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makes a mistake local variable 'randc' referenced before assignment dicsord.py.

Mystic

I made mistakes. Here's the code.

async def Mishka(ctx):
    global x
    r = (random.randint(1,1000))
    if r <= 799>800:
        randc = (random.choice(rand2))
    elif r >= 801<950:
        randc = (random.choice(rand1))
    elif r >= 951<1000:
        randc = (random.choice(rand3))
    await ctx.send(randc) ```

E
emerson last edited by

Condition x < y < z equivalent condition x < y and y < z♪ What is a sign of comparison there, the point is, how this is divided into two conditions, the rule is the same. Therefore, your first condition is never met at all, and the remaining right is always fulfilled. In general, it looks like you didn't have the conditions, and the variable didn't do anything before it was sent.
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keyboard = types.InlineKeyboardMarkup() keyboard.add(types.InlineKeyboardButton("Да", callback_data="yes")) keyboard.add(types.InlineKeyboardButton("Нет", callback_data="no")) Everything else is right.
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Let us continue the English lesson initiated by aleksandr barakinnumber means number, numbervariable means variabledoesn't match means no match.number means number, numberparameter means the parameter, the sign of the shellprepared statementi.e. the text of the error, in English translation, means that the number of variables we attach to the request for bind_param does not correspond to the number of parameters(s) in the request.We're going to the code and we're thinking:Number of variables in bind_param: 1 (one)Number of parameters: 0 (l) grandTo address the problem, the number of parameters(s) in the request for conformity with the number of variable functions bind_param, i.e. add a 1-kilometre to the request. By removing the variable from the request. Since the nature of the requests prepared is to replace variables with parameters.In general, the problem shows the advantage of literacy (i.e. reading) and knowledge of foreign languages before illiteracy (i.e. ignorance of reading) and non-recognition of foreign languages in number, variable and parameter.
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You can do that. (we're computing the average for each dictionary, we're making a list of them, we're computing the average on that list):a = [sum(scor['scores'])/len(scor['scores']) for scor in ocenka] print(sum(a)/len(a)) Or so. (made scores from all vocabularies to one large list and calculate the average):a = sum((scor['scores'] for scor in ocenka), []) print(sum(a)/len(a))
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Because... func inside decorator removed from worker in a global context.let worker = { someMethod() { return 1; }, slow(x) { alert("Called with " + x); return x * this.someMethod(); // (*) } }; function decorator(func) { let cashe = new Map(); return function(x) { if (cashe.has(x)) { return cashe.get(x); } let result = func(x); cashe.set(x, result); return result; }; }; worker.slow = decorator(worker.slow.bind(worker)); // !!! bind console.log(worker.slow(1))
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Collect been helpful. If anyone comes to the same question, that's the solution:from datetime import date as Dtt https://ru.stackoverflow.com/q/1100668 - Copy the grading from here, made a check for the presence of a quote in a given set of "l" lists (Same two types of data that it cannot sort contain points = Date in the format: DD.M.GHG and float respectively).If one point, the random variable(s select_func) = 1, two points select_func = 2. (The whole code goes over l.sort, which is in the reference decision) select_func = 0 for one_l in l: if '.' in one_l[0]: if len(one_l[0].split('.'))<3: select_func = 1 elif 2<len(one_l[0].split('.'))<4 select_func = 2 The inspection function is further challenged (/or sorting (if this date/float) I've got an entrance check: a set of "l" and a variable select_func. l = prover_ochka(l,select_func) First, we're building a two-band from a list of "l"s. def prover_ochka(l,select_func) mda = [] for i in mda: kon = [] kon.append(i[0]) kon.append(i[1]) mda.append(kon) Next, we check the value of selecting_func, which will determine our grading. if select_func == 1: for i in mda: if i[0].isdigit() == True: i[0] = float(i[0]) elif i[0].split('.')[0].isdigit() == True and i[0].split('.')[1].isdigit() == True: i[0] = float(i[0]) else: return None return mda elif select_func == 2: for i in mda: date_i = i[0] i[0] = Dtt(int(date_i[6:]),int(date_i[3:5]),int(date_i[:2])) return mda else: return mda Then we're going back to sorting. ♪ ♪ l = prover_ochka(l,select_func) if l != None: if select_func != 2: l.sort(reverse = reverse) else: l.sort(reverse = reverse, key = lambda x: x[0]) for index, (val, k) in enumerate(l)# продолжение кода, что по ссылке I'm sorry, I can't drop the whole code because I don't know if there's going to be any problems in the course.In general, I am comfortable with this decision because the key elements(s) are in any case checked at the data entry (they will always be in the right ring). If you don't have one, it's not a big problem, but the code might be useful.
What is the join() function in the threading module?
Software Programming • python python 3.x join thread multithreading • • Laycee

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O .join simply pauses the current thread (from where it was called) until the target thread (the thread to which it is tied) is finished.It is a way to synchronize the program again, ensuring that no parallel processing is taking place.What he No makes it the most important: it does not transmit any signal, makes any request, or causes the target thread to be terminated. The program continues to run normally, the target thread continues its processing – and when it is terminated, the join function returns.In other words, the call minhathread.join() is the equivalent of "pause the program here and expect mythread to end."If you need a way to warn another thread that it should end, you have to take a look at the thread and queues events - and your code that runs on the thread has to look at these messages and close the execution. In Python there is no way an external signal closes a thread running, without her code predicting it. (in other languages there is also no safe way to do this, only "violent stops")The join method also accepts a timeout parameter in seconds: it is the maximum time the execution will wait for the closure of the other thread - after that time the call returns, and you have to check if the target thread is still active by calling the method .is_alive hers.Below, I play with two threads in an interactive session, and demonstrate the use of a "threading" object. Event" to stop a thread, and the "join" in a second thread that awaits the end of the first: In [227]: import threading In [228]: evento = threading.Event() In [229]: def contador(): ...: i = 0 ...: while True: ...: parar_agora = evento.wait(1) ...: if parar_agora: ...: break ...: i += 1 ...: print(i) ...: In [230]: def espera_contador(t): ...: t.join() ...: print("O contador parou") ...: In [231]: t1 = threading.Thread(target=contador); t2 = threading.Thread(target=lambda: espera_contador(t1)) In [232]: t1.start(); t2.start() ...: 1 2 3 4 5 6 In [233]: evento.set() O contador parou (The use of lambda in lambda: espera_contador(t1) was made to pass the "t1" parameter to the call, since "target" does not accept parameters - then I remembered that I could have used the "args" parameter when creating the thread: t1 = threading.Thread(target=contador); t2 = threading.Thread(target=espera_contador, args=[t1]))
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The object(s) of the class is in the only copy, the methods may alter it, read, do anything. The method does not make a copy of the variable, which is the variable class only if the variable is not declared inside the body of the method. If variable static, it exists in the only copy for all copies of this class.
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In Python, the masses (or more list or leaf) are not declared as size. For example, items = [] Initially empty, however, after the programme has been implemented, its size may be len(items) = 5 or len(items) = 10♪Call attention. https://docs.python.org/3/library/stdtypes.html#list ♪
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Of the problems there are two, which affect notably the functioning of your neuronal network:1. Few dataNeuronal networks need a large amount of data (depends on the problem) there are cases in which with something more than a thousand is sufficient (pocos), if the dependent variable is easily predictable, but always the more data better, in fact most neural networks in real life have millions of data.More data, easier it will be to the neural network generalize and capture the varianza of your data properly, assigning weights to neurons and corresponding activations according to the type of data.2.Do not standardize the dependent variableIn this case you are standardizing the dependent variable (what you want to predict), which makes no sense, the reason for standardizing the independent variables (predictives) is that the difference of magnitudes between different variables, does not lead the neuronal to ponder some variables more important than others by the simple fact of having larger numerical magnitudesas this causes the neuronal network capture me. the variance of your data.In this case the dependent variable is not predicting anything, is being predicted by the other variablesSo you shouldn't standardize it.
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Before working with Money You need to initiate it.As in the design cycle try-catch You don't have an initialisation of the variable. try It didn't work. pricelist.append I don't know what it is.
Function makes a mistake.
Software Programming • javascript regular expression • • carriann

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Use this regex to test the e-mail:/^[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,6}$/i At least you don't have a advantage at the end. Besides, according to your regularity as a whole, the address allows sense of the type. "asdf@lkj@lkj@.s"♪function validateEmail(field) { field = field.toLowerCase(); if (field == "") { return "Не введен адрес электронной почты.\n"} else if (!(/^[a-z0-9._%+-]+@[a-z0-9.-]+\.[a-z]{2,6}$/.test(field))) {return "Адрес электронной почты имеет неверный формат.\n"} return "" }
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Graphic large LBP image histogram?
Software Programming • python matplotlib intelligenc languagee art • • jonetta

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If I am not mistaken, what is done by representing the histogram after applying LBP It's in the eje x represent the value of each pixel and in the eje y the number of absolute pixels or their percentage on the total.The problem does not depend on the size of the image, there are images of 640 x 400 that cause the same error. The problem is that you're going through a two-methon array to create a dimention histogram. If you want to use a 1D histogram you should pass a unidimencinal array that is what you expect to receive matplotlib. When receiving a 2D array tries to transform it into an 1D array and that causes the WarningWhich, on the other hand, doesn't prevent me from finally graphing something.I mean lbp is a form matrix:lbp = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] To fix it you can use the method https://docs.scipy.org/doc/numpy/reference/generated/numpy.ravel.html of NumPy. This way we get the unidimentional array:lbp = [1, 2, 3, 4, 5, 6, 7, 8, 9,] The code would be:import matplotlib.pyplot as plt import numpy as np from skimage.feature import local_binary_pattern from skimage import io, util lbp_radius = 3 lbp_n_points = 8 * lbp_radius lbp_METHOD = 'nri_uniform' image = io.imread('prueba2.tif', as_grey=True) image = util.img_as_ubyte(image) lbp = local_binary_pattern(image, lbp_n_points, lbp_radius, lbp_METHOD) plt.hist(lbp.ravel(),59,[0,59]) plt.ylim([0,500]) plt.show() For example, with your code the following image of 640x400 gives me the same mistake as you: The result using my modification does not generate that Warning and we get:

makes a mistake local variable 'randc' referenced before assignment dicsord.py.

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