正確的方向可能是不使用filter，而是使用zip： db = [3, 5, 7, 3, 2, 7, 9]dc = [1, 0, 1, 0, 1, 0, 1]dd = [b**2 for b,c in zip(db, dc) if not c] Output: [25,9,49] using filter 這需要找到一個(gè)共同點(diǎn)，這里是索引，但代碼要差得多。。。 list(map(lambda x: db[x]**2, filter(lambda y: dc[y]==0, range(len(dc)))))

你的實(shí)際代碼 users[str(user.id)]["submitted"] + 'earnings'with open("mainbank.json",'w') as f: json.dump(users,f) 不起作用，因?yàn)樗鼪]有更新任何內(nèi)容。嘗試 # I suppose the + sign is to update the empty string ""users[str(user.id)]["submitted"].append('earnings')with open("mainbank.json",'w') as f: json.dump(users,f) 如果您碰巧需要將提交的信息作為列表填寫，請(qǐng)嘗試： # with an empty list{"809849410283372605": {"wallet": 0, "submitted": [], "bank": 0}}users[str(user.id)]["submitted"].append('earnings')with open("mainbank.json",'w') as f: json.dump(users,f) 如果需要更改""空字符串的值，只需使用： users[str(user.id)]["submitted"][0] = 'earnings'with open("mainbank.json

您可以在這兩個(gè)列表上運(yùn)行一個(gè)循環(huán)，并構(gòu)造一個(gè)remaining_items列表，從中隨機(jī)采樣- c = [] for constraints, items in zip(a, b): num_items, *required_items = constraints pick_list = list(_ for _ in range(1, items[0] + 1) if _ not in required_items) num_items_to_pick = num_items - len(required_items) remaining_items = random.sample(pick_list, num_items_to_pick) print(required_items + remaining_items) c.append(required_items + remaining_items) Output [[1, 2, 3], [2, 3, 1], [1, 4, 8, 3, 6], [5, 3, 1], [1, 3, 6, 7, 18, 2, 14, 21, 19, 15], [1, 2]]

我不知道你到底想做什么，但這個(gè)錯(cuò)誤是一致的。 new_list = [item for sublist in input_list for item in sublist] 試試下面的代碼，它應(yīng)該可以工作，如果你可以張貼你的要求更清楚，這將更容易理解和回答。 def subtract_months(input_list): output_list = [] #TODO: implement your code here new_list = [list(item) for sublist in input_list for item in sublist] print(new_list) for i in range(len(new_list)): if new_list[i][2] > new_list[i][1]: new_list[i][2] = new_list[i][1] - new_list[i][2] return new_list input_list = [[(2000,1,14),(2020,5,3)]]print(subtract_months(input_list))

我將在,上拆分，然后在迭代結(jié)果時(shí)，通過使用if語句以不同的方式處理*的任何項(xiàng)。 {{447}加上一個(gè){447}的結(jié)果，然后把@447的結(jié)果加到一個(gè)475}的后面。 fruits = "mango*banana*grapes,orange*apple"ListOfFruits = fruits.split(',')storage = ""for f in ListOfFruits: if '*' not in f: storage+= ("({}) \n({})\n".format(f,f)) else: tab_fruits = f.split('*') p = ['\t'*i + '('+x+')' +'\n' for i,x in enumerate(tab_fruits)] p.extend(reversed(p)) storage+=''.join(p) finalStorage = storage + "\n"print(finalStorage) Ouput (mango) (banana) (grapes) (grapes) (banana)(mango)(orange) (apple) (apple)(orange)

如果要進(jìn)行語言轉(zhuǎn)換，可以執(zhí)行以下操作： data=[ { 'English': 'swaziland', 'Nepali': '????? ????????' }, { 'English': 'varieties', 'Nepali': '??????????' }, { 'English': 'arbor', 'Nepali': 'arbor' }, { 'English': 'bibek', 'Nepali': '?????' }, { 'English': 'configurations', 'Nepali': '???????' }, ]english_name = input('Give me an English word: ')found = Falsefor item in data: if item['English'] == english_name: print(item['Nepali']) found = True breakif found == False: print('Not found') 找不到您鍵入的單詞時(shí)的結(jié)果： python3 test.pyGive me an English word: askNot found 找到您鍵入的單詞時(shí)的結(jié)果： python3 test.pyGive me an English word: bibek???

這是因?yàn)椋谝淮蝘的值是1，所以它去掉了12，現(xiàn)在i變成了2，第二個(gè)元素現(xiàn)在是23而不是34，所以它刪除了23。 要修復(fù)此錯(cuò)誤，可以執(zhí)行以下操作： a=[12,34,23]for i in a: a.remove(0)print(a)

一個(gè)2D列表會(huì)有行和列，比如mXn，您的列表是2X4，但是您只檢查這2行。 fruits = [["Banana", "Apple", "Watermelon", "Apple"], ["Apple", "Banana", "Watermelon", "Lemon"]]print(fruits)findFruit = input("Fruit: ")replaceFruit = input("Replace: ")for i in range(len(fruits)): # len(fruits) is 2 for j in range(len(fruits[i])): # len(fruits[i]) is 4 if fruits[i][j] == findFruit: fruits[i][j] = replaceFruitprint(fruits)

通常，我建議使用outputlst = list(set(lst1 + lst2))，但由于兩端的空格略有不同，您可能可以創(chuàng)建一個(gè)列表或集合（already_contained在下面的代碼中），以檢查是否刪除了尾部空格的字符串，并將其用作檢查lst2中的字符串是否應(yīng)傳遞給outputlst lst1 =["a: ","b: ","c:"]lst2 =[" b:"," a:","f:","g:","c: "]already_contained = set([c.replace(':', '').strip() for c in lst1])outputlst = lst1 + [c for c in lst2 if c.replace(':', '').strip() not in already_contained]

假設(shè)列表I已排序，可以使用bisect_left為i中的每個(gè)元素在列表I中獲取插入點(diǎn)，然后對(duì)列表進(jìn)行切片。它使用二進(jìn)制搜索。 有了它，您可以： from bisect import bisect_leftValues = [I[:bisect_left(I, e)] for e in i]New_i = [e + len(Values[j]) for j, e in enumerate(i)] print(Values): [[8, 11], [8, 11], [8, 11, 19], [8, 11, 19], [8, 11, 19], [8, 11, 19, 37, 40]] print(New_i): [17, 19, 30, 31, 34, 46] 順便說一句，我強(qiáng)烈建議不要在變量名中使用I和i。

首先收集所有“是”值的總和： dictsum = {i[0]: [sum(x[1] for x in example if x[0]==i[0] and x[-1]=="Yes"), sum(x[2] for x in example if x[0]==i[0] and x[-1]=="Yes")] for i in example}>>> dictsum{2: [10, 11], 4: [-13, 48]} 使用max和所需的key函數(shù)更新每個(gè)鍵的第0個(gè)和第1個(gè)列表元素： output = {k: [v[0]+max([l[1] for l in example if l[0]==k and l[-1]=="Non"], key=abs), \ v[1]+max([l[2] for l in example if l[0]==k and l[-1]=="Non"], key=abs)] \ for k,v in dictsum.items()}>>> output{2: [40, -11], 4: [-10, -42]}

看看這個(gè)。 embeding_2 = ['GSSSSSSSSSSS', 'AGGGGGSSSSSS']res = []for i in embeding_2: res.append( [i[:j+1] for j in range(len(i))] )print(res)

pandas在這方面非常擅長(zhǎng)： import pandas as pd # set up DataFrametransformed_data = (pd.DataFrame(data) # convert to numeric .apply(pd.to_numeric, errors='coerce') # replace null with 0 # trying to cast as integer if possible .fillna(0, downcast='infer') # convert to numpy array .to_numpy() ) output: array([[ 1, 12345, 1, 0, 0], [ 1, 34567, 1, 1, 0]])

為您提供一些建議，讓您了解如何繼續(xù)進(jìn)行： 使用您提供的字符串列表（我想在"spinach"后面缺少一個(gè)逗號(hào)） strings = [ "fruits", "apple", "{\n \"color\": \"red\",\n \"harvest\": \"ready\"\n}\n", "{\n \"color\": \"green\",\n \"harvest\": \"not ready\"\n}\n", "veggies", "spinach", "{\n \"color\": \"green\",\n \"harvest\": \"not ready\"\n}\n", "{\n \"color\": \"light green\",\n \"harvest\": \"not ready\"\n}\n"] this import jsonitems = []for string in strings: try: items.append(json.loads(string)) except: items.append(string) 為您（print(items)）提供以下列表： ['fruits', 'apple', {'color': 'red', 'harvest': 'ready'}, {'color': 'green', 'harvest': 'not ready'}

你可以使用下面的列表和聽寫理解 dict = [{'complete': True, 'volume': 2290, 'time': '2021-01-15', 'mid': {'o': '0.77540', 'h': '0.77584', 'l': '0.77292', 'c': '0.77440'}}, {'complete': True, 'volume': 2312, 'time': '2021-01-15', 'mid': {'o': '0.77436', 'h': '0.77521', 'l': '0.77206', 'c': '0.77206'}}]res = [{k:v for k, v in i['mid'].items() if k in 'hl'} for i in dict]print(res) Output [{'h': '0.77584', 'l': '0.77292'}, {'h': '0.77521', 'l': '0.77206'}]