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.
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6 Answers
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)
Comments
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
2 Comments
ecatmur
Like the higher order diff, I'd forgotten about that!
Belgin Fish
hey unfortunately I don't have access to numpy :( great answer though :)
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
1 Comment
Eric
A list comprehension would be worse, because it generates a list which is then thrown away (like your invocation of
range). However, none of these answers are list comprehensions - they're generator comprehensions.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
>>>
Comments
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
Comments
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]))
