How to smooth list of numerical values in Python?-CodePudding

I have a collection of lists of integer values in python like the following:

[0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 1, 2, 1]

Now I would like to have a somewhat "smoothed" sequence where each value with the same preceding and following value (which both differ from the central value in question) is replaced with this preceeding-following value. So my list above becomes:

[0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 1, 1, 1]

(The order or procession is from left to right, just to reconcile possible conflicting groupings.)

How could I achieve list?

Bonus: same as above with possible parametrization how many preceeding-following values must occur to change the central value (2-2 or 3-3 instead of just 1-1).

CodePudding user response：

arr = [0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 1, 2, 1]

res = [arr[0]]
i = 0

for i in range(1,len(arr)):

    if res[i-1] not in arr[i:i 2]:
        res.append(arr[i])

    else:
        res.append(res[i-1] )
print(res)

CodePudding user response：

A straightforward loop should do the trick:

_list = [0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 1, 2, 1]

for i in range(1, len(_list)-1):
    if _list[i-1] == _list[i 1]:
        _list[i] = _list[i-1]

print(_list)

Output:

[0, 0, 0, 0, 0, 0, 0, 2, 1, 1, 1, 1, 1]

CodePudding user response：

To allow the number of preceding / following values to be changed, you can create a 'pad' the list and iterate through a moving window on the padded list to check if all surrounding values are the same.

def smooth(lst, values=1, padding=None):
    padded = [padding] * values   lst   [padding] * values
    for i, n in enumerate(lst):
        surrounding = set(padded[i:i values]   padded[i values 1:i values*2 1])
        if len(surrounding) == 1:
            yield surrounding.pop()
        else:
            yield n

print(list(smooth([0, 0, 1, 0, 1, 0, 0, 2, 1, 1, 1, 2, 1])))  # [0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 1, 1, 1]

If your input list may contain None, choose a different padding parameter when calling the generator.