max -5 years ago 296
Python Question

# check if all elements in a list are identical

I need the following function:

Input: a

list

Output:

• True
if all elements in the input list evaluate as equal to each other using the standard equality operator;

• False
otherwise.

Performance: of course, I prefer not to incur any unnecessary overhead.

I feel it would be best to:

• iterate through the list

• and
AND
all the resulting Boolean values

But I'm not sure what's the most Pythonic way to do that.

EDIT:

Thank you for all the great answers. I rated up several, and it was really hard to choose between @KennyTM and @Ivo van der Wijk solutions.

The lack of short-circuit feature only hurts on a long input (over ~50 elements) that have unequal elements early on. If this occurs often enough (how often depends on how long the lists might be), the short-circuit is required. The best short-circuit algorithm seems to be @KennyTM
checkEqual1
. It pays, however, a significant cost for this:

• up to 20x in performance nearly-identical lists

• up to 2.5x in performance on short lists

If the long inputs with early unequal elements don't happen (or happen sufficiently rarely), short-circuit isn't required. Then, by far the fastest is @Ivo van der Wijk solution.

General method:

def checkEqual1(iterator):
try:
iterator = iter(iterator)
first = next(iterator)
return all(first == rest for rest in iterator)
except StopIteration:
return True

One-liner:

def checkEqual2(iterator):
return len(set(iterator)) <= 1

Also one-liner:

def checkEqual3(lst):
return lst[1:] == lst[:-1]

The difference between the 3 versions are that:

1. In checkEqual2 the content must be hashable.
2. checkEqual1 and checkEqual2 can use any iterators, but checkEqual3 must take a sequence input, typically concrete containers like a list or tuple.
3. checkEqual1 stops as soon as a difference is found.
4. Since checkEqual1 contains more Python code, it is less efficient when many of the items are equal in the beginning.
5. Since checkEqual2 and checkEqual3 always perform O(N) copying operations, they will take longer if most of your input will return False.
6. checkEqual2 and checkEqual3 can't be easily changed to adopt to compare a is b instead of a == b.

timeit result, for Python 2.7 and (only s1, s4, s7, s9 should return True)

s1 = [1] * 5000
s2 = [1] * 4999 + [2]
s3 = [2] + [1]*4999
s4 = [set([9])] * 5000
s5 = [set([9])] * 4999 + [set([10])]
s6 = [set([10])] + [set([9])] * 4999
s7 = [1,1]
s8 = [1,2]
s9 = []

we get

| checkEqual1 | checkEqual2 | checkEqual3  | checkEqualIvo | checkEqual6502 |
|-----|-------------|-------------|--------------|---------------|----------------|
| s1  | 1.19   msec | 348    usec | 183     usec | 51.6    usec  | 121     usec   |
| s2  | 1.17   msec | 376    usec | 185     usec | 50.9    usec  | 118     usec   |
| s3  | 4.17   usec | 348    usec | 120     usec | 264     usec  | 61.3    usec   |
|     |             |             |              |               |                |
| s4  | 1.73   msec |             | 182     usec | 50.5    usec  | 121     usec   |
| s5  | 1.71   msec |             | 181     usec | 50.6    usec  | 125     usec   |
| s6  | 4.29   usec |             | 122     usec | 423     usec  | 61.1    usec   |
|     |             |             |              |               |                |
| s7  | 3.1    usec | 1.4    usec | 1.24    usec | 0.932   usec  | 1.92    usec   |
| s8  | 4.07   usec | 1.54   usec | 1.28    usec | 0.997   usec  | 1.79    usec   |
| s9  | 5.91   usec | 1.25   usec | 0.749   usec | 0.407   usec  | 0.386   usec   |

Note:

# http://stackoverflow.com/q/3844948/
def checkEqualIvo(lst):
return not lst or lst.count(lst[0]) == len(lst)

# http://stackoverflow.com/q/3844931/
def checkEqual6502(lst):
return not lst or [lst[0]]*len(lst) == lst
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