codecademy   def median data       new list   sorted data      if len new list  2  gt  0          return new list len new list  2      elif len new list  2    0          return  new list  len new list  2     new list  len new list  2 -1    2 0  print median  1 2 3 4 5 9

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It is very simple   def median alist        to find median you will have to sort the list first     sList   sorted alist      first   0     last   len sList -1     midpoint    first   last   2     return midpoint   And you can use the return value like this median   median anyList

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def median x       x   sorted x      listlength   len x       num   listlength  2     if listlength 2  0          middlenum    x num  x num-1   2     else          middlenum   x num      return middlenum

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Python 3 4 has statistics median      Return the median  middle value  of numeric data       When the number of data points is odd  return the middle data point    When the number of data points is even  the median is interpolated by taking the average of the two middle values    gt  gt  gt  median  1  3  5   3  gt  gt  gt  median  1  3  5  7   4 0    Usage   import statistics  items    6  1  8  2  3   statistics median items    gt  gt  gt  3   It s pretty careful with types  too   statistics median map float  items     gt  gt  gt  3 0  from decimal import Decimal statistics median map Decimal  items     gt  gt  gt  Decimal  3

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A simple function to return the median of the given list  def median lst       lst sort      Sort the list first     if len lst    2    0     Checking if the length is even           Applying formula which is sum of middle two divided by 2         return  lst len lst     2    lst  len lst  - 1     2     2     else            If length is odd then get middle value         return lst len lst     2   Some examples with the medain function   gt  gt  gt  median  9  12  20  21  34  80      Even 20 5  gt  gt  gt  median  9  12  80  21  34      Odd 21  If you want to use library you can just simply do   gt  gt  gt  import statistics  gt  gt  gt  statistics median  9  12  20  21  34  80      Even 20 5  gt  gt  gt  statistics median  9  12  80  21  34      Odd 21

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Here s the tedious way to find median without using the median function    def median  arg       order arg      numArg   len arg      half   int numArg 2      if numArg 2   half          print  arg half-1  arg half   2      else          print int arg half     def order tup       ordered    tup i  for i in range len tup        test ordered      while test ordered            test ordered      print ordered    def test ordered       whileloop   0      for i in range len ordered -1           print i          if  ordered i  gt ordered i 1                print str ordered i       is greater than     str ordered i 1                original   ordered i 1              ordered i 1  ordered i              ordered i  original             whileloop   1  run the loop again if you had to switch values     return whileloop

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You can use the list sort to avoid creating new lists with sorted and sort the lists in place   Also you should not use list as a variable name as it shadows  python s own list   def median l       half   len l     2     l sort       if not len l    2          return  l half - 1    l half     2 0     return l half

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import numpy as np def get median xs           mid   len xs     2    Take the mid of the list         if len xs    2    1    check if the len of list is odd             return sorted xs  mid   if true then mid will be median after sorting         else               return 0 5   sum sorted xs  mid - 1 mid   1               return 0 5   np sum sorted xs  mid - 1 mid   1    if false take the avg of mid print get median  7  7  3  1  4  5    print get median  1 2 3  4 5

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I defined a median function for a list of numbers as  def median numbers       return  sorted numbers  int round  len numbers  - 1    2 0      sorted numbers  int round  len numbers  - 1     2 0       2 0

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The sorted   function is very helpful for this  Use the sorted function to order the list  then simply return the middle value  or average the two middle values if the list contains an even amount of elements   def median lst       sortedLst   sorted lst      lstLen   len lst      index    lstLen - 1     2         if  lstLen   2           return sortedLst index      else          return  sortedLst index    sortedLst index   1   2 0

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I had some problems with lists of float values  I ended up using a code snippet from the python3 statistics median and is working perfect with float values without imports  source  def calculateMedian list       data   sorted list      n   len data      if n    0          return None     if n   2    1          return data n    2      else          i   n    2         return  data i - 1    data i     2

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You can try the quickselect algorithm if faster average-case running times are needed  Quickselect has average  and best  case performance O n   although it can end up O n    on a bad day   Here s an implementation with a randomly chosen pivot   import random  def select nth n  items       pivot   random choice items       lesser    item for item in items if item  lt  pivot      if len lesser   gt  n          return select nth n  lesser      n -  len lesser       numequal   items count pivot      if numequal  gt  n          return pivot     n -  numequal      greater    item for item in items if item  gt  pivot      return select nth n  greater    You can trivially turn this into a method to find medians   def median items       if len items    2          return select nth len items   2  items       else          left    select nth  len items -1     2  items          right   select nth  len items  1     2  items           return  left   right    2   This is very unoptimised  but it s not likely that even an optimised version will outperform Tim Sort  CPython s built-in sort  because that s really fast  I ve tried before and I lost

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A more generalized approach for median  and percentiles  would be   def get percentile data  percentile         Get the number of observations     cnt len data        Sort the list     data sorted data        Determine the split point     i  cnt-1  percentile       Find the  floor  of the split point     diff i-int i        Return the weighted average of the value above and below the split point     return data int i    1-diff  data int i  1   diff     Data data  1 2 3 4 5    For the median print get percentile data data  percentile  50      gt  3 print get percentile data data  percentile  75      gt  4    Note the weighted average difference when an int is not returned by the percentile print get percentile data data  percentile  51      gt  3 04

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def midme list1        list1 sort       if len list1  2 gt 0              x   list1 int  len list1  2        else              x     list1 int  len list1  2  -1    list1 int   len list1  2       2     return x   midme  4 5 1 7 2

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In case you need additional information on the distribution of your list  the percentile method will probably be useful  And a median value corresponds to the 50th percentile of a list   import numpy as np a   np array  1 2 3 4 5 6 7 8 9   median value   np percentile a  50    return 50th percentile print median value

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Just two lines are enough  def get median arr               Calculate the median of a sequence       param arr  list      return  int or float             arr sort       return arr len arr   2  if len arr    2 else  arr len arr   2    arr len arr   2-1   2

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I posted my solution at Python implementation of  quot median of medians quot  algorithm   which is a little bit faster than using sort     My solution uses 15 numbers per column  for a speed  5N which is faster than the speed  10N of using 5 numbers per column   The optimal speed is  4N  but I could be wrong about it   Per Tom s request in his comment  I added my code here  for reference   I believe the critical part for speed is using 15 numbers per column  instead of 5      bin pypy     TH  stackoverflow  2016-01-20  linear time  median of medians  algorithm   import sys  random   items per column   15   def find i th smallest  A  i        t   len A      if t  lt   items per column             if A is a small list with less than items per column items  then                      1  do sort on A           2  find i-th smallest item of A                   return sorted A  i      else            1  partition A into columns of k items each  k is odd  say 5            2  find the median of every column           3  put all medians in a new list  say  B                   B     find i th smallest k   len k  - 1  2  for k in  A j  j   items per column   for j in range 0 len A  items per column               4  find M  the median of B                   M   find i th smallest B   len B  - 1  2              5  split A into 3 parts by M     lt  M         M    and    gt  M             6  find which above set has A s i-th smallest  recursively                    P1     j for j in A if j  lt  M           if i  lt  len P1                return find i th smallest  P1  i          P3     j for j in A if j  gt  M           L3   len P3          if i  lt   t - L3                return M         return find i th smallest  P3  i -  t - L3       How many numbers should be randomly generated for testing    number of numbers   int sys argv 1       create a list of random positive integers   L     random randint 0  number of numbers  for i in range 0  number of numbers        Show the original list     print L     This is for validation     print sorted L  int  len L  - 1  2       This is the result of the  median of medians  function    Its result should be the same as the above    print find i th smallest  L   len L  - 1    2

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What I did was this  def median a       a sort       if len a    2    int          return a len a    2      else          return  a len a    2    a  len a    2  - 1     2  Explanation  Basically if the number of items in the list is odd  return the middle number  otherwise  if you half an even list  python automatically rounds the higher number so we know the number before that will be one less  since we sorted it  and we can add the default higher number and the number lower than it and divide them by 2 to find the median

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Works with python-2 x    def median lst       n   len lst      s   sorted lst      return  sum s n  2-1 n  2 1   2 0  s n  2   n   2  if n else None      gt  gt  gt  median  -5  -5  -3  -4  0  -1   -3 5     numpy median      gt  gt  gt  from numpy import median  gt  gt  gt  median  1  -4  -1  -1  1  -3   -1 0     For python-3 x  use statistics median    gt  gt  gt  from statistics import median  gt  gt  gt  median  5  2  3  8  9  -2   4 0

[python] Finding median of list in Python

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