SciPy has added an inverse Box-Cox transformation. https://docs.scipy.org/doc/scipy/reference/generated/scipy.special.inv_boxcox.html scipy.special.inv_boxcox scipy.special.inv_boxcox(y, lmbda) = Compute the inverse of the Box-Cox transformation. Find x such that: y = (x**lmbda – 1) / lmbda if lmbda != 0 log(x) if lmbda == 0 Parameters: y : array_like Data to be transformed. lmbda : array_like Power parameter of the Box-Cox … Read more
Here is a more elegant and scalable solution, imo. It’ll work for any nxn matrix and you may find use for the other methods. Note that getMatrixInverse(m) takes in an array of arrays as input. Please feel free to ask any questions. def transposeMatrix(m): return map(list,zip(*m)) def getMatrixMinor(m,i,j): return [row[:j] + row[j+1:] for row in … Read more
Just assign using the logical NOT operator ! like you tend to do in your condition statements (if, for, while…). You’re working with a boolean value already, so it’ll flip true to false (and vice versa): myBool = !myBool;
In case of many-to-many relation through intermediary table; “Cascade” says whether a record will be created/updated in the child table. Whereas “Inverse” says whether a record will be created/updated in the intermediary table e.g. Assume below scenario 1 student can have multiple phones. So Student class has property for Set of phones. Also 1 Phone … Read more
You use look ahead assertions to check if a string contains a word or not. If you want to assure that the string contains “Android” at some place you can do it like this: ^(?=.*Android).* You can also combine them, to ensure that it contains “Android” at some place AND “Mobile” at some place: ^(?=.*Android)(?=.*Mobile).* … Read more
As Matthieu says, the only case where you wouldn’t want to set inverse = true is where it does not make sense for the child to be responsible for updating itself, such as in the case where the child has no knowledge of its parent. Lets try a real world, and not at all contrived … Read more
The inverse is XOR! If you have: c = a^b; You can get a or b back if you have the other value available: a = c^b; // or b^c (order is not important) b = c^a; // or a^c For example if a = 5, b = 3 (and thus c = 6 as … Read more