[python] Check if all elements in a list are identical

I need a function which takes in a list and outputs True if all elements in the input list evaluate as equal to each other using the standard equality operator and False otherwise.

I feel it would be best to iterate through the list comparing adjacent elements and then AND all the resulting Boolean values. But I'm not sure what's the most Pythonic way to do that.

Regarding using reduce() with lambda. Here is a working code that I personally think is way nicer than some of the other answers.

reduce(lambda x, y: (x[1]==y, y), [2, 2, 2], (True, 2))

Returns a tuple where the first value is the boolean if all items are same or not.

You can convert the list to a set. A set cannot have duplicates. So if all the elements in the original list are identical, the set will have just one element.

if len(set(input_list)) == 1:
    # input_list has all identical elements.

For what it's worth, this came up on the python-ideas mailing list recently. It turns out that there is an itertools recipe for doing this already:¹

def all_equal(iterable):
    "Returns True if all the elements are equal to each other"
    g = groupby(iterable)
    return next(g, True) and not next(g, False)

Supposedly it performs very nicely and has a few nice properties.

1. Short-circuits: It will stop consuming items from the iterable as soon as it finds the first non-equal item.
2. Doesn't require items to be hashable.
3. It is lazy and only requires O(1) additional memory to do the check.

^{1In other words, I can't take the credit for coming up with the solution -- nor can I take credit for even finding it.}

A solution faster than using set() that works on sequences (not iterables) is to simply count the first element. This assumes the list is non-empty (but that's trivial to check, and decide yourself what the outcome should be on an empty list)

x.count(x[0]) == len(x)

some simple benchmarks:

>>> timeit.timeit('len(set(s1))<=1', 's1=[1]*5000', number=10000)
1.4383411407470703
>>> timeit.timeit('len(set(s1))<=1', 's1=[1]*4999+[2]', number=10000)
1.4765670299530029
>>> timeit.timeit('s1.count(s1[0])==len(s1)', 's1=[1]*5000', number=10000)
0.26274609565734863
>>> timeit.timeit('s1.count(s1[0])==len(s1)', 's1=[1]*4999+[2]', number=10000)
0.25654196739196777

Doubt this is the "most Pythonic", but something like:

>>> falseList = [1,2,3,4]
>>> trueList = [1, 1, 1]
>>> 
>>> def testList(list):
...   for item in list[1:]:
...     if item != list[0]:
...       return False
...   return True
... 
>>> testList(falseList)
False
>>> testList(trueList)
True

would do the trick.

lambda lst: reduce(lambda a,b:(b,b==a[0] and a[1]), lst, (lst[0], True))[1]

The next one will short short circuit:

all(itertools.imap(lambda i:yourlist[i]==yourlist[i+1], xrange(len(yourlist)-1)))

Check if all elements equal to the first.

np.allclose(array, array[0])

This is another option, faster than len(set(x))==1 for long lists (uses short circuit)

def constantList(x):
    return x and [x[0]]*len(x) == x

I'd do:

not any((x[i] != x[i+1] for i in range(0, len(x)-1)))

as any stops searching the iterable as soon as it finds a True condition.

Or use diff method of numpy:

import numpy as np
def allthesame(l):
    return np.unique(l).shape[0]<=1

And to call:

print(allthesame([1,1,1]))

Output:

True

This is a simple way of doing it:

result = mylist and all(mylist[0] == elem for elem in mylist)

This is slightly more complicated, it incurs function call overhead, but the semantics are more clearly spelled out:

def all_identical(seq):
    if not seq:
        # empty list is False.
        return False
    first = seq[0]
    return all(first == elem for elem in seq)

The simple solution is to apply set on list

if all elements are identical len will be 1 else greater than 1

lst = [1,1,1,1,1,1,1,1,1]
len_lst = len(list(set(lst)))

print(len_lst)

1


lst = [1,2,1,1,1,1,1,1,1]
len_lst = len(list(set(lst)))
print(len_lst)

2

>>> a = [1, 2, 3, 4, 5, 6]
>>> z = [(a[x], a[x+1]) for x in range(0, len(a)-1)]
>>> z
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
# Replacing it with the test
>>> z = [(a[x] == a[x+1]) for x in range(0, len(a)-1)]
>>> z
[False, False, False, False, False]
>>> if False in z : Print "All elements are not equal"

Or use diff method of numpy:

import numpy as np
def allthesame(l):
    return np.all(np.diff(l)==0)

And to call:

print(allthesame([1,1,1]))

Output:

True

def allTheSame(i):
    j = itertools.groupby(i)
    for k in j: break
    for k in j: return False
    return True

Works in Python 2.4, which doesn't have "all".

There is also a pure Python recursive option:

def checkEqual(lst):
    if len(lst)==2 :
        return lst[0]==lst[1]
    else:
        return lst[0]==lst[1] and checkEqual(lst[1:])

However for some reason it is in some cases two orders of magnitude slower than other options. Coming from C language mentality, I expected this to be faster, but it is not!

The other disadvantage is that there is recursion limit in Python which needs to be adjusted in this case. For example using this.

Can use map and lambda

lst = [1,1,1,1,1,1,1,1,1]

print all(map(lambda x: x == lst[0], lst[1:]))

If you're interested in something a little more readable (but of course not as efficient,) you could try:

def compare_lists(list1, list2):
    if len(list1) != len(list2): # Weed out unequal length lists.
        return False
    for item in list1:
        if item not in list2:
            return False
    return True

a_list_1 = ['apple', 'orange', 'grape', 'pear']
a_list_2 = ['pear', 'orange', 'grape', 'apple']

b_list_1 = ['apple', 'orange', 'grape', 'pear']
b_list_2 = ['apple', 'orange', 'banana', 'pear']

c_list_1 = ['apple', 'orange', 'grape']
c_list_2 = ['grape', 'orange']

print compare_lists(a_list_1, a_list_2) # Returns True
print compare_lists(b_list_1, b_list_2) # Returns False
print compare_lists(c_list_1, c_list_2) # Returns False

The simplest and most elegant way is as follows:

all(x==myList[0] for x in myList)

(Yes, this even works with the empty list! This is because this is one of the few cases where python has lazy semantics.)

Regarding performance, this will fail at the earliest possible time, so it is asymptotically optimal.

You can do:

reduce(and_, (x==yourList[0] for x in yourList), True)

It is fairly annoying that python makes you import the operators like operator.and_. As of python3, you will need to also import functools.reduce.

(You should not use this method because it will not break if it finds non-equal values, but will continue examining the entire list. It is just included here as an answer for completeness.)

You can use .nunique() to find number of unique items in a list.

def identical_elements(list):
    series = pd.Series(list)
    if series.nunique() == 1: identical = True
    else:  identical = False
    return identical



identical_elements(['a', 'a'])
Out[427]: True

identical_elements(['a', 'b'])
Out[428]: False

A set comparison work:

len(set(the_list)) == 1

Using set removes all duplicate elements.