How to find all positions of the maximum value in a list?

Question

I have a list:

a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50,
             35, 41, 49, 37, 19, 40, 41, 31]

max element is 55 (two elements on position 9 and 12)

I need to find on which position(s) the maximum value is situated. Please, help.

Answer 1

a.index(max(a))

will tell you the index of the first instance of the largest valued element of list a.

Answer 2

>>> m = max(a)
>>> [i for i, j in enumerate(a) if j == m]
[9, 12]

Answer 3

The chosen answer (and most others) require at least two passes through the list.
Here's a one pass solution which might be a better choice for longer lists.

Edited: To address the two deficiencies pointed out by @John Machin. For (2) I attempted to optimize the tests based on guesstimated probability of occurrence of each condition and inferences allowed from predecessors. It was a little tricky figuring out the proper initialization values for max_val and max_indices which worked for all possible cases, especially if the max happened to be the first value in the list — but I believe it now does.

def maxelements(seq):
    ''' Return list of position(s) of largest element '''
    max_indices = []
    if seq:
        max_val = seq[0]
        for i,val in ((i,val) for i,val in enumerate(seq) if val >= max_val):
            if val == max_val:
                max_indices.append(i)
            else:
                max_val = val
                max_indices = [i]

    return max_indices

Answer 4

I came up with the following and it works as you can see with max, min and others functions over lists like these:

So, please consider the next example list find out the position of the maximum in the list a:

>>> a = [3,2,1, 4,5]

Using the generator enumerate and making a casting

>>> list(enumerate(a))
[(0, 3), (1, 2), (2, 1), (3, 4), (4, 5)]

At this point, we can extract the position of max with

>>> max(enumerate(a), key=(lambda x: x[1]))
(4, 5)

The above tells us, the maximum is in the position 4 and his value is 5.

As you see, in the key argument, you can find the maximum over any iterable object by defining a lambda appropriate.

I hope that it contributes.

PD: As @PaulOyster noted in a comment. With Python 3.x the min and max allow a new keyword default that avoid the raise exception ValueError when argument is empty list. max(enumerate(list), key=(lambda x:x[1]), default = -1)

Answer 5

Also a solution, which gives only the first appearance, can be achieved by using numpy:

>>> import numpy as np
>>> a_np = np.array(a)
>>> np.argmax(a_np)
9

Answer 6

I can't reproduce the @SilentGhost-beating performance quoted by @martineau. Here's my effort with comparisons:

=== maxelements.py ===

a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50,
             35, 41, 49, 37, 19, 40, 41, 31]
b = range(10000)
c = range(10000 - 1, -1, -1)
d = b + c

def maxelements_s(seq): # @SilentGhost
    ''' Return list of position(s) of largest element '''
    m = max(seq)
    return [i for i, j in enumerate(seq) if j == m]

def maxelements_m(seq): # @martineau
    ''' Return list of position(s) of largest element '''
    max_indices = []
    if len(seq):
        max_val = seq[0]
        for i, val in ((i, val) for i, val in enumerate(seq) if val >= max_val):
            if val == max_val:
                max_indices.append(i)
            else:
                max_val = val
                max_indices = [i]
    return max_indices

def maxelements_j(seq): # @John Machin
    ''' Return list of position(s) of largest element '''
    if not seq: return []
    max_val = seq[0] if seq[0] >= seq[-1] else seq[-1]
    max_indices = []
    for i, val in enumerate(seq):
        if val < max_val: continue
        if val == max_val:
            max_indices.append(i)
        else:
            max_val = val
            max_indices = [i]
    return max_indices

Results from a beat-up old laptop running Python 2.7 on Windows XP SP3:

>\python27\python -mtimeit -s"import maxelements as me" "me.maxelements_s(me.a)"
100000 loops, best of 3: 6.88 usec per loop

>\python27\python -mtimeit -s"import maxelements as me" "me.maxelements_m(me.a)"
100000 loops, best of 3: 11.1 usec per loop

>\python27\python -mtimeit -s"import maxelements as me" "me.maxelements_j(me.a)"
100000 loops, best of 3: 8.51 usec per loop

>\python27\python -mtimeit -s"import maxelements as me;a100=me.a*100" "me.maxelements_s(a100)"
1000 loops, best of 3: 535 usec per loop

>\python27\python -mtimeit -s"import maxelements as me;a100=me.a*100" "me.maxelements_m(a100)"
1000 loops, best of 3: 558 usec per loop

>\python27\python -mtimeit -s"import maxelements as me;a100=me.a*100" "me.maxelements_j(a100)"
1000 loops, best of 3: 489 usec per loop

Answer 7

You can also use the numpy package:

import numpy as np
A = np.array(a)
maximum_indices = np.where(A==max(a))

This will return an numpy array of all the indices that contain the max value

if you want to turn this to a list:

maximum_indices_list = maximum_indices.tolist()

Answer 8

a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 
         55, 23, 31, 55, 21, 40, 18, 50,
         35, 41, 49, 37, 19, 40, 41, 31]

import pandas as pd

pd.Series(a).idxmax()

9

That is how I usually do it.

Answer 9

>>> max(enumerate([1,2,3,32,1,5,7,9]),key=lambda x: x[1])
>>> (3, 32)

Answer 10

@shash answered this elsewhere

A Pythonic way to find the index of the maximum list element would be

position = max(enumerate(a), key=lambda x: x[1])[0]

Which does one pass. Yet, it is slower than the solution by @Silent_Ghost and, even more so, @nmichaels:

for i in s m j n; do echo $i;  python -mtimeit -s"import maxelements as me" "me.maxelements_${i}(me.a)"; done
s
100000 loops, best of 3: 3.13 usec per loop
m
100000 loops, best of 3: 4.99 usec per loop
j
100000 loops, best of 3: 3.71 usec per loop
n
1000000 loops, best of 3: 1.31 usec per loop

Answer 11

Just one line:

idx = max(range(len(a)), key = lambda i: a[i])

Answer 12

Similar idea with a list comprehension but without enumerate

m = max(a)
[i for i in range(len(a)) if a[i] == m]

Answer 13

Here is the max value and the indexes it appears at:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> for i, x in enumerate(a):
...     d[x].append(i)
... 
>>> k = max(d.keys())
>>> print k, d[k]
55 [9, 12]

---

Later: for the satisfaction of @SilentGhost

>>> from itertools import takewhile
>>> import heapq
>>> 
>>> def popper(heap):
...     while heap:
...         yield heapq.heappop(heap)
... 
>>> a = [32, 37, 28, 30, 37, 25, 27, 24, 35, 55, 23, 31, 55, 21, 40, 18, 50, 35, 41, 49, 37, 19, 40, 41, 31]
>>> h = [(-x, i) for i, x in enumerate(a)]
>>> heapq.heapify(h)
>>> 
>>> largest = heapq.heappop(h)
>>> indexes = [largest[1]] + [x[1] for x in takewhile(lambda large: large[0] == largest[0], popper(h))]
>>> print -largest[0], indexes
55 [9, 12]

Answer 14

If you want to get the indices of the largest n numbers in a list called data, you can use Pandas sort_values:

pd.Series(data).sort_values(ascending=False).index[0:n]

Answer 15

Here's a simple single-pass solution.

import math
nums = [32, 37, 28, 30, 37, 25, 55, 27, 24, 35, 55, 23, 31]

max_val = -math.inf
res = []

for i, val in enumerate(nums):
    if(max_val < val):
        max_val = val
        res = [i]
    elif(max_val == val):
        res.append(i)
print(res)

Answer 16

This code is not as sophisticated as the answers posted earlier but it will work:

m = max(a)
n = 0    # frequency of max (a)
for number in a :
    if number == m :
        n = n + 1
ilist = [None] * n  # a list containing index values of maximum number in list a.
ilistindex = 0
aindex = 0  # required index value.    
for number in a :
    if number == m :
        ilist[ilistindex] = aindex
        ilistindex = ilistindex + 1
    aindex = aindex + 1

print ilist

ilist in the above code would contain all the positions of the maximum number in the list.

Answer 17

import operator

def max_positions(iterable, key=None, reverse=False):
  if key is None:
    def key(x):
      return x
  if reverse:
    better = operator.lt
  else:
    better = operator.gt

  it = enumerate(iterable)
  for pos, item in it:
    break
  else:
    raise ValueError("max_positions: empty iterable")
    # note this is the same exception type raised by max([])
  cur_max = key(item)
  cur_pos = [pos]

  for pos, item in it:
    k = key(item)
    if better(k, cur_max):
      cur_max = k
      cur_pos = [pos]
    elif k == cur_max:
      cur_pos.append(pos)

  return cur_max, cur_pos

def min_positions(iterable, key=None, reverse=False):
  return max_positions(iterable, key, not reverse)

>>> L = range(10) * 2
>>> L
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> max_positions(L)
(9, [9, 19])
>>> min_positions(L)
(0, [0, 10])
>>> max_positions(L, key=lambda x: x // 2, reverse=True)
(0, [0, 1, 10, 11])

Answer 18

You can do it in various ways.

The old conventional way is,

maxIndexList = list() #this list will store indices of maximum values
maximumValue = max(a) #get maximum value of the list
length = len(a)       #calculate length of the array

for i in range(length): #loop through 0 to length-1 (because, 0 based indexing)
    if a[i]==maximumValue: #if any value of list a is equal to maximum value then store its index to maxIndexList
        maxIndexList.append(i)

print(maxIndexList) #finally print the list

Another way without calculating the length of the list and storing maximum value to any variable,

maxIndexList = list()
index = 0 #variable to store index
for i in a: #iterate through the list (actually iterating through the value of list, not index )
    if i==max(a): #max(a) returns a maximum value of list.
        maxIndexList.append(index) #store the index of maximum value
index = index+1 #increment the index

print(maxIndexList)

We can do it in Pythonic and smart way! Using list comprehension just in one line,

maxIndexList = [i for i,j in enumerate(a) if j==max(a)] #here,i=index and j = value of that index

All my codes are in Python 3.