C++：仅在具有指定索引的元素中查找数组中的最小值

Question

I couldn't find a solution to the following problem:

I am trying to find the minimum in a C++ array, but without looking at all elements inside, only among elements that have a specified index. Eg, if I am given an index k, and an array int* mylist = int[n] , then I want to look for the minimal element in mylist only in a "sublist" of elements, where eg the index of the respective elements i fulfills i%n = k . In Python, this can be easily solved by

min([mylist[i]  for i in range(n) if i%n==k]),

but afaik, there is no equivalent in C++. Also, lambda functions do not do the trick, as far as I understood.

Any idea how to do this efficiently? Any advice would be appreciated!

Answer 1

This is an example of why standard algorithms work with iterators rather than directly with containers/collections. It allows you to write an iterator that embodies the indices you care about.

The problem is that writing an iterator is more work than most people would like, or consider justified for this simple of a task. Boost has some utilities to make it easier to implement iterators, such as a filter iterator , but as far as I know these are applied based only on the content of the collection, not the indices.

The code involved in creating an index_filter would be fairly trivial (eg, should be easy to extrapolate from the code for filter_iterator linked above), but there's more boiler-plate there than most people would like. Unless you're doing things like this pretty regularly, it may be difficult to justify.

Answer 2

With range-v3 , you may do something similar to:

for (auto e : v | ranges::view::drop(3) | ranges::view::stride(5)) {
    std::cout << e << std::endl;   
}

so you iterate over index 5 * k + 3 .

And you can call ranges::min_element on resulting view:

auto view = v | ranges::view::drop(3) | ranges::view::stride(5);
auto it = ranges::min_element(view);
std::cout << *it << std::endl;

Demo

Answer 3

i % n = k will be a either:

1. k if k < n
2. [] empty that is, if k > n

This is because i varies from 0 to n

Hence there will be only one or no elements.

For clarity: n = 50; print([i%n for i in range(n)]) n = 50; print([i%n for i in range(n)]) , gives:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49]

That is, unique values between 0 to n-1