Python find longest list in list

Identifying the Longest List within a List of Lists in Python

Consider transforming the entire content into a generator to prevent the creation of a list of lists. If you prefer having a dictionary of lists, it is within your discretion to make the inputs more organized.

How to the find the longest match from a set of lists in Python?

def longest_match(main_list, *other_lists, **options): try: if options: ordered = options['ordered'] else: raise Exception except: ordered = False best_match = 0 longest_match = 0 for index, list in enumerate(other_lists): current_match = 0 if not ordered: while True: if list[:current_match] == main_list[:current_match]: current_match += 1 else: if current_match > longest_match: longest_match = current_match best_match = index break else: for i, letter in enumerate(list): current_match = i if letter == main_list[0]: current_match += 1 while True: if list[i:current_match] == main_list[:current_match]: current_match += 1 else: if current_match > longest_match: longest_match = current_match best_match = index break return other_lists[best_match] print longest_match(my_list, my_other_list_1, my_other_list_2, ordered=True) #kwarg odered=True will allow for mid list comprehension 

I haven’t tried it yet on my phone, but the following code should get the job done!

Certainly, I am willing to assist you! Please take this!

def matcher_algorithm_large(original_list, **kwargs): match_dict = <> for i in kwargs: counter=0 for j in kwargs[i]: if j == original_list[kwargs[i].index(j)]: counter += 1 else: break match_dict[i]=counter for key,value in match_dict.items(): print('<> has match <>'.format(key,value)) 

This essentially enables you to easily insert them in this manner:

my_list = [1, 7, 3, 4, 2, 9, 6, 5, 8] my_other_list_1 = [1, 7, 3, 4, 2] my_other_list_2 = [1, 7, 4, 9, 8] matcher_algorithm_large(my_list,my_other_list_1=my_other_list_1,my_other_list_2=my_other_list_2) 

After running the program, you can expect to see the resulting output to be as follows:

my_other_list_1 has match 5 my_other_list_2 has match 2 

I trust that this is beneficial to you. I’ve adjusted it to allow the incorporation of multiple lists. To streamline the inputs, you could use a dictionary of lists, but the decision is yours. If you prefer, I could gladly create a script for that. 🙂

Finding longest list in list of lists in Python, Modified 4 years, 4 months ago. Viewed 25k times. 1. I have the following code to find the longest list in a list of lists: max_len = max ( [len (path) for path in paths]) [path for path in paths if len (path) == max_len] [0] Each path is a list of integers. paths doesn’t have to be a list. So if I have, for example:

How do I find the longest list in a list of lists?

This is a universal condition that examines a list in search of a specific element based on a predetermined objective.

select_element(Goal, [Head | Tail], Selected) :- select_element(Goal, Tail, Head, Selected). select_element(_Goal, [], Selected, Selected). select_element(Goal, [Head | Tail], Current, FinalSelected) :- call(Goal, Head, Current, Selected), select_element(Goal, Tail, Selected, FinalSelected). 

Lets say you define a predicate

get_bigger_number(N1, N2, N) :- N is max(N1, N2). 

?- select_element(get_bigger_number, [5, 1, -2, 10, 3.2, 0], Selected). Selected = 10 

To replace get_bigger_number/3 , simply define a predicate as get_longer_list(L1, L2, L) .

It may not be efficient to use a broad descriptor such as select_element/3 . It is advisable to refrain from repeatedly computing the length of a list as this operation can be slow in Prolog, particularly when implemented in the standard manner.

Please consider my aproach.

longest([L], L) :- !. longest([H|T], H) :- length(H, N), longest(T, X), length(X, M), N > M, !. longest([H|T], X) :- longest(T, X), !. 

?- longest([[1]], N). N = [1] ; ?- longest([[1],[2]], N). N = [2] . ?- longest([[1],[2], [3,3,3], [2]], N). N = [3, 3, 3] ; ?- longest([[1],[2], [3,3,3], [2]], N). N = [3, 3, 3]. ?- longest([[1],[2], [3,3,3], [2], [4,4,4,4]], N). N = [4, 4, 4, 4] . ?- longest([[1],[2], [3,3,3], [2], [4,4,4,4]], N). N = [4, 4, 4, 4] ; 

Our definition of longest/2 relies on the utilization of max_of_by/3 meta-predicate in conjunction with length/2 .

longest(Xss,Ys) :- max_of_by(Ys,Xss,length). 

?- longest([[1],[2]],Xs). % we expect multiple solutions Xs = [1] ; Xs = [2]. % we _get_ multiple solutions ?- longest([[2,1,3],[7,5],[1,8,2,3,1],[2,7,1,4]],Xs). Xs = [1,8,2,3,1]. % succeeds deterministically 

Python’s most efficient way to choose longest string in, I have a list of variable length and am trying to find a way to test if the list item currently being evaluated is the longest string contained in the list. And I am using Python 2.6.1 For example:

Finding longest run in a list

This uncomplicated approach involves counting the number of repetitions by iterating through the list in reverse order.

last_num = None result = [] for num in reversed(x_list): if num != last_num: # if the number changed, reset the counter to 1 counter = 1 last_num = num else: # if the number is the same, increment the counter counter += 1 result.append(counter) # reverse the result result = list(reversed(result)) 

Achieving this can be done through the utilization of itertools .

from itertools import groupby, chain x_list = [1, 1, 2, 3, 3, 3] gen = (range(len(list(j)), 0, -1) for _, j in groupby(x_list)) res = list(chain.from_iterable(gen)) 

Explanation

• Utilize itertools.groupby initially to categorize similar objects in the list.
• Generate a range object that performs a countdown from the number of consecutive items’ length to 1, for each entry in your groupby .
• To circumvent creating a list of lists, transform the entire thing into a generator.
• Employ the itertools.chain to link together the intervals from the source.

@Aran-Fey’s solution outperforms the performance of the solution being discussed. While the time complexity of itertools.groupby is linear, it heavily relies on costly __next__ calls, which do not scale as efficiently as simple for loops. For the pseudo-code, refer to the itertools documentation mentioned in groupby .

For those who prioritize performance, it is recommended to stick with the for loop.

A cumulative count function can be created in Numpy to perform a reverse count on contiguous groups.

import numpy as np def cumcount(a): a = np.asarray(a) b = np.append(False, a[:-1] != a[1:]) c = b.cumsum() r = np.arange(len(a)) return r - np.append(0, np.flatnonzero(b))[c] + 1 

and then generate our result with

a = np.array(x_list) cumcount(a[::-1])[::-1] array([2, 1, 1, 3, 2, 1]) 

Python — Find the length of the longest list that a, Try this: def find_max_list(lst): lst_len = max((‘a’ in x, len(x)) for x in lst) print(lst_len) This piece of code creates a list of tuples where the first element is True or False (depending on ‘a’ being present or not in the list) and the second element is just the length of a list.

Find the length of the longest list that a contains a certain character in another list

def find_max_list(lst): lst_len = max(('a' in x, len(x)) for x in lst) print(lst_len) 

The code generates a list of tuples containing a Boolean value indicating the presence of ‘a’ in the list and the length of the list. The maximum value is calculated while considering that False < True , and the evaluation of first elements occurs first.

Rephrased MSDTHOT — Another method is to eliminate lists that do not have the letter ‘a’.

def find_max_list(lst): lst_len = max(len(x) for x in lst if 'a' in x) print(lst_len) 

import collections def find_max_list(a_list): most_common = [] for i in a_list: a_counter = collections.Counter(i) most_common.append(a_counter.most_common()) maxfinal = [a[:1] for a in most_common] return maxfinal 

Find longest consecutive sublist from list using python, Browse other questions tagged python-2.7 or ask your own question. The Overflow Blog Code completion isn’t magic; it just feels that way (Ep. 464)

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Python Find Longest List in List

💬 Programming Challenge: Given a list of lists (nested list). Find and return the longest inner list from the outer list of lists.

Also, you’ll learn how to solve a variant of this challenge.

💬 Bonus challenge: Find only the length of the longest list in the list of lists.

So without further ado, let’s get started!

Method 1: max(lst, key=len)

Use Python’s built-in max() function with a key argument to find the longest list in a list of lists. Call max(lst, key=len) to return the longest list in lst using the built-in len() function to associate the weight of each list, so that the longest inner list will be the maximum.

def get_longest_list(lst): return max(lst, key=len) print(get_longest_list([[1], [2, 3], [4, 5, 6]])) # [4, 5, 6] print(get_longest_list([[1, [2, 3], 4], [5, 6], [7]])) # [1, [2, 3], 4] print(get_longest_list([[[1], [2], [3]], [4, 5, [6]], [7, 8, 9, 10]])) # [7, 8, 9, 10]

A beautiful one-liner solution, isn’t it? 🙂 Let’s have a look at a slight variant to check the length of the longest list instead.

Method 2: len(max(lst, key=len))

To get the length of the longest list in a nested list, use the len(max(lst, key=len)) function. First, you determine the longest inner list using the max() function with the key argument set to the len() function. Second, you pass this longest list into the len() function itself to determine the maximum.

Here’s an analogous example:

def get_length_of_longest_list(lst): return len(max(lst, key=len)) print(get_length_of_longest_list([[1], [2, 3], [4, 5, 6]])) # 3 print(get_length_of_longest_list([[1, [2, 3], 4], [5, 6], [7]])) # 3 print(get_length_of_longest_list([[[1], [2], [3]], [4, 5, [6]], [7, 8, 9, 10]])) # 4

Method 3: max(len(x) for x in lst)

A Pythonic way to check the length of the longest list is to combine a generator expression or list comprehension with the max() function without key. For instance, max(len(x) for x in lst) first turns all inner list into length integer numbers and passes this iterable into the max() function to get the result.

Here’s this approach on the same examples as before:

def get_length_of_longest_list(lst): return max(len(x) for x in lst) print(get_length_of_longest_list([[1], [2, 3], [4, 5, 6]])) # 3 print(get_length_of_longest_list([[1, [2, 3], 4], [5, 6], [7]])) # 3 print(get_length_of_longest_list([[[1], [2], [3]], [4, 5, [6]], [7, 8, 9, 10]])) # 4

A good training effect can be obtained by studying the following tutorial on the topic—feel free to do so!

Method 4: Naive For Loop

A not so Pythonic but still fine approach is to iterate over all lists in a for loop, check their length using the len() function, and compare it against the currently longest list stored in a separate variable. After the termination of the loop, the variable contains the longest list.

def get_longest_list(lst): longest = lst[0] if lst else None for x in lst: if len(x) > len(longest): longest = x return longest print(get_longest_list([[1], [2, 3], [4, 5, 6]])) # [4, 5, 6] print(get_longest_list([[1, [2, 3], 4], [5, 6], [7]])) # [1, [2, 3], 4] print(get_longest_list([[[1], [2], [3]], [4, 5, [6]], [7, 8, 9, 10]])) # [7, 8, 9, 10] print(get_longest_list([])) # None

So many lines of code! 😅 At least does the approach also work when passing in an empty list due to the ternary operator used in the first line.

If you need a refresher on the ternary operator, you should check out our blog tutorial.

⭐ Note: If you need the length of the longest list, you could simply replace the last line of the function with return len(longest) , and you’re done!

Summary

You have learned about four ways to find the longest list and its length from a Python list of lists (nested list):

• Method 1: max(lst, key=len)
• Method 2: len(max(lst, key=len))
• Method 3: max(len(x) for x in lst)
• Method 4: Naive For Loop

I hope you found the tutorial helpful, if you did, feel free to consider joining our community of likeminded coders—we do have lots of free training material!

👉 Also, check out our tutorial on finding the general maximum of a list of lists—it’s a slight variation!

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