Replace group of value in an array

Question

I have an array, lets say this:

arr = ["60", "DD", "81", "01" , "01", "29", "B8", "1B", "00", "30", "2C", "46", "00", "0A", "81", "02" , "0D", "25", "00", "37", "41", "31", "00", "C2", "7F", "06", "00", "17", "94", "1A", "00", "48", "06", "05", "00", "5C", "7F", "3E", "87", "FF", "0F", "B8", "0A", "38", "0C"]

I am trying to replace every occurance of "81", "01" with "81" and "81", "02" with "82" . I tried but it not replacing the values appropriately. Here is my code.

import numpy as np
values = np.array(arr)
searchval = ["81", "01"]
N = len(searchval)
possibles = np.where(values == searchval[0])[0]

solns = []
for p in possibles:
    check = values[p:p+N]
    if np.all(check == searchval):
        arr.pop(p+1)
        solns.append(p)

print(solns)

It would be great if someone can help me solving this. Thank you.

Answer 1

Given your two character strings, you could convert the list to string and do replacements with str.replace then split to return the transformed list:

s = ' '.join(arr)
s = s.replace('81 01', '81')
s = s.replace('81 02', '82')
print s.split()
# ['60', 'DD', '81', '01', '29', 'B8', '1B', '00', '30', '2C', '46', '00', '0A', '82', '0D', '25', '00', '37', '41', '31', '00', 'C2', '7F', '06', '00', '17', '94', '1A', '00', '48', '06', '05', '00', '5C', '7F', '3E', '87', 'FF', '0F', 'B8', '0A', '38', '0C']

Not very efficient but qiute concise and readable.

Answer 2

For an approach that works just with the list, and would generalize to other kinds of values:

solns = [arr[0]]
for i, entry in enumerate(arr[:-1]):
    if entry == '81':
        if arr[i+1] == '02':
            solns[-1] = '82'
        elif not arr[i+1] == '01':
            solns.append(arr[i+1])
    else:
        solns.append(arr[i+1])

Or, if you prefer:

solns = [arr[0]]
for i, entry in enumerate(arr[:-1]):
    if entry == '81':
        if arr[i+1] == '02':
            solns[-1] = '82'
            continue
        elif arr[i+1] == '01':
            continue
    solns.append(arr[i+1])