Efficiently sorting a numpy array in descending order?
temp[::-1].sort() sorts the array in place, whereas np.sort(temp)[::-1] creates a new array.
temp[::-1].sort() sorts the array in place, whereas np.sort(temp)[::-1] creates a new array.
Here is Java 8 solution
You can’t directly do array2 = array1, because in this case you manipulate the addresses of the arrays (char *) and not of their inner values (char). What you, conceptually, want is to do is iterate through all the chars of your source (array1) and copy them to the destination (array2). There are several ways to … Read more
There’s Not that it does anything different than you’d do on your own (it just loops through every element and sets it to null). It’s not native in that it’s pure Java code that performs this, but it is a library function if maybe that’s what you meant. This of course doesn’t allow you to resize the array (to … Read more
You’re accessing memory beyond the range of the allocated array The first element is [0]. The last is [seatNum-1]
Try this: See Ruby’s Enumerable Documentation (note: the 0 base case is needed so that 0 will be returned on an empty array instead of nil)
Assuming student::name is a char array or a pointer to char, the following expression compares pointers to char, after decaying sName from char[28] to char*. Given that you want to compare the strings container in these arrays, a simple option is to read the names into std::string and use bool operator==: This will work for … Read more
In newer version of numpy (>1.8.x), numpy.meshgrid() provides a much faster implementation: @pv’s solution numpy.meshgrid() use to be 2D only, now it is capable of ND. In this case, 3D: Note that the order of the final resultant is slightly different.
Is there a numpy method which is equivalent to the builtin pop for python lists? Popping obviously doesn’t work on numpy arrays, and I want to avoid a list conversion
char* and char[] are different types, but it’s not immediately apparent in all cases. This is because arrays decay into pointers, meaning that if an expression of type char[] is provided where one of type char* is expected, the compiler automatically converts the array into a pointer to its first element. Your example function printSomething … Read more