您可以使用zip和all： x = [0,1,0,0]y = [1,1,0,1]z = [0,1,1,1]output = [all(zipped) for zipped in zip(x, y, z)]print(output) # [False, True, False, False] 如果您確實想得到[0,1,0,0]，請使用int(all(zipped))而不是all(zipped)（但這種顯式重鑄在大多數情況下是多余的）。

使用collections.Counter對元素進行計數，然后在結果上循環以僅保留計數大于1的元素，sum： my_list = [[123, 456], [111, 111], [222, 222], [333, 333], [123, 456], [222, 222], [123, 456]]from collections import Counterc = Counter(map(tuple, my_list))number = sum(v>1 for v in c.values()) output: 2 注意。您需要將子列表轉換為元組，以便它們可以散列并按Counter計數

您可以使用遞歸： def n_l(a, b): return [[] if a == 1 else n_l(a-1, b) for _ in range(b)]print(n_l(1, 3))print(n_l(2, 3)) Output: [[], [], []][[[], [], []], [[], [], []], [[], [], []]]

您可以使用itertools.groupby： from itertools import groupbydef summary_of_list(lst): return [(k, sum(1 for _ in g)) for k, g in groupby(lst)]print(summary_of_list(['a','b','a','a'])) # [('a', 1), ('b', 1), ('a', 2)]print(summary_of_list(['a','a','b','b','b','d'])) # [('a', 2), ('b', 3), ('d', 1)] （我相信您對['a','a','b','b','b','d']的預期輸出[('a',2), ('b',1), ('d',1)]有一個typo.）

如果你想堅持只使用列表，那么你可以在一個單獨的列表中跟蹤計數，然后在最后合并這兩個列表。 sizes = [30, 34, 30, 20, 34]unique = []tally = []for i in sizes: if i not in unique: unique.append(i) tally.append(1) else: idx = unique.index(i) tally[idx] += 1result = []for i in range(len(unique)): result.append(unique[i]) result.append(tally[i])print(result) # [30, 2, 34, 2, 20, 1] 如果你同意使用字典，這是更好/更有效的方法，那么你可以這樣做： sizes = [30, 34, 30, 20, 34]unique = {}for i in sizes: if i in unique: unique[i] += 1 else: unique[i] = 1result = [j for i in unique.items() for j in i] print(result) # [30, 2, 34, 2, 20, 1]

permutations允許您指定排列列表的長度： 如果包含空集： permlist = []for i in range(len(mylist) + 1): permlist += itertools.permutations(mylist, i) 如果排除空集： permlist = []for i in range(len(mylist)): permlist += itertools.permutations(mylist, i+1)

這應該有效： animals= [['dog', 'snake', 'snake'], ['dog', 'dog', 'cat'], ['snake', 'snake', 'cat']]animal_to_count = ['dog', 'cat']results = []for sublist in animals: results_per_animal_count = [] for count_animal in animal_to_count: counter = 0 for animal in sublist: if animal == count_animal: counter += 1 results_per_animal_count.append(counter) results.append(results_per_animal_count) results的打印輸出是 [[1, 0], [2, 1], [0, 1]] EDIT 如果要獲取聚合列表，可以使用： animals= [['dog', 'snake', 'snake'], ['dog', 'dog', 'cat'], ['snake', 'snake', 'cat']]animal_to_count = ['dog', 'cat']results = []for count_animal in anima

我假設您正在使用list（）built-in方法，并將其用作循環中的變量，這會導致錯誤。您可以執行與此相同的任務 #Function definitiondef nlc(n): '''It checks how many lists are in the list, and whether the number of elements in each list are the same.''' sl = len(n) print("Number of Lists is", sl) lengths = [] for element in n: lengths.append(len(element)) #appending length of each sublist if len(set(lengths)) == 1: #checking if all elements are same in a list. It means that all lengths are equal print("Lengths Match") else: print("Lengths Not Equal; Check Lists")nl1 = [[1, 2, 3, 4, 5], [3, 4, 5, 6, 7], [2, 4, 6, 8, 10]]nl2 = [[1, 2, 3, 4, 5], [3,

def findDuplicated(*args): return list(set.intersection(*[set(j) for j in [i for i in args]]))findDuplicated([1,2,3,4], [4,5,6,7], [4,8,9,20], [4,567,756,456,423])

您可以使用enumerate： int_list = [ int(x) for count,x in enumerate(input().split(" ")) if count < 20]

把這個添加到任何你想禁用的View中，比如NavigationLink，這樣用戶就不能去那個View .disabled(results.count >= 10)

l1=[7,1,3,4,1,7,6,3]chosen_number = 3number_indexes = [i for i, j in enumerate(l1) if j == chosen_number]items_in_between = max(number_indexes) - min(number_indexes) - 1print(items_in_between)

您可以使用Counter來執行此操作： from collections import Counterlist1 = ["hello", "goodbye", "hello"]counts = Counter(list1)message = ", ".join(f"{counts[key]}x {key}" for key in counts)print(message) 這段代碼使用f-strings和list comphrensions。

要展平列表，可以使用itertools.chain。然后你必須找到一些隨機索引，在那里你可以拆分列表。你可以使用random.sample。 from itertools import chainimport randoma_list = [[2, 0], [7], [4, 3], [1, 5, 6]]number_of_chunks = 4# flatten the listdata = list(chain.from_iterable(a_list))print(data)# get indexes where to split the listindexes = sorted(random.sample(list(range(1, len(data))), number_of chunks-1))print(indexes)# add start end end of the list indexesindexes = [0] + indexes + [None]print(indexes)# with zip we get pairs of start and end points for the chunks# and then create the chunks in a list comprehensionresult = [data[s:e] for s, e in

基于你提問的方式，聽起來你只是想要一個實用的基本python解決方案，所以我給你一個簡單的解決方案。 與其將列表視為所有獨立的實體，不如簡單地使用一個由non-duplicate數字組成的大集群。您可以利用僅包含唯一值的set屬性繼續并將所有列表聚集在一起： # Throws all contents of lists into a set, converts it back to list, and sortselems = sorted(list({*l1, *l2, *l3}))# elems = [1, 2, 3, 4, 6, 10, 12, 13, 15] 如果有要執行此操作的列表： lists = [l1, l2, l3]elems = [][elems.extend(l) for l in lists]elems = sorted(list(set(elems)))# elems = [1, 2, 3, 4, 6, 10, 12, 13, 15] 如果要保留副本： elems = sorted([*l1, *l2, *l3])# and respectivelyelems = sorted(elems) 從那里，你可以迭代地進行分離。明確地： 遍歷元素one-by-one。如果下一個元素的間距有效，請將其添加到正在構建的列表中。 當遇到invalidly-spaced