if my list is in the following format:
a=[1,3,4,5,8,10]
How can I show that the elements are not increasing by 1? In other words, if my list is
a=[1,2,3,4,5,6]
the function should say that it is sequential i.e. each element is greater than the previous element by one. Thanks in advance.
One simple way to do the testing is
if all(x[i] == x[0]+i for i in range(1, len(x))):
...
>>> a=[1,2,3,4,5,6]
>>> all(i==j for i,j in enumerate(a, a[0]))
True
>>> a=[1,3,4,5,8,10]
>>>
>>> all(i==j for i,j in enumerate(a, a[0]))
False
a [0]. It would complain with floats but the OP is consifering integers..You can use zip and set:
def check_sequential(a):
return set([p - q for p, q in zip(a[1:], a)]) == {1}
print check_sequential([1,3,4,5,8,10])
print check_sequential([1,2,3,4,5])
Output:
False
True
Use numpy.
Differentiate it
Use Max.
min too in case the sequence isn't increasingFor short lists, you can just create a sequential range yourself and compare them
>>> a = [1,2,3,4,5,6]
>>> a == range(a[0], a[-1] + 1) # in Python3, you need list(range(...))
True
>>> a = [1,3,4,5,8,10]
>>> a == range(a[0], a[-1] + 1)
False