我最后写了这样一句话

from itertools import starmap, pairwise
all(starmap(eq, (pairwise(x)))

这是一种简单的方法:

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

这稍微复杂一点，它会引起函数调用开销，但语义更清楚地说明:

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

>>> 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"

也许我低估了问题的严重性?检查列表中唯一值的长度。

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")

可以使用map和lambda吗

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

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

最佳答案

Twitter上有一个不错的帖子，介绍了实现all_equal()函数的各种方法。

给定一个列表输入，最好的提交是:

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

其他方法

下面是线程的结果:

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)