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Why is the verification of the presence of a multiplicity element on the list taking place at different times?

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charlenef last edited by

Why is the verification of the presence of a multiplicity element on the list taking place at different times?
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shizuka last edited by
Because on the list (massium) to verify that there is an element, we need to look at all the elements, that is, spend O(n) time (line time)
a multiples to understand the element index where the element is stored can calculate the xash from the element
```
index = hash(value)
```
In this case, the search for the element will be the constant time O(1)
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Found the answer!In importing the database, we need to use coding. latin1Then Python3 needs to be implemented to decode these lines. str.encode('latin1').decode('utf8')♪This article has prompted a decision: http://jalada.co.uk/2011/12/07/solving-latin1-and-utf8-errors-for-good-in-ruby.html :Latin1 is a character encoding used by MySQL. People mistakenly think it is equivalent to ISO-8859-1 but it isn't, it’s actually CP-1252 (also known as Windows-1252). CP-1252 is a superset of ISO-8859-1 with some additional characters (recently including the I symbol). I found it. https://dev.mysql.com/doc/refman/5.5/en/charset-charsets.html latin1and used it to connect to MySQL with Connector/Python.
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The data structure should be separated from the indicator to the next Node♪ Then we transfer the data structure to the node added function. And the list itself will be making nodes to its discretion. Doing auxiliary functions that return to the first and last of the list. (All right.) It'll make it easier to do that.// g++ -Wall -Wextra -std=c++11 actions.cpp -o actions # include <string> using std::wstring; # include <iostream> using std::wcout; struct Data { wstring Adres; double square; }; class Actions{ public: Actions(void){} ~Actions(void); void insBack(Data);//Добавление в элемента в конец void inFront(Data d){start = new Node{d,start};}//Добавление элемента в начало struct Node{ Data d ; Node * next; }; Node * first(void){return start;} Node * last(void); private: Node *start{nullptr};//Указатель на начало списка }; Actions::Node * Actions::last(void){ Node * i = start ; Node * r = nullptr ; while(i){ r = i ; i=i->next; } return r ; } Actions::~Actions(void){ while(start){ Node * next = start->next; delete start ; start = next ; } } void Actions::insBack(Data d){ Node * l =last(); Node **ps; if(l) ps = &(l->next); else ps = & start; (*ps)= new Node{d,nullptr};} int main(void){ setlocale(LC_ALL,""); Actions a ; a.inFront(Data{L"Фрунзе",12}); a.insBack(Data{L"Ленина",7}); for(Actions::Node * i = a.first();i;i=i->next) wcout<<L"адрес = "<<i->d.Adres<<L" , квартал = "<<i->d.square<<std::endl; }
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Turns out the requests work so he doesn't need that line. post_data = urllib.urlencode( req ) But in this case, it needs to be left to create a signature. And one line of the code will change:r = requests.post( url, data = post_data, headers = headers ) Here.r = requests.post( url, data = req, headers = headers )
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You can do that.struct NameEqualTo { NameEqualTo(const std::string& name) : name_(name) {} bool operator()(const librery& l) const {return l.name == name_;} std::string name_; }; bibl.remove_if(NameEqualTo("John")); If you use the boost, you can write it shorter:bibl.remove_if(boost::bind(&librery::name, _1) == "John"); You can just walk with your hands:for (std::list<librer>::iterator it = bibl.begin(); it != bibl.end();) { if (it->name == "John") { it = bibl.erase(it); } else { ++it; } }
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Why don't you just give it to Record? And Record designer checks if there's anything. In this case, it's all very simple.class Record: def __init__(self, data): self.id = data['id'] self.type = data['type'] self.number = data['number'] self.video = data['video'] self.audio = data['audio'] keys = ['id', 'type', 'number', 'video', 'audio'] temp = {} for key in keys: print('enter value for ', key) value = input('enter value: ') temp[key] = value obj= Record(temp) You can go further and write a smart designer.keys = ['id', 'type', 'number', 'video', 'audio'] class Record(object): def init(self, data): for key in keys: setattr(self, key, data[key]) temp = {} for key in keys: print('enter value for ', key) value = input('enter value: ') temp[key] = value obj= Record(temp)
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As @crlsktr points out in his comment, bookstores like PIL are designed for that! Code of Example!from PIL import Image import numpy as np w, h = 512, 512 data = np.zeros((h, w, 3), dtype=np.uint8) data[0:256, 0:256] = [255, 0, 0] # Habrá rojo en la esquina superior izquierda im = Image.fromarray(data, 'RGB') im.save('ejemplo.jpg') im.show() # Para que la muestre
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The reference is because you use the method. send_message instead of method send_photo details of method https://core.telegram.org/bots/api#sendphoto and https://github.com/eternnoir/pyTelegramBotAPI to your library.
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I think it's easier to solve the problem using two iterators instead of just one:void homogeniza(std::list<int>& C, int hmin, int hmax) { for( auto it = C.begin(); it != C.end(); ++it) { auto it2 = std::next(it,1); if( it2 != C.end() ) { for( ; it2 != C.end() && ( *it + hmin > *it2 ); ) it2 = C.erase(it2); if( it2 != C.end() ) { if( *it + hmax < *it2 ) C.insert(it2,*it+hmax); } } } } Using only one iterator complicates the code as you have to be playing all the time with scrolls on the same iterator... you hear, it is also possible to do it with one only:void homogeniza(std::list<int>& C, int hmin, int hmax) { for( auto it = C.begin(); it != C.end(); ++it) { int valActual = *it; std::advance(it,1); if( it != C.end() ) { for( ; it != C.end() && ( valActual + hmin > *it ); ) it = C.erase(it); if( it != C.end() ) { if( valActual + hmax < *it) it = C.insert(it,valActual+hmax); } } std::advance(it,-1); } }
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The answer was found in my article, and the problem with the erroneousness decided to list all fields in the meta class, the total meth class:class PhotoAdminForm(forms.ModelForm): class Meta: model = Photo widgets = {'image':MultiFileInput} fields="all"
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Baseline:data = \ [['var1', 'val1'], ['var2', 'val2'], ['var1', 'val4'], ['var3', 'val1'], ['var4', 'val1'], ['var4', 'val15']] Use your decision. https://docs.python.org/3/library/itertools.html#itertools.groupby :from itertools import groupby res = [[a, "+".join(el[1] for el in b)] for a,b in groupby(sorted(data, key=lambda x: x[0]), key=lambda x: x[0])] result:In [26]: res Out[26]: [['var1', 'val1+val4'], ['var2', 'val2'], ['var3', 'val1'], ['var4', 'val1+val15']]
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With https://docs.python.org/3.6/library/itertools.html#itertools.chain you can use the expansion of elements:list(itertools.chain(*A)) If not, use a double understanding of lists:[item for lista in A for item in lista] Done A=[[1,2,3],[4,5]] in both cases we get [1, 2, 3, 4, 5].
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It's like calling a cycle. Here, instead of the two invested. for two filter c lambdabut it's basically a cycle of complexity. O(n^2):print(list(filter(lambda x: any(filter(lambda y: y in x and x != y, b)), a))) Conclusion:['500бухт', '700м'] P.S. By the way, MaxU's right-- there's not much inside the word "m." It's better if it's true to use regulars, or at least check through. inand through endswithif the line is from b should be at the end of the line, and before that, the number can be checked without regular intersections. isdigit:print(list( filter(lambda x: any( filter(lambda y: x.endswith(y) and x[:-len(y)].isdigit() and x != y, b) ), a) ))
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import asyncio from concurrent.futures import ThreadPoolExecutor async def gen_values(file): for s in file: s1, s2 = s.split() yield s1, s2 def write(w, i_1, i_2, k_1, k_2): w.poly([[[float(i_1), float(k_1)], [float(i_2), float(k_1)], [float(i_2), float(k_2)], [float(i_1), float(k_2)], [float(i_1), float(k_1)]]]) w.record(f'{i_1}_{k_1}_{i_2}_{k_2}') async def main(): w = shapefile.Writer(f'shapefiles/test/Indonesia') w.field('Latitude', 'C') loop = asyncio.get_event_loop() with ThreadPoolExecutor(max_workers=96) as pool: with open('long_1_double.txt', 'r') as long: with open('short_1_double.txt', 'r') as short: async for i1, i2 in gen_values(long): async for k1, k2 in gen_values(short): await loop.run_in_executor(pool, lambda: write(w, i1, i2, k1, k2)) if __name__ == '__main__': asyncio.run(main()) If you make it work, you can try it. from concurrent.futures import ProcessPoolExecutor instead ThreadPoolExecutorRuled the code
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Find out how many certain elements are in the vicinity for another certain element on the list?
Software Programming • python list count • • rosemadder

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use numpy # исходная матрица m = np.array(m) # делаем маску шире и выше на 2, чтобы окружить нулями mask = np.zeros((m.shape[0]+2,m.shape[1]+2)) # и такой же размерности массив под суммы s = m.copy() # середину маски заполняем 1 на местах мин :) mask[1:-1,1:-1] = (m=='x').astype(int) // подвигаем верх-вниз и влево-вправо for axis in [0,1]: for sign in [1,-1]: s += np.roll(mask.copy(),sign,axis) // а теперь по диагоналям for x in [1,-1]: for y in [1,-1]: s += np.roll(mask.copy(),(x,y), (0,1)) // осталось оформить ответ s = s[1:-1,1:-1] np.where(m=='x', 'x', s.astype(int).astype(str)) array([['x', '1', '0'], ['2', '3', '2'], ['1', 'x', 'x'], ['1', '2', '2']], dtype='<U11') You can't use it.def count(m, row, col): if m[row][col] == 'x' : return 'x' # слайс по рядам temp = m[(row-1 if row else 0):row+2] # в каждом ряду в слайсе по колонке считаем количество "х", # а потом всё это складываем return str(sum(map(lambda x: sum(i=='x' for i in x[(col-1 if col else 0):col+2]), temp)))
How to determine the number of different elements on the orderly list?
Software Programming • python algorithm • • emmalee

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Enough with one cycle: consider the difference between the two adjacent elements. If not 0, they're different. It's if the numbers are. If not the numbers, just compare the neighboring elements.But not at all.n = len(set(spisok))

Why is the verification of the presence of a multiplicity element on the list taking place at different times?

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