CODEWITHSUNDEEP
pythonpython-3.xlistmax
Mozart lycee
Mozart lycee

Reputation: 23

How do I get the longest list and the overall max value from a list of lists?

Given a list of lists, I would like to find the longest list, as well as the overall max value from all of the lists. Being restricted to not use max(), how can I go about doing this?

My attempt is shown below:

def newMax(table):
    a=-1
    if len(table) == 0:
        print("The list is empty")
    for i in range(len(table)*2):
        a+=1
        if i+1 >= len(table):
            print("Big number")
            break
        else:
            if len(table[a]) < len(table[a+1]):
                maxs=a+1
                if maxs < len(table[a+2]):
                    maxs = a + 2
                    print("test",table[maxs])
                    if maxs < len(table[a + 3]):
                        maxs = a + 3
                        print("test", table[maxs])

newMax([[1,3,1,5,1,51,15],[0,3,5,4,5],[1,2,3],[1,3,1,5]])

Upvotes: 0

Views: 188

Answers (3)

Yukun Li
Yukun Li

Reputation: 254

Not sure have you learn recursion yet , which this is an recursion based solution, since I`m not sure how many level could have in your input list

input = [[1,3,1,5,1,51,15],[0,3,5,4,5],[1,2,3],[1,3,1,5]]

def findMaxRec(input):
    max = -999999999
    for i in input:
        if isinstance(i,list):
            temp = findMaxRec(i)
            if temp > max:
                max = temp
        else:
            if i > max:
                max = i
    return  max

print(findMaxRec(input))

Upvotes: 0

Wim Lavrijsen
Wim Lavrijsen

Reputation: 3778

Put all entries in a single collection, then take the max of that. This way, all the looping happens on the C-side, so efficient:

def newMax(table):
    s = set()
    for l in table:
        s.update(l)
    return max(s)

Based on the comments above and below, to get all the individual maxes and not use the builtin max(), while staying linear in time and not writing a super-nested function:

import heapq

def newMax(table):
    maxvalues = list()
    for l in table:
        heapq.heapify(l)
        maxvalues.append(heapq.nlargest(1, l)[0])
    mv2 = maxvalues[:]
    heapq.heapify(mv2)
    return heapq.nlargest(1, mv2)[0], maxvalues

One more as the goal/requirements seems to have changed again in the comments:

def newMax(table):
    length_max = -1
    max_length_list = None
    value_max = -2**31
    for l in table:
        if len(l) > length_max:
            length_max = len(l)
            max_length_list = l
        for i in l:
            if i > value_max:
                value_max = i
    return value_max, max_length_list

Upvotes: 1

CH.
CH.

Reputation: 606

To find

  1. the largest value in all the lists, and
  2. the longest list,

simply iterate through all the lists while keeping track of the max encountered so far.

def newMax(table):

    maxValue = table[0][0]
    longestList = table[0]
    longestListLen = len(table[0])

This sets the largest value encountered so far to the first element of the first list, and sets the longest list to to be the first list. We then iterate through the rest of the lists to see if there are larger values / longer lists.

    for row in table: # iterate through all lists
        for val in row: # iterate through each value in each list
            if val > maxValue: # encountered a greater value
                maxValue = val # let's save the new biggest value

        rowLen = len(row) # get this list's length
        if len(row) > longestListLen: # encountered a longer list
            longestListLen = rowLen # save this list as the longest list
            longestList = row

Now you have the max value of all lists in maxValue and the longest list in longestList.

Upvotes: 2

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