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Break the line on Python.

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Duquan last edited by

There's a line like that. sdd,L12,W45,H78,7cm,K,D6
We need to break by 12.457.7.7 per individual variables to enter the base.
It's an eletary mission. ♪ We need to get to Python, never work for Python, thank you.
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carriann last edited by
I'm not sure I got the question right, but the line on the sequence of values breaks the split function:
```
           t=s.split(',')
```
And then I'd be with help and if I checked if the text contained a number, and if so, I would record it in the body.
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The following is a good idea.pred = torch.mean(predicted, feed_dict={x: np.array([tmp])}) predictedis passed with an argument,predictedis not defined in the source code
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//Строка которую нужно разбить: "4 5 29 54 4 0 -214 542 -64 1 -3 6 -6" function highAndLow(numbers) { var num = String(numbers).trim(); num = String(numbers).split(' '); function getMaxValue(num) { var max = +num[0]; for (var i = 0; i < num.length; i++) { if (max < +num[i]) max = +num[i]; } return max; } return parseInt(getMaxValue(num)); // вывод: 9, а не 542, так как строка // разделяется на символы, а не на отдельные числа } console.log(highAndLow("4 5 29 54 4 0 -214 542 -64 1 -3 6 -6"))
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I finally solved the problem.from oauth2client.client import flow_from_clientsecrets flow = flow_from_clientsecrets('cred_json_file_from_google_api_console', scope='https://www.googleapis.com/auth/contacts', redirect_uri='http://example.com/auth_return') auth_uri = flow.step1_get_authorize_url() #переходим по сгенерированной ссылке, переадресовываемся на страницу example.com/auth_return?code={code} credentials = flow.step2_exchange({code}) scope='https://www.googleapis.com/auth/contacts' auth2token = gdata.gauth.OAuth2Token(client_id=credentials.client_id, client_secret=credentials.client_secret, scope=scope, access_token=credentials.access_token, refresh_token=credentials.refresh_token, user_agent=None) gd_client = gdata.contacts.client.ContactsClient( source='airtable2gcontacts', auth_token=auth2token) print(gd_client.get_contacts()) #все работает
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Total negative numbers:b = sum(x for x in a if type(x) == int and x < 0) print(b) Total negative numbers c filter:d = sum(filter(lambda x: isinstance(x, int) and x < 0 , a )) print(d) The numbers are 0-1.c = sum(1 for x in a if x<1 and x>0) print(c)
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The solution to the forehead is not very effective, because the password can be generated in several passages, but the shortest solution, and the length of the bulb password will be minimal:import random password = '' while True: dictionary = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!&?' password = ''.join(random.choices(dictionary, k=random.randint(8, 16))) if any(l.isdigit() for l in password) and any(l.isupper() for l in password) and any(l.islower() for l in password) and any(l in '!&?' for l in password): break print(password) And that's how one-way algorithm looks:import random dictionary = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!&?' dictionaries = ['abcdefghijklmnopqrstuvwxyz', 'ABCDEFGHIJKLMNOPQRSTUVWXYZ', '0123456789', '!&?'] password = [random.choice(d) for d in dictionaries] + random.choices(dictionary, k=random.randint(8, 16) - 4) random.shuffle(password) password = ''.join(password) print(password) Or like that, to get rid of duplication of vocabularies:import random dictionaries = ['abcdefghijklmnopqrstuvwxyz', 'ABCDEFGHIJKLMNOPQRSTUVWXYZ', '0123456789', '!&?'] password = [random.choice(d) for d in dictionaries] + random.choices(''.join(dictionaries), k=random.randint(8, 16) - 4) random.shuffle(password) password = ''.join(password) print(password)
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The solution is simply to generate a new vector of predictions from the vector of probability of belonging to the reference class.To get the probability vectors, just do:y_scores=logistic.predict_proba(X_test) In the case of data from the example that have two classes, y_scores has two columns. The second column is that presents the vector of probability of belonging to the reference class. Now we can generate a new vector of predictions:pred2=pd.Series(y_scores[:,1]).map(lambda x: 1 if x > threshold else 0) Where threshold is the value chosen by the researcher. To see the difference, compare now classification report with the two prediction vectors:from sklearn.metrics import classification_report print(classification_report(y_test, pred)) Return: precision recall f1-score support 0 0.72 0.97 0.82 90 1 0.73 0.19 0.30 42 accuracy 0.72 132 macro avg 0.72 0.58 0.56 132 weighted avg 0.72 0.72 0.66 132 And:threshold=0.3 pred2=pd.Series(y_scores[:,1]).map(lambda x: 1 if x > threshold else 0) print(classification_report(y_test, pred2)) Return: precision recall f1-score support 0 0.86 0.56 0.68 90 1 0.46 0.81 0.59 42 accuracy 0.64 132 macro avg 0.66 0.68 0.63 132 weighted avg 0.73 0.64 0.65 132 Note that change increased model sensitivity from 0.19 to 0.81
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Unless the text in the bank is already marked with HTML, you will have this "trip" on the screen. If you are looking to present the text in your original spacing layout and lines, even if it is not marked with HTML, wrap it with the https://developer.mozilla.org/pt-BR/docs/Web/HTML/Element/pre .It is quite common to process the text of the bank before displaying it. For example, you can load the text in a vector, divided into paragraphs. Then print each element of this vector (a paragraph) with the https://developer.mozilla.org/pt-BR/docs/Web/HTML/Element/p :$paragrafos = explode("\n\n", $texto); foreach($paragrafos as $paragrafo) { echo '<p>' . $paragrafo . '</p>'; } Take a look at https://www.php.net/manual/pt_BR/function.explode.php and https://www.php.net/manual/pt_BR/control-structures.foreach.php .
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std::string s("Welcome to my home!"); s.replace(s.begin(), s.end(), ' ', '\n');
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This code defines how many methods exist to select 3 animals from the animal list.in this case:AA B C D You can choose.AAB, AAC, AAD, ABC, ABD, ACD, BCD I mean. 7 different optionsIn fact, for each animal, the bulbs are computed with other animals animals, and it's all sha works out.
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I'm guessing you're gonna need one of the keys that you've got, the other one as a value.$keys_string = 'Тут строка с индексом [0] из исходного массива'; $values_string = 'Тут строка с индексом [1] из исходного массива'; $keys = explode(',', $keys_string); // Это массив, который вы хотели получить $values = explode(',', $values_string); $data = array_combine($keys, $values); print_r($data); Result:Array ( ["Datasheets"] => "http://www.knowles.com/eng/content/download/3165/37817/version/2/file/bf-1861-000.pdf" ["Image"] => "http://media.digikey.com/Photos/Knowles%20Acoustics%20Photos/BF-1861-000.jpg" ["Digi-Key Part Number"] => "423-1158-ND" ["Manufacturer Part Number"] => "BF-1861-000" ["Manufacturer"] => "Knowles" ["Description"] => "ACOUSTIC DAMPER 1500 OHMS" ["Quantity Available"] => 1067 ["Factory Stock"] => 0 ["Unit Price (USD)"] => "1.67000" ["@ qty"] => 0 ["Minimum Quantity"] => 1 ["Series"] => "BF" ["Accessory Type"] => "Damper" ["Material"] => "Metal Ferrule Housing" ["Color"] => "Green" ["For Use With/Related Products"] => "Hearing Aids" ) http://ideone.com/QSHPMw I don't understand why your line wasn't broken correctly.
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Another solution:string retorno = $"<h3>Codigo: {codigoProduto} <br> Nome: {nome} <br> Valor: {valor} </h3>";

Break the line on Python.

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