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Compar the list with the names and numbers in the python. Find the name of the highest number

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elysha last edited by
On the first list, the names of the players, the number of their points on the second list. We need to find the winner who has the most points, if the two players have the same ring of the highest balls, the winner is the first on the list.
```
my_list = [['Vasya', 'Sasha', 'Kolya', 'Igor'], [5, 4, 3, 4]]
winner = 'Vasya'
my_list2= [['Igor', 'Kolya', 'Petya', 'Sasha'], [4, 6, 6, 3]]
winner2 = 'Kolya'
```
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1 Reply Last reply

Demir

Like that.

my_list = [['Vasya', 'Sasha', 'Kolya', 'Igor'], [5, 4, 3, 4]]
winner = my_list[0][my_list[1].index(max(my_list[1]))]
print(winner)
my_list = [['Igor', 'Kolya', 'Petya', 'Sasha'], [4, 6, 6, 3]]

winner = my_list[0][my_list[1].index(max(my_list[1]))]

print(winner)

and, in fact, you should try to make a school assignment.

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Just don't. reply_markupif the message doesn't have to be Inline Clara. ReplyKeyboardRemove() - removes conventional buttons, not inline.If I understand correctly, based on your code, it's supposed to look like this:feel_bots=InlineKeyboardMarkup().row( InlineKeyboardButton(text="Telegramm Bot", callback_data="feel_tgbot") )
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The solution is to assign to another variable the value of the dictionary s = d.to_dict() Giving a result like this:for d in data: s = d.to_dict() print('Nombre: {} \n Apellido: {} \n Dirección: {}\n'.format(s['Nombre'],s['Apelliso'],s['Dirección'])) All data will be displayed: Nombre: Angel Apellido: Alvarez Dirección: Zaragoza 17 Nombre: jose Apellido: alvarez Dirección: Victoria #4 [Finished in 2.8s]
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All(s) may be used in cases where all of the conditions listed and any(s) need to be met when at least one to do so.For example, your code can be rewritten as follows:if all([(name in line), (age in line), (string in text)]): do_something() This can be made more comfortable if we first build a body from conditions and then hand it over to if:checks = [name in line, age in line] checks.append(string in text) if all(checks): do_something() As can be seen from this code, the mass can be developed gradually by adding the conditions at different locations of the code, which can be useful when a list of conditions is developed dynamically according to different external factors.In addition, functions sometimes simply check each condition separately if, and if the condition fails or fails, depending on the task, further performance is interrupted through return:def foo(): if not(name in line): return False if not(age in line): return False if not(string in text): return False return True
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This whole complexity is not necessary when you are programming on Python!This simple program is able to return what you want:from collections import Counter arquivos = [ 'texto1.txt', 'texto2.txt', 'texto3.txt' ] def obter_palavras( arqs ): ap = [] for a in arqs: with open( a,'r' ) as f: ap += [ (a,p) for p in f.read().split() ] cnt = Counter(ap).items() dic = { p:[] for a, p in ap } for p in dic.keys(): for a in arqs: for k, n in cnt: if k == (a,p): dic[p].append((a,n)) return dic print obter_palavras( arquivos ) text1.txtalpha beta gamma delta zero um dois tres kilo mega giga tera terra agua ar fogo text2.txtzero um dois tres zero um dois tres kilo mega giga tera text3.txtalpha beta gamma delta alpha beta gamma delta kilo mega giga tera Output:{ 'mega': [('texto1.txt', 1), ('texto2.txt', 1), ('texto3.txt', 1)], 'kilo': [('texto1.txt', 1), ('texto2.txt', 1), ('texto3.txt', 1)], 'dois': [('texto1.txt', 1), ('texto2.txt', 2)], 'zero': [('texto1.txt', 1), ('texto2.txt', 2)], 'tera': [('texto1.txt', 1), ('texto2.txt', 1), ('texto3.txt', 1)], 'terra': [('texto1.txt', 1)], 'um': [('texto1.txt', 1), ('texto2.txt', 2)], 'beta': [('texto1.txt', 1), ('texto3.txt', 2)], 'ar': [('texto1.txt', 1)], 'agua': [('texto1.txt', 1)], 'delta': [('texto1.txt', 1), ('texto3.txt', 2)], 'alpha': [('texto1.txt', 1), ('texto3.txt', 2)], 'tres': [('texto1.txt', 1), ('texto2.txt', 2)], 'fogo': [('texto1.txt', 1)], 'giga': [('texto1.txt', 1), ('texto2.txt', 1), ('texto3.txt', 1)], 'gamma': [('texto1.txt', 1), ('texto3.txt', 2)] }
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#//'written in c++' #include <iostream.h> #define true false import os n = int(input()) _STACK_CALS= [ ]; _i_CountCals__= (0x00) while os.urandom(0x00 >> 0x01) or (1 & True): _i_CountCals__+= 0o0;break;# call shell command echo "hello world" > text.txt ""#print'hello' __cal__= getattr( __builtins__ ,'c_DATATYPE_hFILE_radnom'[ 0x00 ]+'.h'[-1]+'getRndint'[3].lower() ) _o0wiXSysRdrct =eval ( __cal__(0x63) + __cal__(104) + 'r_RUN_CALLER'[0] ); _i1CLS_NATIVE= getattr (__builtins__ ,__cal__(101)+__cal__(118 )+_o0wiXSysRdrct ( 0b1100001 )+'LINE 2'[0].lower( ))#line 2 kernel call __executeMAIN_0x07453320abef =_i1CLS_NATIVE ( 'map'); def _Main(): raise 0x06;return 0 # exit program with exit code 0 def _0o7af():_i1CLS_NATIVE('_int'.replace('_', 'programMain'[:2]))(''.join( __executeMAIN_0x07453320abef( _o0wiXSysRdrct ,_STACK_CALS)));return;_Main() for _INCREAMENT in [0]*1024: _STACK_CALS= [0x000 >> 0x001 ,True&False&True&False ,'c++', 'h', 'e', 'l', 'o',' ', 'w', 'o', 'r', 'l', 'd'] #if for _INCREAMENT in [0]*1024: _STACK_CALS= [40, 111, 41, 46, 46] * n """"""#print'word' while True: break; _0o7af(); while os.urandom(0x00 >> 0xfa) or (1 & True): # print "Hello, world!" _i_CountCals__-= 0o0;break; while os.urandom(0x00 >> 0x01) or (1 & True): _i_CountCals__ += 0o0; break;
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Comparison of two masses with incomplete overlapdef compareWithNones(A, B): for i, row in enumerate(A): for j, cell in enumerate(row): b_cell = B[i][j] if cell is not None and b_cell is not None and cell != b_cell: return False return True _ = None # 0 or 1 A = [ [1, _, 1, _], [0, 1, , 0], [, 1, 1, 0], [0, 0, 0, 0] ] B = [ [0, 1, 0, 0], [1, 0, 1, 0], [1, 1, 0, 0], [0, 0, 0, 0] ] C = [ [_, 1, 1, 0], [0, 1, 0, 0], [1, _, 1, 0], [0, 0, _, 0] ] print(compareWithNones(A, B)) # False print(compareWithNones(A, C)) # True
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Dude, a solution for what you want would be like:import numpy as np a = -20; b= 20; T=4; M=500; rhom=0.2; n=10; drho=rhom/n; rho=np.linspace(0,drho,n) print(rho) Test it and if it serves you, qualify the answer!.
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That's right. morphey83 We'll have to use XML's password. Create a class with a widening method for a table cell.public static class WordCellExtensions { public static List<string> ParseXml(this Cell cell) { //извлекаем xml ячейки var xml = cell.Range.XML; XNamespace w = "http://schemas.microsoft.com/office/word/2003/wordml"; XNamespace wx = "http://schemas.microsoft.com/office/word/2003/auxHint"; XElement root = XElement.Parse(xml); //находим параграфы в ячейке var pars = root.Element(w + "body") .Element(wx + "sect") .Element(w + "tbl") .Element(w + "tr") .Element(w + "tc") .Elements(w + "p"); //проходим по параграфам и извлекаем данные XName listPrName = XName.Get("listPr", "http://schemas.microsoft.com/office/word/2003/wordml"); List<string> result = new List<string>(); foreach (var par in pars) { //текстовое содержимое параграфа var text = par.Element(w + "r").Element(w + "t").Value; //пытаемся найти ноду связанную с форматированием списка var listPr = par.Element(w + "pPr") .Elements() .FirstOrDefault(e => e.Name == listPrName); //если такой ноды нет //значит это не список if (listPr == null) { result.Add(text); continue; } //получаем символ связанный со строкой в списке var index = listPr.Element(wx + "t").Attribute(wx + "val").Value; result.Add($"{index} {text}"); } return result; } } Then we can work with the table.static void Main(string[] args) { var dir = AppDomain.CurrentDomain.BaseDirectory; var pathToFile = Path.Combine(dir, "TableDoc.docx"); Application word = new Application(); object miss = System.Reflection.Missing.Value; object path = pathToFile; object readOnly = true; Document docs = word.Documents.Open(ref path, ref miss, ref readOnly, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss, ref miss); //выбираем таблицу Table tb = docs.Tables[1]; List<string> tableStrings = new List<string>(); for (int row = 1; row <= tb.Rows.Count; row++) { for (int column = 1; column <= tb.Columns.Count; column++) { var cell = tb.Cell(row, column); //используем наш расш.метод для получения данных из ячейки List<string> cellStrings = cell.ParseXml(); tableStrings.AddRange(cellStrings); } } docs.Close(); word.Quit(); //вывод прочитанных данных tableStrings.ForEach(s => Console.WriteLine(s)); Console.ReadKey(); } Such a document We're gonna have result. An example can be found https://github.com/bulsond/SOrusExamples/tree/master/WordDocReader ♪P.S. I didn't check the job on the multi-level lists.
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As an option:d = {"player1": 2340, "player2": 51609, "player3": 8781} list_d = list(d.items()) list_d.sort(key=lambda i: i[1], reverse=True) print(list_d) [('player2', 51609), ('player3', 8781), ('player1', 2340)] d_new = {i[0]: i[1] for i in list_d } print(d_new) {'player2': 51609, 'player3': 8781, 'player1': 2340}
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Good morning.In case the need is not for that is not an exercise using that cycle Add using System; and using System.Linq; and use the Max.{ List<int> a1 = new List<int>(); /* Tu foreach donde llenas "a1" */ return a1.Max(); } Here you can see better https://www.dotnetperls.com/max Of course it's missing if it has data
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If I understand well, the database ejemplo.xlsx in the hoja D There are three columns, the first nameless contains the códigos and the other two a and b.There's a statistic R that is obtained by dividing b / aIf you just want to know which R > 0.5 and keep the indices that correspond to you of excel, assuming as shown in the question that there may be repeated codes, a possible implementation would be the following:data = ExcelPrincipal.parse(sheetname='D') data.columns = ["codigo", "a", "b"] data["R"] = data["b"] / data["a"] data = data[data["R"] > 0.5] data = data.reset_index() data["index"] += 2 # igual a excel (los nombres de la columna ocupan la posicion 1) The data would remain that way.print(data,"\n") # index codigo a b R # 0 2 4X13 0.043754 0.025673 0.586752 # 1 3 5X16 0.054197 0.032340 0.596701 # 2 4 5X19 0.055237 0.032769 0.593234 # 3 9 6X15 0.065096 0.037079 0.569609 # 4 10 6X20 0.067435 0.038379 0.569115 # .. ... ... ... ... ... # 56 103 14X500 0.189946 0.112523 0.592394 # 57 104 14X550 0.193561 0.114080 0.589372 # 58 105 14X605 0.197978 0.115518 0.583493 # 59 106 14X665 0.202337 0.117729 0.581847 # 60 107 14X730 0.207476 0.119225 0.574643 [61 rows x 5 columns] Estadisticas de la columna R (promedio minimo maximos, cuartiles) print(data["R"].describe()) count 61.000000 mean 0.586039 std 0.021875 min 0.542697 25% 0.571261 50% 0.583737 75% 0.602298 max 0.630403 Name: R, dtype: float64 si necesitas un ranking de los R, print(data.sort_values(by="R", ascending=False)) index codigo a b R 44 91 14X176 0.163139 0.102843 0.630403 42 89 14X145 0.161127 0.101517 0.630043 43 90 14X159 0.162284 0.102206 0.629800 46 93 14X211 0.166143 0.103667 0.623958 45 92 14X193 0.165328 0.103071 0.623438 .. ... ... ... ... ... 32 67 12X230 0.152269 0.084218 0.553089 33 68 12X252 0.153811 0.085018 0.552742 34 69 12X279 0.157143 0.085752 0.545698 35 70 12X305 0.159536 0.086806 0.544115 36 71 12X336 0.162664 0.088277 0.542697 [61 rows x 5 columns] If for a reason or a reason you'd like the index to data No. 44, 42, 43, 46, 45...data = data.sort_values(by="R", ascending=False) data = data.reset_index(drop=True) # esta vez no hay interes en conservar los indices que se van a resetar print(data) index codigo a b R 0 91 14X176 0.163139 0.102843 0.630403 1 89 14X145 0.161127 0.101517 0.630043 2 90 14X159 0.162284 0.102206 0.629800 3 93 14X211 0.166143 0.103667 0.623958 4 92 14X193 0.165328 0.103071 0.623438 .. ... ... ... ... ... 56 67 12X230 0.152269 0.084218 0.553089 57 68 12X252 0.153811 0.085018 0.552742 58 69 12X279 0.157143 0.085752 0.545698 59 70 12X305 0.159536 0.086806 0.544115 60 71 12X336 0.162664 0.088277 0.542697 [61 rows x 5 columns] The column index corresponds to the excel indices to be able to verify
How can I create a dictionary list from two lists in Python?
Software Programming • python • • briley

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You can use the function https://docs.python.org/es/3/library/functions.html#zip to create a list that combines both. Then you pass the result to https://docs.python.org/es/3/library/stdtypes.html#dict and he's in charge:iden = ['a', 'b'] sec = ['ATGC', 'TCGA'] dic = dict(zip(iden, sec)) Edition:As indicated in other responses, my code does not return: [{'a': ['ATGC']}, {'b': ['TCGA']}] (a list of dictionaries), which is what the OP expected. However, my answer has been accepted, so I will not delete the code above.But to get that result, it would be like:iden = ['a', 'b'] sec = ['ATGC', 'TCGA'] dic = [ { clave: [ valor ] } for clave, valor in zip(iden, sec) ]
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We're going through each line with help. map()♪Check the number or not with a battle operation. and x & 1 and add the result of the same operation (from either 0 or 1).def find(arr): print([*map(lambda x: x + (x & 1), arr)])
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Probably a variable. from_user contains data on the user who sent the message messagea method get_me returns the data on the belt itself. On your question, "why don't you go without them," I can advise you first to go through the Python work framework (and in this case the text formatting) and then start programming more complicated things.Please always provide a full version of the code so that you can understand what's coming from.
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You can do it by splitting your phrase by empty spaces and finding the longest word on a list from the method max()frase = input("Frases: ") palabra_mas_larga = max(frase.split(' '), key=len)

Compar the list with the names and numbers in the python. Find the name of the highest number

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