javascript,arrays,algorithm,intersection,set-intersection

The first challenge is to make the function correct. Once it's correct, we can worry about the speed. There are a few things which could trip-up a function like this: NaN Bad limits Repeated numbers Only 1 input array (or none at all) Your original function can handle repeated numbers,...

this ought to do what you want import random as RND import string import itertools as IT mock some data fnx = lambda: set(RND.sample(string.ascii_uppercase, 7)) S = [fnx() for c in range(5)] generate an index list of the sets in S so the sets can be referenced more concisely below...

python,python-2.7,set,set-intersection

No, it's not. CPython's set intersection implementation works by parallel iteration over the two sets, in hash order. Matching hashes are further tested for equality. If you have a set of small contiguous ints, they'll all hash to themselves, so everything will work out fine. But if the sets are...

java,set,set-intersection,set-theory

What would be the best way to do this in Java? The first part you are describing is a powerset (as I edited your question to include last week). You would then get the intersection for each set of sets in the powerset. Because you are doing a powerset...

python,dictionary,set,iteration,set-intersection

You are making it way too complicated for yourself. You appear to be creating default values for missing keys, so the following is far simpler: for key in dict_A.viewkeys() | dict_A.viewkeys(): some_function(dict_A.get(key, 0), dict_B.get(key, 0)) using the dict.get() function to substitute a default for a missing key. Note that I...

What about using this double sapply? l <- list(a,b,c,d) sapply(seq_len(length(l)), function(x) sapply(seq_len(length(l)), function(y) length(intersect(unlist(l[x]), unlist(l[y]))))) [,1] [,2] [,3] [,4] [1,] 4 0 0 0 [2,] 0 4 1 0 [3,] 0 1 3 0 [4,] 0 0 0 2 Interpretation: e.g. the element [1,2] of the matrix shows you how...

python,list,python-2.7,set,set-intersection

Use a Counter to count how many times each word appears and pick the ones that appear enough: import collections import itertools counts = collections.Counter(itertools.chain.from_iterable(words)) threshold = len(words) - allowed_number_of_times_missing results = {word for word in counts if counts[word] >= threshold} If words can appear repeatedly in a list, you...

First of all, when you are programming in C++, you can just use: struct Face { uint a,b,c; }; Here's a simple strategy for implementing operator< that works for the algorithms and containers in the standard library. struct Face { uint a,b,c; bool operator<(Face const& rhs) const { if (...

algorithm,data-structures,time-complexity,set-intersection

Take one of the sets and convert it to a hash map. Iterate the other set, and for each member try to look up the element in the hash map. If you find it, add the result to a heap; if the size of the heap grows to one greater...

c++,vector,iterator,containers,set-intersection

The important requirement for an output iterator is that it be valid and write-able for the range [out, out+size of output). Passing c.begin() will lead to the values being overwritten which only works if the container c holds enough elements to overwrite. Imagine that c.begin() returns a pointer to an...

arrays,fortran,fortran90,fortran95,set-intersection

I am not quite sure what you mean. If you want to compare two arrays index by index, you can do that by just using ==, like so: INTEGER :: a(4), b(4) LOGICAL :: inter(4) a = (/ 1, 2, 3, 4 /) b = (/ 4, 2, 3, 1...

java,set,set-intersection,set-union

this works :) SetInt Class public class SetInt { int[] setArray = new int[52]; public void add(int a) { if (a < setArray.length - 2) { setArray[a] = a; } } /* * add a to the set in the correct position * * @param a the value to be...