How to get local min in a list?

user_8085980 0
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Can anyone suggest a method to get local min in a list?

For example, I have a list [50, 50, 49, 49, 49, 50, 50, 50, 49, 49, 49, 48, 48, 48, 47, 47, 48, 48, 13, 12, 11, 10, 11, 12].

I want to get the answer [49, 47, 10], and the index.

I had tried, but I get the answer like [49, 50, 47, 48, 10, 11, 12].

Can someone help me to revise the code?

Thanks a lot.

for i in range(len(list)):
    if (i != 0) & (i+1 != len(list)):
        if (list[i] == list[i + 1]):
            repeat += 1
        elif (list[i] < list[i - 1]):
            repeat = 1
            continue
        elif (list[i] < list[i + 1]):
            group_list.append(list[i])
list python
2 Answers

You could loop through the list and find places where it changes from decreasing to increasing.

def local_minima(lst):
    mins = []
    decreasing = True
    for i in range(len(lst) - 1):
        a = lst[i]
        b = lst[i + 1]
        if not decreasing and b < a:
            decreasing = True
        elif decreasing and b > a:
            mins.append(a)
            decreasing = False
    return mins

>>> local_minima([50, 50, 49, 49, 49, 50, 50, 50, 49, 49, 49, 48, 48, 48, 47, 47, 48, 48, 13, 12, 11, 10, 11, 12])
[49, 47, 10]
user_55870010

Note: Please avoid using list as a name for your list. list is a predefined keyword in Python. You can use or to check if you are at the beginning or end of the list and at the same time compare values to their left and right. Assuming your list name is l, I think you want this:

mins = [x for i,x in enumerate(l) if ((i==0) or (l[i-1]>=x)) and ((i==len(l)-1) or (x<l[i+1]))]

output:

[49, 47, 10]
user_49759810
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