不管怎样，这是最近在python想法邮件列表中出现的。事实证明，已经有一个itertools配方可以做到这一点

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)

据说它的性能非常好，有一些不错的属性。

短路:一旦发现第一个不相等的项，它就会停止从可迭代对象中消费项目。 不要求项是可哈希的。 它是懒惰的，只需要O(1)个额外的内存来进行检查。

换句话说，我不能把提出这个解决方案的功劳占为己有，甚至我也不能把找到它的功劳占为己有。

你可以:

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

python让你导入operator.and_这样的操作符是相当烦人的。从python3开始，还需要导入functools.reduce。

(您不应该使用此方法，因为如果它发现不相等的值，它不会中断，而是会继续检查整个列表。这里只是作为完整性的回答。)

您可以将列表转换为集合。集合不能有重复项。所以如果原始列表中的所有元素都是相同的，那么集合就只有一个元素。

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

检查是否所有元素都等于第一个。

np。allclose(数组,数组[0])

出现使用itertools。Groupby(参见itertools食谱):

from itertools import groupby

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

或不带groupby:

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

您可以考虑使用许多其他的一行程序:

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 Comparing against the input list without the first item def all_equal3(lst): return lst[:-1] == lst[1:] 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) Comparing against a list of the first element repeated def all_equal_6502(lst): return not lst or [lst[0]]*len(lst) == lst

但它们也有一些缺点，即:

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. 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. In all_equal2 the content must be hashable. A list of lists will raise a TypeError for example. all_equal2 (in the worst case) and all_equal_6502 create a copy of the list, meaning you need to use double the memory.

在Python 3.9中，使用perfplot，我们得到这些计时(越低的Runtime [s]越好):

我想做的事:

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

as any一旦找到True条件就会停止搜索可迭代对象。