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How to find the the sum of the total negative numbers of the mass and find the numbers of the numbers between 0 and 1 Python

C
chanisef last edited by
There's such a thing.
```
a = [-1.1, -1, 0.1, 1, -2.1, -2, 0.2, 2, -3.1, -3, 0.3]
```
from the cycle for (and no numpy) We need to get out of here.
1. Total negative numbers: (-1)+(-2)+(-3) = -6
2. Number of chips in range 0..1: 0.1, 0.2, 0.3 = 3
I'm a newcomer. Python I can only find the sum of all the negative numbers, but I can't deny them.
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1 Reply Last reply

emerson

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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for(int n = a; n <= b; ++n) { if (n%3 != 0 && n%2 != 0) { f += n; cout << "Число " << n << "\n"; } } cout << "Сумма равна " << f << "\n";
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Your biggest mistake is a complete division.1 / i when i>1 Output 0. Write the division ass += 1.0/i But still, your program will work in a really big way for a long time. For example, for 12, it takes 100,000 : float Maybe not enough.So I'd change the algorithm myself - for chisel to 5-10, I'd still think a row, and for the big ones, I'd use it. https://ru.wikipedia.org/wiki/%D0%93%D0%B0%D1%80%D0%BC%D0%BE%D0%BD%D0%B8%D1%87%D0%B5%D1%81%D0%BA%D0%BE%D0%B5_%D1%87%D0%B8%D1%81%D0%BB%D0%BE#%D0%9F%D1%80%D0%B8%D0%B1%D0%BB%D0%B8%D0%B6%D1%91%D0%BD%D0%BD%D0%BE%D0%B5_%D0%B2%D1%8B%D1%87%D0%B8%D1%81%D0%BB%D0%B5%D0%BD%D0%B8%D0%B5 ♪Well, for a < 3 There's only 11 terations, so I may have run a little ahead:
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Classic methodfunction calc(a) { var count={}, res=0, q; for (q=0; q<a.length; ++q) { count[a[q]] = ~~count[a[q]] + 1; } for (q in count) { if (count.hasOwnProperty(q) && count[q] > 1) { res += count[q]; } } return res; } calc([1,11,13,121,13,11,11,11,7,9,9]) // 8 ES6function calc(a) { let count = Object.create(null); for (let x of a) { = ~~count[x] + 1; } return a.length - Object.keys(count).filter(x => count[x]>1).length; } calc([1,11,13,121,13,11,11,11,7,9,9]) // 8 PS: I'm allowed to rewrite something on ES6.
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Your mistake in an unconditional assignment in the block:else { p = m; m = 0; } We need to check that. p less than m♪ For example:else { if (p < m) p = m; m = 0; }
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Simultaneous number - natural (target positive) number with a flat Two different natural dividers, one and one.In order to verify the mere number or not, it is sufficient to verify whether this number is divided by the number to the square root of this number. If it splits, it's not simple, otherwise it's simple.To determine whether the number is a simple method that defines a simple number or not:private static bool IsPrimeNumber(int number) { int sqrtNumber = (int) (Math.Sqrt(number)); for (int i = 2; i <= sqrtNumber; i++) { if (number % i == 0) return false; } return true; } Now, by the very word, the code in which we call the method:int count = 0; int maxNumber = 1000;// максимальное число, до которого будем считать (можно менять) // можно переделать на ввод с клавиатуры for (int num = 2; i <= maxNumber; num++) { if (IsPrimeNumber(num)) { count++; // Console.Write(num.ToString()+","); // можно вывести на экран } } // Console.WriteLine(); // отступ просто для красоты, если выводим на экран Console.WriteLine("Количество простых чисел в диапазоне от 0 до " + maxNumber); Console.WriteLine("Простых чисел: " + count); Result: In your case, the condition is wrong. Here's a learning page: http://ru.wikihow.com/%D0%BF%D1%80%D0%BE%D0%B2%D0%B5%D1%80%D0%B8%D1%82%D1%8C,-%D1%8F%D0%B2%D0%BB%D1%8F%D0%B5%D1%82%D1%81%D1%8F-%D0%BB%D0%B8-%D1%87%D0%B8%D1%81%D0%BB%D0%BE-%D0%BF%D1%80%D0%BE%D1%81%D1%82%D1%8B%D0%BC
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If you mean simply optimizing your division algorithm by attempt there are some important improvements you can make and they are very simple:Calculate the rest of the division only with the odds and 2. Note that possible factors will be primes and there are no pair cousins by definition (less than 2).Nor is it necessary to check from 2 to nJust do it until the square root of n. This is because of being a composite number there is always, at least, a factor that is less or equal to the square root. If we find a divisor we already know it's compound.You can also use a generator next to any.Your function may remain:def es_primo(n): # Comprobamos si n es 2 (unico primo par) if n == 2: return True # Comprobamos si es menor de 2 o es par if n < 2 or not n % 2: return False # Comprobamos si es divisible entre cualquier entero impar entre 3 y sqrt(n) return not any(n % i == 0 for i in range(3, int(n**0.5) + 1, 2)) The key is in range, the first parameter (start) is 3, the second (stop) is the square root of n plus 1 and the third parameter (step) is 2. This makes the range generates an odd number sequence that starts in 3 and goes from two to two square root n. Stop. n = 70 generate (3, 5, 7).n**0.5 is equivalent to making the square root of n.any return True at the time that any of the conditions are met, in this case at the time that some remainder is 0.For large numbers it will be a slow method obviously. In these cases there are multiple primality tests that we can implement that can be more or less complex, efficient or deterministic or probabilities ( Fermat test, Miller-Rabin test, AKS, etc). If by chance your purpose is to get a list of prime numbers from 2 to a given number, an Eratosthenian broom is very easy to implement and gives good results whenever n Don't be overly high.
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Igor, thank you!In the cross_validation class there was a y parameter (Samples to split in K folds):class sklearn.cross_validation.StratifiedKFold(y, n_folds=3, shuffle=False, random_state=None)[source]That's what I missed.If that's what it takes, it's all up. # simulate splitting a dataset of 25 observations into 5 folds from sklearn.model_selection import KFold kf = KFold(n_splits=5, random_state=None, shuffle=False) Vec = np.arange(0,26) # print the contents of each training and testing set print('{} {:^61} {}'.format('Iteration', 'Training set observations', 'Testing set observations')) for iteration, data in enumerate(kf.split(Vec), start=1): print('{:^9} {} {!s:^25}'.format(iteration, data[0], data[1]))

How to find the the sum of the total negative numbers of the mass and find the numbers of the numbers between 0 and 1 Python

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