# 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.

``````if is_min_level:
return values.index(min(values))
else:
return values.index(max(values))
``````

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).

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.

`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.

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

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)
``````

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)
``````
``````seq=[1.1412, 4.3453, 5.8709, 0.1314]
seq.index(min(seq))
``````

Will give you first index of minimum.

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)
``````

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>3)&(x<60))[0]
idx
array([2, 3, 4, 5])
x[idx]
array([ 4,  7,  4, 56])
``````

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 encountered, 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]
>(0, 17)

max_val_index[0]
>(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)]
``````

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]
``````

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])
# => (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.

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 ] )
``````

Simple as that :

``````stuff = [2, 4, 8, 15, 11]

index = stuff.index(max(stuff))
``````

I was also interested in this and compared some of the suggested solutions using perfplot (a pet project of mine).

It turns out that

``````min(range(len(a)), key=a.__getitem__)
``````

is the fastest method for small and large lists.

(In former versions, `np.argmin` used to take the cake.)

Code for generating the plot:

``````import numpy as np
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 np.argmin(a)

b = perfplot.bench(
setup=lambda n: np.random.rand(n).tolist(),
kernels=[
min_enumerate,
min_enumerate_itemgetter,
getitem,
np_argmin,
],
n_range=[2**k for k in range(15)],
)
b.show()
``````

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.

Use a numpy array and the argmax() function

`````` a=np.array([1,2,3])
b=np.argmax(a)
print(b) #2
``````

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 µs ± 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 µs ± 78.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
``````

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)
``````

``````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.

Pandas has now got a much more gentle solution, try it:

`df[column].idxmax()`

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.

``````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``````

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