我不确定它是pythonic的
from itertools import takewhile,izip
x = [[3,2,1], [3,2,1,4,5], [3,2,1,8,9], [3,2,1,5,7,8,9]]
def allsame(x):
return len(set(x)) == 1
r = [i[0] for i in takewhile(allsame ,izip(*x))]
查找列表的最长公共前缀的Python方法是什么?
我不确定它是pythonic的
from itertools import takewhile,izip
x = [[3,2,1], [3,2,1,4,5], [3,2,1,8,9], [3,2,1,5,7,8,9]]
def allsame(x):
return len(set(x)) == 1
r = [i[0] for i in takewhile(allsame ,izip(*x))]