Identify if list has consecutive elements that are equal in python

Question

I\'m trying to identify if a large given list has consecutive elements that are the same.

So let\'s say

lst = [1, 2, 3, 4, 5, 5, 6]

Answer 1

You can use itertools.groupby() and a generator expression within any() ^*:

>>> from itertools import groupby
>>> any(sum(1 for _ in g) > 1 for _, g in groupby(lst))
True

Or as a more Pythonic way you can use zip(), in order to check if at least there are two equal consecutive items in your list:

>>> any(i==j for i,j in zip(lst, lst[1:])) # In python-2.x,in order to avoid creating a 'list' of all pairs instead of an iterator use itertools.izip()
True

Note: The first approach is good when you want to check if there are more than 2 consecutive equal items, otherwise, in this case the second one takes the cake!

---

^{* Using sum(1 for _ in g) instead of len(list(g)) is very optimized in terms of memory use (not reading the whole list in memory at once) but the latter is slightly faster.}

Answer 2

You can use a simple any condition:

lst = [1, 2, 3, 4, 5, 5, 6]
any(lst[i]==lst[i+1] for i in range(len(lst)-1))
#outputs:
True

any return True if any of the iterable elements are True

Answer 3

If you're looking for an efficient way of doing this and the lists are numerical, you would probably want to use numpy and apply the diff (difference) function:

>>> numpy.diff([1,2,3,4,5,5,6])
array([1, 1, 1, 1, 0, 1])

Then to get a single result regarding whether there are any consecutive elements:

>>> numpy.any(~numpy.diff([1,2,3,4,5,5,6]).astype(bool))

This first performs the diff, inverts the answer, and then checks if any of the resulting elements are non-zero.

Similarly,

>>> 0 in numpy.diff([1, 2, 3, 4, 5, 5, 6])

also works well and is similar in speed to the np.any approach (credit for this last version to heracho).

Answer 4

My solution for this if you want to find out whether 3 consecutive values are equal to 7. For example, a tuple of intList = (7, 7, 7, 8, 9, 1):

for i in range(len(intList) - 1):
        if intList[i] == 7 and intList[i + 2] == 7 and intList[i + 1] == 7:
            return True
    return False

Answer 5

Here a more general numpy one-liner:

number = 7
n_consecutive = 3
arr = np.array([3, 3, 6, 5, 8, 7, 7, 7, 4, 5])
#                              ^  ^  ^ 
 
np.any(np.convolve(arr == number, v=np.ones(n_consecutive), mode='valid') 
       == n_consecutive)[0]

This method always searches the whole array, while the approach from @Kasramvd ends when the condition is first met. So which method is faster dependents on how sparse those cases of consecutive numbers are. If you are interested in the positions of the consecutive numbers, and have to look at all elements of the array this approach should be faster (for larger arrays (or/and longer sequences)).

idx = np.nonzero(np.convolve(arr==number, v=np.ones(n_consecutive), mode='valid') 
                             == n_consecutive)
# idx = i: all(arr[i:i+n_consecutive] == number)

If you are not interested in a specific value but at all consecutive numbers in general a slight variation of @jmetz's answer:

np.any(np.convolve(np.diff(arr), v=np.ones(n_consecutive-1), mode='valid') == 0)

Answer 6

A simple for loop should do it:

def check(lst):
    last = lst[0]
    for num in lst[1:]:
        if num == last:
            return True
        last = num
    return False


lst = [1, 2, 3, 4, 5, 5, 6]
print (check(lst)) #Prints True

Here, in each loop, I check if the current element is equal to the previous element.