Getting the index of the returned max or min item using max min on a list

Question

I m using Python s max and min functions on lists for a minimax algorithm  and I need the index of the value returned by max   or min    In other words  I need to know which move produced the max  at a first player s turn  or min  second player  value  for i in range 9       new board   current board new board with move  i   3  i   3   player       if new board          temp   min max new board  depth   1  not is min level            values append temp   if is min level      return min values  else      return max values   I need to be able to return the actual index of the min or max value  not just the value

User · Answer

You can find the min max index and value at the same time if you enumerate the items in the list  but perform min max on the original values of the list  Like so   import operator min index  min value   min enumerate values   key operator itemgetter 1   max index  max value   max enumerate values   key operator itemgetter 1     This way the list will only be traversed once for min  or max

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I think the answer above solves your problem but I thought I d share a method that gives you the minimum and all the indices the minimum appears in   minval   min mylist  ind    i for i  v in enumerate mylist  if v    minval    This passes the list twice but is still quite fast  It is however slightly slower than finding the  index of the first encounter of the minimum  So if you need just one of the minima  use Matt Anderson s solution  if you need them all  use this

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What about this   a  1 55 2 36 35 34 98 0  max index dict zip a range len a     max a      It creates a dictionary from the items in a as keys and their indexes as values  thus dict zip a range len a     max a   returns the value that corresponds to the key max a  which is the index of the maximum in a  I m a beginner in python so I don t know about the computational complexity of this solution

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I was also interested in this and compared some of the suggested solutions using perfplot  a pet project of mine    Turns out that numpy s argmin   numpy argmin x    is the fastest method for large enough lists  even with the implicit conversion from the input list to a numpy array       Code for generating the plot     import numpy import operator import perfplot   def min enumerate a       return min enumerate a   key lambda x  x 1   0    def min enumerate itemgetter a       min index  min value   min enumerate a   key operator itemgetter 1       return min index   def getitem a       return min range len a    key a   getitem      def np argmin a       return numpy argmin a    perfplot show      setup lambda n  numpy random rand n  tolist        kernels           min enumerate          min enumerate itemgetter          getitem          np argmin                 n range  2  k for k in range 15        logx True      logy True

User · Answer

if is min level      return values index min values   else      return values index max values

User · Answer

Simple as that    stuff    2  4  8  15  11   index   stuff index max stuff

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Possibly a simpler solution would be to turn the array of values into an array of value index-pairs  and take the max min of that  This would give the largest smallest index that has the max min  i e  pairs are compared by first comparing the first element  and then comparing the second element if the first ones are the same   Note that it s not necessary to actually create the array  because min max allow generators as input   values    3 4 5   m i    max  v i  for i v in enumerate values   print  m i    5  2

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After you get the maximum values  try this   max val   max list  index max   list index max val    Much simpler than a lot of options

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Say that you have a list values    3 6 1 5   and need the index of the smallest element  i e  index min   2 in this case   Avoid the solution with itemgetter   presented in the other answers  and use instead  index min   min range len values    key values   getitem      because it doesn t require to import operator nor to use enumerate  and it is always faster benchmark below  than a solution using itemgetter     If you are dealing with numpy arrays or can afford numpy as a dependency  consider also using  import numpy as np index min   np argmin values    This will be faster than the first solution even if you apply it to a pure Python list if    it is larger than a few elements  about 2  4 elements on my machine  you can afford the memory copy from a pure list to a numpy array   as this benchmark points out    I have run the benchmark on my machine with python 2 7 for the two solutions above  blue  pure python  first solution   red  numpy solution  and for the standard solution based on itemgetter    black  reference solution   The same benchmark with python 3 5 showed that the methods compare exactly the same of the python 2 7 case presented above

User · Answer

If you want to find the index of max within a list of numbers  which seems your case   then I suggest you use numpy   import numpy as np ind   np argmax mylist

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list  1 1412  4 3453  5 8709  0 1314  list index min list     Will give you first index of minimum

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Dont have high enough rep to comment on existing answer   But for https   stackoverflow com a 11825864 3920439 answer  This works for integers  but does not work for array of floats  at least in python 3 6  It will raise TypeError  list indices must be integers or slices  not float

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Pandas has now got a much more gentle solution  try it  df column  idxmax

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Just a minor addition to what has already been said  values index min values   seems to return the smallest index of min  The following gets the largest index       values reverse        values index min values     len values  - 1    len values      values reverse     The last line can be left out if the side effect of reversing in place does not matter   To iterate through all occurrences      indices          i   -1     for   in range values count min values           i   values i   1   index min values     i   1       indices append i    For the sake of brevity  It is probably a better idea to cache min values   values count min  outside the loop

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Use numpy module s function numpy where  import numpy as n x   n array  3 3 4 7 4 56 65 1     For index of minimum value       idx   n where x  x min    0    For index of maximum value   idx   n where x  x max    0    In fact  this function is much more powerful  You can pose all kinds of boolean operations For index of value between 3 and 60   idx   n where  x gt 3  amp  x lt 60   0  idx array  2  3  4  5   x idx  array   4   7   4  56

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Assuming you have a following list my list    1 2 3 4 5 6 7 8 9 10  and we know that if we do max my list  it will return 10 and min my list  will return 1  Now we want to get the index of the maximum or minimum element we can do the following   x000D   x000D  my list    1 2 3 4 5 6 7 8 9 10   max value   max my list    returns 10 max value index   my list index max value    retuns 9   to get an index of minimum value  min value   min my list    returns 1 min value index   my list index min value    retuns 0 x000D   x000D   x000D

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https   docs python org 3 library functions html max  If multiple items are maximal  the function returns the first one encountered  This is consistent with other sort-stability preserving tools such as sorted iterable  key keyfunc  reverse True  0   To get more than just the first use the sort method   import operator  x    2  5  7  4  8  2  6  1  7  1  8  3  4  9  3  6  5  0  9  0   min   False max   True  min val index   sorted  list zip x  range len x      key   operator itemgetter 0   reverse   min    max val index   sorted  list zip x  range len x      key   operator itemgetter 0   reverse   max     min val index 0   gt  0  17   max val index 0   gt  9  13   import ittertools  max val   max val index 0  0   maxes    n for n in itertools takewhile lambda x  x 0     max val  max val index

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Say you have a list such as   a    9 8 7    The following two methods are pretty compact ways to get a tuple with the minimum element and its index  Both take a similar time to process  I better like the zip method  but that is my taste   zip method  element  index   min list zip a  range len a       min list zip a  range len a       7  2   timeit min list zip a  range len a      1 36   s    107 ns per loop  mean    std  dev  of 7 runs  1000000 loops each    enumerate method  index  element   min list enumerate a    key lambda x x 1    min list enumerate a    key lambda x x 1    2  7   timeit min list enumerate a    key lambda x x 1   1 45   s    78 1 ns per loop  mean    std  dev  of 7 runs  1000000 loops each

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Why bother to add indices first and then reverse them  Enumerate   function is just a special case of zip   function usage  Let s use it in appropiate way   my indexed list   zip my list  range len my list     min value  min index   min my indexed list  max value  max index   max my indexed list

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I think the best thing to do is convert the list to a numpy array and use this function     a   np array list  idx   np argmax a

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As long as you know how to use lambda and the  key  argument  a simple solution is   max index   max  range  len my list     key   lambda index   my list  index

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A simple way for finding the indexes with minimal value in a list if you don t want to import additional modules   min value   min values  indexes with min value    i for i in range 0 len values   if values i     min value    Then choose for example the first one   choosen   indexes with min value 0

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Use a numpy array and the argmax   function   a np array  1 2 3    b np argmax a   print b   2

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This is simply possible using the built-in enumerate   and max   function and the optional key argument of the max   function and a simple lambda expression   theList    1  5  10  maxIndex  maxValue   max enumerate theList   key lambda v  v 1       gt   2  10    In the docs for max   it says that the key argument expects a function like in the list sort   function  Also see the Sorting How To   It works the same for min    Btw it returns the first max min value

[python] Getting the index of the returned max or min item using max()/min() on a list

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