python - Check if all elements in a list are identical

Question

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.

Answer 1

Use itertools.groupby (see the itertools recipes):

from itertools import groupby

def all_equal(iterable):
    g = groupby(iterable)
    return next(g, True) and not next(g, False)

or without groupby:

def all_equal(iterator):
    iterator = iter(iterator)
    try:
        first = next(iterator)
    except StopIteration:
        return True
    return all(first == x for x in iterator)

---

There are a number of alternative one-liners you might consider:

1. Converting the input to a set and checking that it only has one or zero (in case the input is empty) items

   def all_equal2(iterator):
       return len(set(iterator)) <= 1
   

2. Comparing against the input list without the first item

   def all_equal3(lst):
       return lst[:-1] == lst[1:]
   

3. Counting how many times the first item appears in the list

   def all_equal_ivo(lst):
       return not lst or lst.count(lst[0]) == len(lst)
   

4. Comparing against a list of the first element repeated

   def all_equal_6502(lst):
       return not lst or [lst[0]]*len(lst) == lst
   

But they have some downsides, namely:

1. all_equal and all_equal2 can use any iterators, but the others must take a sequence input, typically concrete containers like a list or tuple.
2. all_equal and all_equal3 stop as soon as a difference is found (what is called "short circuit"), whereas all the alternatives require iterating over the entire list, even if you can tell that the answer is False just by looking at the first two elements.
3. In all_equal2 the content must be hashable. A list of lists will raise a TypeError for example.
4. all_equal2 (in the worst case) and all_equal_6502 create a copy of the list, meaning you need to use double the memory.

On Python 3.9, using perfplot, we get these timings (lower Runtime [s] is better):

Answer 2

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

Answer 3

[edit: This answer addresses the currently top-voted itertools.groupby (which is a good answer) answer later on.]

Without rewriting the program, the most asymptotically performant and most readable 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.)

This will fail at the earliest possible time, so it is asymptotically optimal (expected time is approximately O(#uniques) rather than O(N), but worst-case time still O(N)). This is assuming you have not seen the data before...

(If you care about performance but not that much about performance, you can just do the usual standard optimizations first, like hoisting the myList[0] constant out of the loop and adding clunky logic for the edge case, though this is something the python compiler might eventually learn how to do and thus one should not do it unless absolutely necessary, as it destroys readability for minimal gain.)

If you care slightly more about performance, this is twice as fast as above but a bit more verbose:

def allEqual(iterable):
    iterator = iter(iterable)
    
    try:
        firstItem = next(iterator)
    except StopIteration:
        return True
        
    for x in iterator:
        if x!=firstItem:
            return False
    return True

If you care even more about performance (but not enough to rewrite your program), use the currently top-voted itertools.groupby answer, which is twice as fast as allEqual because it is probably optimized C code. (According to the docs, it should (similar to this answer) not have any memory overhead because the lazy generator is never evaluated into a list... which one might be worried about, but the pseudocode shows that the grouped 'lists' are actually lazy generators.)

If you care even more about performance read on...

---

sidenotes regarding performance, because the other answers are talking about it for some unknown reason:

... if you have seen the data before and are likely using a collection data structure of some sort, and you really care about performance, you can get .isAllEqual() for free O(1) by augmenting your structure with a Counter that is updated with every insert/delete/etc. operation and just checking if it's of the form {something:someCount} i.e. len(counter.keys())==1; alternatively you can keep a Counter on the side in a separate variable. This is provably better than anything else up to constant factor. Perhaps you can also use python's FFI with ctypes with your chosen method, and perhaps with a heuristic (like if it's a sequence with getitem, then checking first element, last element, then elements in-order).

Of course, there's something to be said for readability.

Answer 4

Convert your input into a set:

len(set(the_list)) <= 1

Using set removes all duplicate elements. <= 1 is so that it correctly returns True when the input is empty.

This requires that all the elements in your input are hashable. You'll get a TypeError if you pass in a list of lists for example.

Answer 5

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.

Answer 6

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.}

Answer 7

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

Answer 8

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)

Answer 9

Check if all elements equal to the first.

np.allclose(array, array[0])

Answer 10

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.

Answer 11

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.

Answer 12

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.

Answer 13

>>> 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"

Answer 14

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".

Answer 15

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

Answer 16

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:]))

Answer 17

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

Answer 18

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

Answer 19

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.)

Answer 20

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)))

Answer 21

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

Answer 22

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.

Answer 23

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

Answer 24

Maybe I'm underestimating the problem? Check the length of unique values in the list.

lzt = [1,1,1,1,1,2]

if (len(set(lzt)) > 1):
    uniform = False
elif (len(set(lzt)) == 1):
    uniform = True
elif (not lzt):
    raise ValueError("List empty, get wrecked")

Answer 25

Here is a code with good amount of Pythonicity, and balance of simplicity and obviousness, I think, which should work also in pretty old Python versions.

def all_eq(lst):
    for idx, itm in enumerate(lst):
        if not idx:   # == 0
            prev = itm
        if itm != prev:
            return False
        prev = itm
    return True