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.
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.
1In other words, I can't take the credit for coming up with the solution -- nor can I take credit for even finding it.
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
There was a nice Twitter thread on the various ways to implement an all_equal() function.
Given a list input, the best submission was:
t.count(t[0]) == len(t)
Here is are the results from the thread:
Have groupby() compare adjacent entries. This has an early-out for a mismatch, does not use extra memory, and it runs at C speed.
g = itertools.groupby(s)
next(g, True) and not next(g, False)
Compare two slices offset from one another by one position. This uses extra memory but runs at C speed.
s[1:] == s[:-1]
Iterator version of slice comparison. It runs at C speed and does not use extra memory; however, the eq calls are expensive.
all(map(operator.eq, s, itertools.islice(s, 1, None)))
Compare the lowest and highest values. This runs at C speed, doesn't use extra memory, but does cost two inequality tests per datum.
min(s) == max(s) # s must be non-empty
Build a set. This runs at C speed and uses little extra memory but requires hashability and does not have an early-out.
len(set(t))==1.
At great cost, this handles NaNs and other objects with exotic equality relations.
all(itertools.starmap(eq, itertools.product(s, repeat=2)))
Pull out the first element and compare all the others to it, stopping at the first mismatch. Only disadvantage is that this doesn't run at C speed.
it = iter(s)
a = next(it, None)
return all(a == b for b in it)
Just count the first element. This is fast, simple, elegant. It runs at C speed, requires no additional memory, uses only equality tests, and makes only a single pass over the data.
t.count(t[0]) == len(t)
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)))
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.
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
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
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
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")
I ended up with this one-liner
from itertools import starmap, pairwise
all(starmap(eq, (pairwise(x)))
More versions using itertools.groupby that I find clearer than the original (more about that below):
def all_equal(iterable)(iterable):
g = groupby(iterable)
return not any(g) or not any(g)
def all_equal(iterable):
g = groupby(iterable)
next(g, None)
return not next(g, False)
def all_equal(iterable)(iterable):
g = groupby(iterable)
return not next(g, False) or not next(g, False)
Here's the original from the Itertools Recipes again:
def all_equal(iterable):
g = groupby(iterable)
return next(g, True) and not next(g, False)
Note that the next(g, True) is always true (it's either a non-empty tuple or True). That means its value doesn't matter. It's executed purely for advancing the groupby iterator. But including it in the return expression leads the reader into thinking that its value gets used there. Since it doesn't, I find that misleading and unnecessarily complicated. My second version above treats the next(g, True) as what it's actually used for, as a statement whose value we don't care about.
My third version goes a different direction and does use the value of the first next(g, False). If there isn't even a first group at all (i.e., if the given iterable is "empty"), then that solution returns the result right away and doesn't even check whether there's a second group.
My first solution is basically the same as my third, just using any. Both solutions read as "All elements are equal iff ... there is no first group or there is no second group."
Benchmark results (although speed is really not my point here, clarity is, and in practice if there are many equal values, most of the time might be spent by the groupby itself, reducing the impact of these differences here):
Python 3.10.4 on my Windows laptop:
iterable = ()
914 ns 914 ns 916 ns use_first_any
917 ns 925 ns 925 ns use_first_next
1074 ns 1075 ns 1075 ns next_as_statement
1081 ns 1083 ns 1084 ns original
iterable = (1,)
1290 ns 1290 ns 1291 ns next_as_statement
1303 ns 1307 ns 1307 ns use_first_next
1306 ns 1307 ns 1309 ns use_first_any
1318 ns 1319 ns 1320 ns original
iterable = (1, 2)
1463 ns 1464 ns 1467 ns use_first_any
1463 ns 1463 ns 1467 ns next_as_statement
1477 ns 1479 ns 1481 ns use_first_next
1487 ns 1489 ns 1492 ns original
Python 3.10.4 on a Debian Google Compute Engine instance:
iterable = ()
234 ns 234 ns 234 ns use_first_any
234 ns 235 ns 235 ns use_first_next
264 ns 264 ns 264 ns next_as_statement
265 ns 265 ns 265 ns original
iterable = (1,)
308 ns 308 ns 308 ns next_as_statement
315 ns 315 ns 315 ns original
316 ns 316 ns 317 ns use_first_any
317 ns 317 ns 317 ns use_first_next
iterable = (1, 2)
361 ns 361 ns 361 ns next_as_statement
367 ns 367 ns 367 ns original
384 ns 385 ns 385 ns use_first_next
386 ns 387 ns 387 ns use_first_any
Benchmark code:
from timeit import timeit
from random import shuffle
from bisect import insort
from itertools import groupby
def original(iterable):
g = groupby(iterable)
return next(g, True) and not next(g, False)
def use_first_any(iterable):
g = groupby(iterable)
return not any(g) or not any(g)
def next_as_statement(iterable):
g = groupby(iterable)
next(g, None)
return not next(g, False)
def use_first_next(iterable):
g = groupby(iterable)
return not next(g, False) or not next(g, False)
funcs = [original, use_first_any, next_as_statement, use_first_next]
for iterable in (), (1,), (1, 2):
print(f'{iterable = }')
times = {func: [] for func in funcs}
for _ in range(1000):
shuffle(funcs)
for func in funcs:
number = 1000
t = timeit(lambda: func(iterable), number=number) / number
insort(times[func], t)
for func in sorted(funcs, key=times.get):
print(*('%4d ns ' % round(t * 1e9) for t in times[func][:3]), func.__name__)
print()
This was a fun one to read through and think about. Thanks everyone! I don't think anything relying on pure count will be reliable for all cases. Also sum could work but only for numbers or length (again resulting in a count scenario).
But I do like the simplicity, so this is what I came up with:
all(i==lst[c-1] for c, i in enumerate(lst))
Alternatively, I do think this clever one by @kennytm would also work for all cases (and is probably the fastest, interestingly). So I concede that it's probably better than mine:
[lst[0]]*len(lst) == lst
A little bonus clever one I think would also work because set gets rid of duplicates (and clever is fun but not generally best practice for maintaining code). And I think the one by @kennytm would still be faster but really only relevant for large lists:
len(set(lst)) == 1
But the simplicity and cleverness of Python is one of my favorite things about the language. And thinking about it a little more, if you have to modify the list in anyway, like I actually do because I'm comparing addresses (and will remove leading/trailing spaces and convert to lower case to remove possible inconsistencies, mine would be more suited for the job). So "better" is subjective as I eluded to by using quotes when I use that word! But you could also cleanup the list beforehand.
Best and good luck!
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