How to find the index of only one minimum value in a list containing more than one minimum value

Question

a = [2, 2, 4, 2, 5, 7]

If one is to find the minimum value in this list (which is 2) the corresponding indexes of 2 in the list a are 0, 1 and 3 respectively.

How to tell python to use the minimum value at the highest index (the number 2 at index 3)?

Answer 1

You can simply reverse the list using a[::-1] and then apply the same technique to find the index. a[::-1].index(min(a)) will give us the index of minimum value from the end, in order to record the index w.r.t to 0th position(starting of list) we can subtract a[::-1].index(min(a)) from len(a) - 1.

a = [2, 2, 4, 2, 5, 7]

print len(a) - a[::-1].index(min(a)) - 1

>>> 3

Answer 2

This can be written as a function indexes() and then take the max() of the resulting iterator:

Example:

def indexes(xs, value):
    for i, x in enumerate(xs):
        if x == value:
            yield i

a = [2, 2, 4, 2, 5, 7]
print max(indexes(a, min(a)))  # 3

Update; Just as a matter of completeness; if you per se wanted the minimum index for a set of values with more than one minimum you'd just swap out max() for min() at the front of the expression. So:

a = [2, 2, 1, 4, 2, 1, 5, 7]
print min(indexes(a, min(a)))  # 2

whilst:

print max(indexes(a, min(a)))  # 5

This is kind of handy and flexible in general :)

Answer 3

>>> -min(zip(a, range(0, -len(a), -1)))[1]
3

Answer 4

use enumerate reversing the list with a start index of 1 calling next in the generator to get the first value, subtracting the index from the length of the list.

a = [2, 2, 3, 2, 5, 7]
mn = min(a)

print(len(a) - next(i for i, ele in enumerate(reversed(a),1) if ele == mn))
 3

Answer 5

Iterate through the enumerated list and use the maximum index :

In [109]: max([ind for ind, val in enumerate(a) if val==min(a)])
Out[109]: 3

Answer 6

Find the smallest, then iterate over the list collecting the indices of the matching values.

smallest = min(items)
indices = [index for index, value in enumerate(items) if value == smallest]
largest_index = max(indices) # or indices[-1]

If you have a very long list then you might want to start from the back and stop once you reach the first item.

largest_index = len(items) - next(index for index, value in \
    enumerate(reversed(items), start=1) if value == smallest)

Answer 7

This might do:

(len(a) -1 )- a[: : -1].index(min(a))

Answer 8

Try this:

a=[2,2,4,2,5,7]

minval=min(a)  #min of a

for l in range(len(a)):

  if a[l]==minval:
    index=l
    continue
print index