Fixed-Length | Scala library for parsing fixed length file format | Parser library

Actually, the loop invariant 0 <= i <= n is not true if n is strictly negative. A loop invariant must hold the first time you reach the loop (this is what the "established" part means, as opposed to the "preserved" part in which you must check that it holds at the end of any loop step, assuming it was true at the beginning of said step), even if you don't end up entering the loop at all (which is obviously the case here if n<0). You must thus add a requires n >=0; in the contract of the function.

EDIT I forgot to mention that another solution is of course to make n and i unsigned (or even better to #include and use size_t): this way, you are guaranteed a positive number without having to write an additional requires)

Insted of doing queue.pop(0) set range in xaxis to length - 1 of your initial data (if type: "numeric"). https://apexcharts.com/docs/options/xaxis/#range

They are compatible. C 2018 6.7.6.2 6 says “For two array types to be compatible, both shall have compatible element types, and if both size specifiers are present, and are integer constant expressions, then both size specifiers shall have the same constant value…” Since it is not true that both size specifiers are integer constant expressions, the requirement to have the same value does not apply.

The pointer compatibility inherits from the pointed-to types.

I think this is what you are looking for?:

You can't improve the computational complexity of your algorithm unless you're able to tweak the output format (see below). In other words, you'll at best be able to improve the speed by a modest percentage (and the percentage will be independent of the size of the input).

I don't see any obvious implementation issues. The one idea I had was to get rid of the large number of append() calls and the overhead that is incurred by gradual list expansions by preallocating the output matrix, but @juanpa.arrivillaga suggests in their comment that append() is in fact very optimized on CPython. If you're on another interpreter, you could try it: you know that the length of the output list for column c will be equal to the sum of all the input numbers in column c. So you can just preallocate each output list by using [0] * sum_of_input_values_at_column_c, and then do proc_list[i] = n instead of proc_list.append(n) (and manually increment i). This does, however, require two passes over the input, so it might not actually be an improvement - your problem is quite memory-intensive as its core computation is extremely simple.

The reason that you can't improve the computational complexity is that it is already optimal: any algorithm needs to spend time on generating its output, so the size of the output is a lower bound for how fast the algorithm can possibly be. And in your case, the size of the output is equal to the sum of the values in your input matrix (and it's generally considered bad when you depend on the input values themselves rather than on the number of input values). And that's the number of iterations that your algorithm spends, so it is optimal. However, if the output of this function is going to reside in memory to be consumed by another function (rather than being written to a file), and you are able to make some adaptations in that function, you could instead output a matrix of generators, where each generator knows that it needs to generate sub_list[k] occurrences of n. Then, the complexity of your algorithm becomes proportional to the size of the input matrix (but consuming the output will still take the same amount of time that it would have taken to generate the full output).

While not quite able to "choose the best (all or nothing) transformer subset of features", we can use scikit's feature selection or dimensionality reduction modules to "choose/simplify the best feature subset across ALL transformers" as an extra step before classification:

The message says to use [] to declare a new empty list. Like this:

This answer is largely based on the software engineering stack thread asking a related question, which itself is based on the KMP algorithm.

The idea is to build a deterministic finite automaton (DFA) whose transitions are the lowercase letters that can encode the property of matching a pattern zero, one or 2+ times. For example, with pattern = 'radar', we have a pattern length n of 5, and so we'll have n + n + 1 states in total in our DFA. It's helpful to think of these as 3 rows of states:

• Row 0, column i is the state corresponding to 0 full matches, and the last i letters seen form our current partial match.
• Row 1, column i is the state corresponding to 1 full match, and the last i letters seen form our current partial match.
• Row 2 has a single state which can't be escaped: We have seen at least 2 full matches already.

Our 'row' can only ever increase. In the code, there's really only one flat list of 2n+1 states, but that's just an implementation detail.

A partial drawing of the DFA graph for 'radar' is shown: almost all edges are not drawn, since every state has 26 outward transitions. Usually at most two transitions do not lead back to the 'mismatch' states at the start of a row (here, 0 and 5), but I also couldn't include some back edges where networkx would draw them overlapping.

To get the answer, we need to count the frequencies of being in each possible state (starting from 0) after seeing length = 10 letters-- the good states are n, n+1 ... 2n-1, which correspond to exactly one full match. To generate the transition matrix, we can use the prefix function from the KMP algorithm, with almost no modifications. The definition of the DFA for the first row is as follows:

I got through the other side. Not proud of the solution, mainly because I'm using ToLocalIterator() and the way I'm passing argument to normalizedTransaction UDF. Thanks @EdElliott, your blog helped me a lot.

Anyway, here it goes: