How do I find ALL the positions of the lowest numbers in a list? (python)

Question

The following code below returns the position of the first lowest number in a list.

xy = [50, 2, 34, 6, 4, 3, 1, 5, 2, 1, 10 ,1] 
t=0
for i in range(len(xy)):
    if xy[i]<xy[t]:        
        t=i
print(t)

out: 6

I would like to get the positions of all the lowest numbers. In this case it should be 6,9,11. How do I go about in base Python?

Answer 1

You could use Python's predefined min function to get the minimum value in the list, then get the indices of the values equal to that minimum using a list comprehension, like this:

xy = [50, 2, 34, 6, 4, 3, 1, 5, 2, 1, 10 ,1]
lowest = min(xy)
positions = [i for i, v in enumerate(xy) if v == lowest]
print(positions) # ==> [6, 9, 11]

Answer 2

This should be the most CPU efficient solution because it runs the list only once. @Mr Geek's solution should be slightly better on memory because his way only produces one list of results.

xy = [50, 2, 34, 6, 4, 3, 1, 5, 2, 1, 10 ,1]
min_val = sys.maxsize
result = []
for index, num in enumerate(xy):
  if num < min_val:
    min_val = num
    result = [index]
  elif num == min_val:
    result.append(index)

result now contains [6, 9, 11]

Whenever we find a new minimum value, we clear the list of results and add the new number. Whenever we find the same number, we add it to the list.