Question

如何找到列表中彼此相邻的最大1s（或任何我想要的元素）？

l = [2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 7, 1, 1, 1]

在这种情况下，我需要一个返回4的函数。

谢谢。

Answer 1

groupby()函数可用于此：

import itertools

l = [2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 7, 1, 1, 1]
print(max([len(list(g))*k for k, g in itertools.groupby(l, lambda x: x == 1)]))

Answer 2

手动：

def makeMyHomework(li):
'''reads a list of int's and looks for the logest run of one int'''
    curVal = None
    curCount = 0
    maxVal = None
    maxCount = -1
    for n in l:
        if curVal == None:
            curVal = n
        if curVal == n:
            curCount +=1
        if curVal != n:
            if curCount > maxCount: # check if current run > max so far
                maxVal = curVal      # store new max value
                maxCount = curCount  # store new max count
            curVal = n     # init the act ones
            curCount = 1   #  -"-

       # print (n, ":", curVal, curCount, maxVal,maxCount)

    # edge case - the max list is the last run            
    if curCount > maxCount:      
        maxVal = curVal
        maxCount = curCount

    return (maxVal, maxCount) # tuple of (value, number of occurences)

l = [2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 1, 2, 7, 1, 1, 1,2,2,2,2,2]

print(makeMyHomework(l))

Python 3：在列表中找到彼此相等的最大元素

2 个答案: