我需要一个函数,它接受一个列表并输出True,如果输入列表中的所有元素使用标准相等运算符计算彼此相等,否则输出False。
我觉得最好是遍历列表,比较相邻的元素,然后与所有结果布尔值。但我不知道最python的方法是什么。
我需要一个函数,它接受一个列表并输出True,如果输入列表中的所有元素使用标准相等运算符计算彼此相等,否则输出False。
我觉得最好是遍历列表,比较相邻的元素,然后与所有结果布尔值。但我不知道最python的方法是什么。
当前回答
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)
其他回答
我怀疑这是“最python化的”,但类似于:
>>> 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
会成功的。
或者使用numpy的diff方法:
import numpy as np
def allthesame(l):
return np.all(np.diff(l)==0)
并呼吁:
print(allthesame([1,1,1]))
输出:
True
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)
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。)
关于使用reduce()和lambda。这里有一个工作代码,我个人认为比其他一些答案更好。
reduce(lambda x, y: (x[1]==y, y), [2, 2, 2], (True, 2))
返回一个元组,其中第一个值是布尔值,如果所有项都相同或不相同。