算法细分的数组＆QUOT;半等于＆QUOT;，均匀子阵 [英] Algorithm for subdividing an array into "semi-equal", uniform sub-arrays

问题描述

由于有N个元素的数组，我在寻找并购（M＆LT; N）相继子阵列，等长或长度相差大多是1。例如，如果N = 12，M = 4，所有子阵列将具有相等的长度N / M = 3。如果N = 100和M = 12，我期望子阵列具有长度8和9，并且两个尺寸应原数组内均匀地小号$ P $垫。这个简单的任务，原来是一点点微妙的实现。我想出了布氏线算法的适应，它看起来像这样，当codeD在C ++：

Given an array with N elements, I am looking for M (M < N) successive sub-arrays with equal lengths or with lengths that differ by mostly 1. For example, if N = 12 and M = 4, all sub-arrays would have equal lengths of N/M = 3. If N = 100 and M = 12, I expect sub-arrays with lengths 8 and 9, and both sizes should be uniformly spread within the original array. This simple task turned to be a little bit subtle to implement. I came up with an adaptation of the Bresenham's line algorithm, which looks like this when coded in C++:

/// The function suggests how an array with num_data-items can be
/// subdivided into successively arranged groups (intervals) with
/// equal or "similar" length. The number of intervals is specified
/// by the parameter num_intervals. The result is stored into an array
/// with (num_data + 1) items, each of which indicates the start-index of
/// an interval, the last additional index being a sentinel item which 
/// contains the value num_data.
///
/// Example:
///
///    Input:  num_data ........... 14,
///            num_intervals ...... 4
///
///    Result: result_start_idx ... [ 0, 3, 7, 10, 14 ]
///

void create_uniform_intervals( const size_t         num_data,
                               const size_t         num_intervals,
                               std::vector<size_t>& result_start_idx )
{
    const size_t avg_interval_len  = num_data / num_intervals;
    const size_t last_interval_len = num_data % num_intervals;

    // establish the new size of the result vector
    result_start_idx.resize( num_intervals + 1L );
    // write the pivot value at the end:
    result_start_idx[ num_intervals ] = num_data;

    size_t offset     = 0L; // current offset

    // use Bresenham's line algorithm to distribute
    // last_interval_len over num_intervals:
    intptr_t error = num_intervals / 2;

    for( size_t i = 0L; i < num_intervals; i++ )
    {
        result_start_idx[ i ] = offset;
        offset += avg_interval_len;
        error -= last_interval_len;
        if( error < 0 )
        {
            offset++;
            error += num_intervals;
        } // if
    } // for
}

这code计算的间隔长度为N = 100，M = 12：8 9 8 8 9 8 8 9 8 8 9 8

This code calculates the interval lengths for N = 100, M=12: 8 9 8 8 9 8 8 9 8 8 9 8

实际的问题是，我不知道究竟怎么叫我的问题，所以我难以寻找它。

The actual question is that I don't know how exactly to call my problem, so I had difficulty searching for it.

• 是否还有其他算法来完成这样的任务？
• 他们如何叫什么名字？也许名字会来，如果我知道应用的其他领域。

我所需要的算法作为用于数据的聚类更大算法的一部分。我想这也可能是实现并行排序有用的（？）。

I needed the algorithm as a part of a bigger algorithm for clustering of data. I think it could also be useful for implementing a parallel sort(?).

推荐答案

如果你的语言都有整数除​​法截断，一个简单的方法来计算截面尺寸我是通过（N * I + N）/ M - （N * I）/ M 。例如，Python程序

If your language has integer division that truncates, an easy way to compute the size of section i is via (N*i+N)/M - (N*i)/M. For example, the python program

  N=100;M=12
  for i in range(M): print (N*i+N)/M - (N*i)/M

输出数字8 8 9 8 8 9 8 8 9 8 8 9. N = 12，M = 5 它输出2 2 3 2 3. N = 12，M = 3 它输出4 4 4。

outputs the numbers 8 8 9 8 8 9 8 8 9 8 8 9. With N=12;M=5 it outputs 2 2 3 2 3. With N=12;M=3 it outputs 4 4 4.

如果您的部分号码从1开始的，而不是从0开始的，前pression是代替（N * I）/ M - （N * IN）/ M 。

If your section numbers are 1-based rather than 0-based, the expression is instead (N*i)/M - (N*i-N)/M.

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