remove last append element jquery
As an alternative, for those who likes more programmatic ways and syntax: $(‘#container-element img’).last().remove();
As an alternative, for those who likes more programmatic ways and syntax: $(‘#container-element img’).last().remove();
In C++03 (for which most of “cplusplus.com”‘s documentation is written), the complexities were unspecified because library implementers were allowed to do Copy-On-Write or “rope-style” internal representations for strings. For instance, a COW implementation might require copying the entire string if a character is modified and there is sharing going on. In C++11, COW and rope … Read more
You can use the concat function for this. Expanding upon the example in your question: variable “foo” { type = “list” default = [ 1,2,3 ] } # assume a value of 4 of type number is the additional value to be appended resource “bar_type” “bar” { bar_field = “${concat(var.foo, [4])}” } which appends to … Read more
Open the file for ‘append’ rather than ‘write’. with open(‘file.txt’, ‘a’) as file: file.write(‘input’)
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
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