Question

问：运行是一系列相邻的重复值。给定一个列表，写一个函数确定最长跑的长度。例如，对于序列[1,2,5,5,3,1,2,4,3,2,2,2,3,3,5,5,6,3,1]，最长跑是4。

我遇到了麻烦，我编写了一个代码，发现最长的运行时间由数字'2'组成，但还没有得到运行的长度为4。

到目前为止，这是我的代码（我已经注释了我正在研究的部分，但是没有注意它）：

# longestrun.py
#   A function to determine the length of the longest run
#   A run is a sequence of adjacent repeated values.

def longestrun(myList):
    result = None
    prev = None
    size = 0
    max_size = 0


    for i in myList:
        if i == prev:
            size += 1
            if size > max_size:
                result = i
                max_size = size
        else:
            size = 0
        prev = i
    return result



def main():
    print("This program finds the length of the longest run within a given list.")
    print("A run is a sequence of adjacent repeated values.")

    myString = input("Please enter a list of objects (numbers, words, etc.) separated by
    commas: ")
    myList = myString.split(',')

    longest_run = longestrun(myList)

    print(">>>", longest_run, "<<<")

main()

请帮助!!! ：（（（

Answer 1

您可以使用itertools.groupby在一行中执行此操作：

import itertools
max(sum(1 for _ in l) for n, l in itertools.groupby(lst))

Answer 2

如果您不想使用itertools和import，这应该有用。

a=[1, 2, 5, 5, 3, 1, 2, 4, 3, 2, 2, 2, 2, 3, 6, 5, 5, 6, 3, 1]

def longestrun(myList):
    result = None
    prev = None
    size = 0
    max_size = 0


    for i in myList:
        if i == prev:
            print (i)
            size += 1
            if size > max_size:
                print ('*******  '+ str(max_size))
                max_size = size 
        else:
            size = 0
        prev = i
    print (max_size+1)    
    return max_size+1


longestrun(a)

Answer 3

另一种方法：

def longestrun(myList):
    sett = set()
    size = 1
    for ind, elm in enumerate(myList):
        if ind > 0:
            if elm == myList[ind - 1]:
                size += 1
            else:
                sett.update([size])
                size = 1
    sett.update([size])
    return max(sett)

myList = [1, 2, 5, 5, 3, 1, 2, 4, 3, 2, 2, 2, 2, 3, 6, 5, 5, 6, 3, 1]
print longestrun(myList)

Answer 4

def longestrun(myList):
    size = 1
    max_size = 0
    for i in range(len(myList)-1):
        if myList[i+1] = myList[i]:
            size += 1
        else: 
            size = 1
        if max_size<size:
            max_size = size
    return size

从.split()中的myList移除main()，您就可以使用此功能了。

Answer 5

def getSublists(L,n):
    outL=[]
    for i in range(0,len(L)-n+1):
        outL.append(L[i:i+n])
    return outL


def longestRun(L):
    for n in range(len(L), 0, -1):
        temp=getSublists(L,n)
        for subL in temp:
            if subL==sorted(subL):
                return len(subL)

Answer 6

作为David Robinson答案的更新，现在（Python 3.4）可以在空序列上返回0（而不是引发ValueError）：

import itertools
max((sum(1 for _ in l) for n, l in itertools.groupby(lst)), default=0)

Python：一个程序，用于查找给定列表中运行时间最长的LENGTH？

6 个答案: