How to calculate a list based on multiple results from items in a list?-CodePudding

I have a list A=[a,b,c,d]. I need to calculate a new list B based on operations between each item in A.

B= [a, b-(a), c-(a (b-a)), d-(a (b-a) (c-(a (b-a)))) ]

Is there a Pythonic way of doing this? List A is not always a 4 item list, so the solution needs to be generalizable to lists of arbitrary length. Thanks in advance.

CodePudding user response：

All your terms cancel out (c-(a (b-a)) simplifies to c - b, d-(a (b-a) (c-(a (b-a)))) simplifies to d - c), so the real algorithm here is that each term is equal to the matching term minus the prior term. This simplifies things dramatically:

B = [A[0]]  # Initial term has no prior to subtract from it
B  = [x - y for x, y in zip(A[1:], A)]  # All other terms computed by subtracting term n - 1 from term n

If you want to one-line this (ignoring imports) you can stick a virtual 0 in to get the results for the first element without explicitly special-casing it:

from itertools import chain  # At top of file

B = [x - y for x, y in zip(A, chain([0], A))]

If you love using map and friends for microoptimizations, you could replace the latter with:

from operator import sub  # At top of file

B = [*map(sub, A, chain([0], A))]

and push all the work to the C layer (no per-element bytecode execution).

CodePudding user response：

Observe that the expression for your list can be simplified to:

B = [a, b-a, c-b, d-c]

With this in mind, we can use a list comprehension:

[y - x for x, y in zip([0]   data, data)]

For example,

data = [1, 2, 7, 6]
result = [y - x for x, y in zip([0]   data, data)]
print(result)

outputs:

[1, 1, 5, -1]

CodePudding user response：

As suggest by the guys on the comments, this is the simplest and fastest solution:

A = [5, 9, 3, 8]
B = [x - y for x, y in zip(A, [0]   A)]

This outputs:

B
[5, 4, -6, 5]

CodePudding user response：

Pythonic way:-

A = [1,2,3,4]
B = [A[0]] [A[i 1]-A[i] for i in range(len(A)-1)]