Predict the next number in any sequence

Yesterday, I came up with a simple method to predict the next value in a sequence.

The method works like this:

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Start with a sequence, say 1,4,9,16,25,36, call it Δ0.

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Now, Δ1 is the difference between every adjacent element in Δ0. (The first element is left unchanged). For our chosen sequence, this is 1,3,5,7,9,11.

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In general, Δn is the difference between every adjacent element in Δn-1.

┌──────────────────┐
│Δ0: 1,4,9,16,25,36│
│Δ1: 1,3,5,7,9,11  │
│Δ2: 1,2,2,2,2,2   │
│Δ3: 1,2,0,0,0,0   │
│Δ4: 1,1,0,0,0,0   │
│Δ5: 1,0,-1,0,0,0  │
│Δ6: 1,-1,-1,1,0,0 │
│...               │
└──────────────────┘

Now we pick first Δn where the sum of the absolute of each value is less than than sum of the absolute values of Δn+1 . In this case it is Δ5: 1,0,-1,0,0,0. Now we duplicate the last value. (1,0,-1,0,0,0,0). Now we repeatedly take the running sums of this list n times, successfully undoing all the "difference between every adjacent element" function, but because we have an extra element, we will have an extra element in this new sequence. In our chosen sequence, this new sequence is 1,4,9,16,25,36,49.

Another example would be the sequence 2,5,3,9,6,2,3(unlike the previous sequence, this one doesn't follow a clear pattern). The sum of the absolutes of this sequence is 30. Δ1 of this sequence is 2,3,-2,6,-3,-4,1. The sum of the absolutes this time is 21. We continue, Δ2 is 2,1,-5,8,-9,-1,5. The sum of the absolutes is 31. Now, we can see that Δ1 is the first with a smaller absolute sum than the next Δ in the series. Now, we duplicate the last value of Δ1, giving 2,3,-2,6,-3,-4,1,1. Since this is Δ1, we take the running sums 1 time giving 2,5,3,9,6,2,3,4 as a final result.

Here's my code.

```
def runningSums(lst):
res = [lst[0]]
for elem in lst[1:]:
res.append(res[-1] + elem)
return res
def antiRunningSums(lst):
res = [lst[0]]
for i in range(1,len(lst)):
res.append(lst[i] - lst[i-1])
return res
de

antiRunningSums

You should use a pairwise helper to raise abstraction:

def antiRunningSums(lst):
    return [b - a for a, b in pairwise(lst)]

pairwise can easily be found in the itertools documentation.

runningSums

I think that referencing the list while you are building it is overcomplicated, I would use a variable to be incremented:

def runningSums(lst):
    s = 0
    for addend in lst:
        s += addend
        yield s

To obtain a list call list on the result of this function.

StackExchange Code Review Q#134202, answer score: 5