Verifying arithmetic sequence python

Question

i'm just wondering how I could check to verify that a list of numbers is arithmetic or not using python, so whether there is a common number in between each item in the list.

Answer 1

This is what I'd write:

all((i - j) == (j - k) for i, j, k in zip(l[:-2], l[1:-1], l[2:]))

You could probably make it more efficient by only computing the differences once, but if you're concerned about efficiency you'd use numpy and write:

np.all((a[:-2] - a[1:-1]) == (a[1:-1] - a[2:]))

or even (saving a slice):

np.all(a[:-2] + a[2:] == 2 * a[1:-1])

Probably the most concise method is to use numpy.diff, as it will automatically convert a list into a numpy array:

np.all(np.diff(l, 2) == 0)

Answer 2

You can use numpy.diff if you have access to numpy:

>>> a = numpy.array(range(1, 15, 2))
>>> numpy.diff(a)
array([2, 2, 2, 2, 2, 2])

So you can do

>>> d = numpy.diff(a)
>>> not numpy.any(d-d[0])
True

or even better

>>> not numpy.any(numpy.diff(a, 2))
True

Answer 3

If you mean an arithmetic sequence as in a series of numbers such that each number is simply the equal to the previous one plus some constant amount (like [1, 3, 5, 7] or [8, 18, 28, 38] but not [1, 2, 4, 8] or [1, 3, 1, 5, 1, 7]) then you probably shouldn't overthink it. It is unlikely that a list comprehension outperforms this:

def is_arithmetic(l):
    delta = l[1] - l[0]
    for index in range(len(l) - 1):
        if not (l[index + 1] - l[index] == delta):
             return False
    return True

Answer 4

Here's a solution that only calculates the difference once:

from itertools import izip

def is_arithmetic(seq):
    gen = (i - j for i, j in izip(seq[:-1], seq[1:]))
    diff = next(gen, None)  # get the first element in the generator
    return all(d == diff for d in gen) # check all the others are equal to it

Or more cryptically:

def is_arithmetic(seq):
    gen = (i - j for i, j in izip(seq[:-1], seq[1:]))
    return all(d == d0 for d in gen for d0 in gen)  # wat

Answer 5

What about examining the length of the set of all differences in the list?

>>> alist = [3,9,15,21,27,33,39,45,51,57]
>>> blist = [-7, -2, 3, 8, 13, 18, 23, 29]
>>> 1 == len(set([alist[x + 1] - alist[x] for x in range(len(alist) - 1)]))
True
>>> 1 == len(set([blist[x + 1] - blist[x] for x in range(len(blist) - 1)]))
False
>>> 

Answer 6

def check_arith(lst):

    l1 = len(lst) - 1
    n= 2

    dif = lst[1] - lst[0]

    while(n<l1):

        if (lst[n+1] - lst[n]) != dif: 
            return False
        else:
            n = n + 1
    return True

print(check_arith([5,10,15, 20, 25]))