It is possible to do it in O(n) time and O(1) space apparently. The paper Stable minimum space partitioning in linear time by Jyrki Katajainen, and Tomi Pasanen claims to be able to do it. Google for stable 0-1 sort.
There is a trivial O(n^2) algorithm, but you can do this in O(n). E.g.: A = [a, b, c, d, e] P = [4, 3, 2, 0, 1] We can swap each element in A with the right element required by P, after each swap, there will be one more element in the right position, … Read more
You probably want to do this: jQuery(“body”).html(“new content”); …where “new content” would ideally only include the markup that would normally appear within the body element and not the rest. That will replace the body element’s contents, whilst leaving anything you have in head (like style sheet information) alone. If you also want to update the … Read more
Get largest sub-array having size 3k+1 Apply cycle leader algorithm to the parts of this sub-array, starting from positions 1, 3, 9, … 3k-1: move element to its proper position in sub-array (even-indexed elements to the left of sub-array, odd-indexed – to the right), the replaced element should be also moved to its proper position, … Read more
From the dropna docstring: Drop the columns where all elements are NaN: df.dropna(axis=1, how=’all’) A B D 0 NaN 2.0 0 1 3.0 4.0 1 2 NaN NaN 5
Simple replacement with sed is okay but not the best possible solution. If there are “extra” spaces between the tabs they will still be there after substitution, so the margins will be ragged. Tabs expanded in the middle of lines will also not work correctly. In bash, we can say instead find . -name ‘*.java’ … Read more
lista = list.sort(lista) This should be lista.sort() The .sort() method is in-place, and returns None. If you want something not in-place, which returns a value, you could use sorted_list = sorted(lista) Aside #1: please don’t call your lists list. That clobbers the builtin list type. Aside #2: I’m not sure what this line is meant … Read more
I’d write it this way: a[i:j] = sorted(a[i:j]) It is not in-place sort either, but fast enough for relatively small segments. Please note, that Python copies only object references, so the speed penalty won’t be that huge compared to a real in-place sort as one would expect.
If you want to apply mathematical operations to a numpy array in-place, you can simply use the standard in-place operators +=, -=, /=, etc. So for example: >>> def foo(a): … a += 10 … >>> a = numpy.arange(10) >>> a array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]) >>> foo(a) >>> a … Read more
This is called the placement new operator. It allows you to supply the memory the data will be allocated in without having the new operator allocate it. For example: Foo * f = new Foo(); The above will allocate memory for you. void * fm = malloc(sizeof(Foo)); Foo *f = new (fm) Foo(); The above … Read more