Here is the code:
import numpy as np
array = np.array([15, 55, 9, 99, 8, 21, 2, 90, 88])
this will output an array([ 15, 55, 9, 99, 8, 21, 2, 90, 88]).
How can I find the first minimum number without sorting then the second minimum number?
I expect the output to be:
first min = 9
second min = 8
You can find the absolute minimum like this:
In [35]: import numpy as np
In [36]: arr = np.array([15, 55, 9, 99, 8, 21, 2, 90, 88])
In [37]: first = np.min(arr)
In [38]: second = np.min(arr[arr != first])
In [39]: first
Out[39]: 2
In [40]: second
Out[40]: 8
To obtain the indices of the local minima, you could use scipy.signal.argrelmin:
In [52]: from scipy.signal import argrelmin
In [53]: idx = argrelmin(arr)[0]
In [54]: idx
Out[54]: array([2, 4, 6], dtype=int64)
In [55]: arr[idx]
Out[55]: array([9, 8, 2])
You could offset the list and zip them:
l0 = [15, 55, 9, 99, 8, 21, 2, 90, 88]
l1 = l0[1:]
l2 = [-1] + l0
[x for x,y,z in zip(l0,l1,l2) if (x < y) & (x < z)]
# Out[32]: [9, 8, 2]
or in one line:
l = [15, 55, 9, 99, 8, 21, 2, 90, 88]
[x for x,y,z in zip(l,l[1:],[-1]+l) if (x < y) & (x < z)]
# Out[32]: [9, 8, 2]
