How to solve array problem to check if element is greater, less than or equal its neighbor-CodePudding

I'm still learning more about programming and have a problem with my code.

I have an array

data = [5,4,4,4,4,3,3,8] the expected result should be P,n,n,n,n,n,v,p. but I'm getting this p,n,n,n,p,n,n,p

#data = [2,1,4,5,5,5,4] expected result p,v,n,n,n,p,v (my code works for this. but the code must be able to solve the two)

I want to assign them to Peak(P) and Valley(V).

where the peak is the high number and Valley is the lower number.

def get_Ps_values(data):
    dt=[]
    details=[]
    n = len(data)
    #Handling first element in the data
    #We check if first element is less or equal to the next element
    if (data[0] < data[1]):
        #if there is, we append the element to the list
        dt.append(data[0])
        details.append('V')
        
    elif (data[0] > data[1]):
        dt.append(data[0])
        details.append('P')
        
    else:
        #this is checking if the first element is a P
        dt.append(data[1])
        details.append('N')
   #To handle other elements in the data that are not first and last element
    for i in range(1, n-1):
            if data[i] == data[i-1]==data[i 1]:
                dt.append(data[i])
                details.append('N')
            elif data[i] < data[i-1]<data[i 1]:
                dt.append(data[i])
                details.append('V')
            elif i 1 > n-1 and data[i 1] > data[i-1]:

                dt.append(data[i])
                details.append('P')
            elif (i == 1 or data[i-1] == data[i]) and (i == n-1 or data[i] > data[i 1]):  # Found peak
                dt.append(data[i])
                details.append('P')
            elif i-1 > n-1 and data[i 1] < data[i-1]:

                dt.append(data[i])
                details.append('V')
            else:
                dt.append(data[i])
                details.append('N')
    
    #Handling last element in the data
    #We check if the last element is greater or equal to the element before it.
    if (data[-1] > data[-2]):
        #if there is, we append the element to the list
        dt.append(data[-1])
        details.append('P')
        
    elif(data[-1] < data[-2]):
        #if there is, we append the element to the list
        dt.append(data[-1])
        details.append('N')
    elif(data[-1] >= data[-2]):
        #if there is, we append the element to the list
        dt.append(data[-1])
        details.append('V')
   
        #this is checking if the last element is a V
        dt.append(data[-1])
        details.append('N')

    return dt, details

Thanks

CodePudding user response：

Assuming:

that a peak is the last point of a stretch of identical values if it is strictly higher than the previous and next stretches (or of only one neighbor on the ends)
that is valley is the same for a strictly lower value compared to the neighbor(s)
all other points being "n"

You can use itertools.groupby to compare the stretches to the previous and next ones:

from itertools import groupby

def pvn(data):
    out = []
    pos = 0
    prev = None
    for k, g in groupby(data):
        g = list(g)
        pos = pos   len(g)
        next_ = data[pos] if pos < len(data) else None
        if (prev is None or k > prev) and (next_ is None or k > next_):
            out.extend(['n']*(len(g)-1) ['p'])
        elif (prev is None or k < prev) and (next_ is None or k < next_):
            out.extend(['n']*(len(g)-1) ['v'])
        else:
            out.extend(['n']*len(g))
        prev = k
    return out

pvn([5, 4, 4, 4, 4, 3, 3, 8])
# ['p', 'n', 'n', 'n', 'n', 'n', 'v', 'p']

pvn([2, 1, 4, 5, 5, 5, 4])
# ['p', 'v', 'n', 'n', 'n', 'p', 'v']