You can assign to a slice: a[:0] = b Demo: >>> a = [1,2,3] >>> b = [4,5,6] >>> a[:0] = b >>> a [4, 5, 6, 1, 2, 3] Essentially, list.extend() is an assignment to the list[len(list):] slice. You can ‘insert’ another list at any position, just address the empty slice at that location: … Read more
In C++03 (for which most of “cplusplus.com”‘s documentation is written), the complexities were unspecified because library implementers were allowed to do Copy-On-Write or “rope-style” internal representations for strings. For instance, a COW implementation might require copying the entire string if a character is modified and there is sharing going on. In C++11, COW and rope … Read more
If I start with a 3×4 array, and concatenate a 3×1 array, with axis 1, I get a 3×5 array: In [911]: x = np.arange(12).reshape(3,4) In [912]: np.concatenate([x,x[:,-1:]], axis=1) Out[912]: array([[ 0, 1, 2, 3, 3], [ 4, 5, 6, 7, 7], [ 8, 9, 10, 11, 11]]) In [913]: x.shape,x[:,-1:].shape Out[913]: ((3, 4), (3, … Read more
Keep in mind that what looks poor asymptotics might actually not be, because you are working in a lazy language. In a strict language, appending to the end of a linked list in this way would always be O(n). In a lazy language, it’s O(n) only if you actually traverse to the end of the … Read more
The builtin functions are cat, vertcat, and horzcat, found on pages 380-381 of the Octave documentation (v 3.8). They are essentially equivalent to what you have though. octave:5> A = [2 3 4]; octave:6> A = [A; 3 4 5] A = 2 3 4 3 4 5 octave:7> B = [4 5 6]; octave:8> … Read more
You could do if item not in mylist: mylist.append(item) But you should really use a set, like this : myset = set() myset.add(item) EDIT: If order is important but your list is very big, you should probably use both a list and a set, like so: mylist = [] myset = set() for item in … Read more
list.append mutates the list itself and returns None. List comprehensions are for storing the result, which isn’t what you want in this case if you want to just change the original lists. >>> x = [[1, 2], [3, 4], [5, 6]] >>> for sublist in x: … sublist.append(‘a’) … >>> x [[1, 2, ‘a’], [3, … Read more
x.extend(y+z) should do what you want or x += y+z or even x = x+y+z
Use dstack: >>> np.dstack((A, B)).shape (480, 640, 4) This handles the cases where the arrays have different numbers of dimensions and stacks the arrays along the third axis. Otherwise, to use append or concatenate, you’ll have to make B three dimensional yourself and specify the axis you want to join them on: >>> np.append(A, np.atleast_3d(B), … Read more
Here’s the trick: subclass ObjectOutputStream and override the writeStreamHeader method: public class AppendingObjectOutputStream extends ObjectOutputStream { public AppendingObjectOutputStream(OutputStream out) throws IOException { super(out); } @Override protected void writeStreamHeader() throws IOException { // do not write a header, but reset: // this line added after another question // showed a problem with the original reset(); } … Read more