大列表，查找列表的所有最小值(python)

Question

给定一个大的波动值列表，如何确定所有本地最小值？不使用numpy。局部最小值意味着列表中的所有值都是函数的低谷。

List_y = [23, 8, -7, 57, 87, 6]

我想：

New_list = [-7, 6]

Answer 1

def local_min(ys):
    return [y for i, y in enumerate(ys)
            if ((i == 0) or (ys[i - 1] >= y))
            and ((i == len(ys) - 1) or (y < ys[i+1]))]


>>> local_min([23, 8, -7, 57, 87, 6])
[-7, 6]
>>> local_min([23, 6, 6, 6, 42])
[6]
>>> local_min([6, 6, 4])
[4]

Answer 2

我非常喜欢分阶段迭代这些问题。

l = [23, 8, -7, -7, 57, 87, 6]

# Remove identical neighbors
# l becomes [23, 8, -7, 57, 87, 6]
l = [x for x,y in zip(l[0:], l[1:]) if x != y] + [l[-1]]

# Append positive infinity to both endpoints
# l becomes [inf, 23, 8, -7, 57, 87, 6, inf]
l = [float("inf")] + l + [float("inf")]

# Retain elements where each of their neighbors are greater than them.
# l becomes [-7, 6]
l = [y for x, y, z in zip(l[0:], l[1:], l[2:]) if x > y < z]

Answer 3

在这里你可以很容易理解数组排序..。

my_array = [10,20,30,5,1,8,2,14,6,29] #main array
if len(my_array) > 0:
    my_array.sort() #sorting array(min to max)
    min = my_array[0] #after sorting get min value
    print(min) #print min value
else:
    print("array is empty")