Matlab 快速搜索排序向量中大于x的最小值_Matlab_Optimization

Matlab 快速搜索排序向量中大于x的最小值

matlab optimization

Matlab 快速搜索排序向量中大于x的最小值,matlab,optimization,Matlab,Optimization,Fast的意思是比O（N）更好，O（N）与find（）的功能一样好。我知道有ismembc和ismembc2，但我认为这两个都不是我想要的。我阅读了文档，他们似乎在搜索一个等于x的成员，但我希望第一个值的索引大于x 现在，如果这两个函数中的任何一个都能做到这一点，有人能举个例子吗，因为我想不出来理想行为： first_greater_than([0, 3, 3, 4, 7], 1) 返回2，第一个值的索引大于1，尽管输入数组显然要大得多当然，二进制搜索并不太难实现，但如果MATLAB已经实

Fast的意思是比O（N）更好，O（N）与find（）的功能一样好。我知道有

ismembc

和

ismembc2

，但我认为这两个都不是我想要的。我阅读了文档，他们似乎在搜索一个等于x的成员，但我希望第一个值的索引大于x

现在，如果这两个函数中的任何一个都能做到这一点，有人能举个例子吗，因为我想不出来

理想行为：

first_greater_than([0, 3, 3, 4, 7], 1)

返回2，第一个值的索引大于1，尽管输入数组显然要大得多

当然，二进制搜索并不太难实现，但如果MATLAB已经实现了，我宁愿使用他们的方法。

由于输入已经排序，自定义二进制搜索应该可以工作（您可能需要对边缘情况进行一些更新，即请求的值小于数组的所有元素）：

function[result，res2]=二进制搜索示例（val）
%//生成示例数据并对其进行排序
N=100000000；
a=兰特（N，1）；
a=排序（a）；
%//运行算法
二进制搜索算法的tic%开始计时
div=1；
idx=楼层（N/div）；
而(1)
div=div*2；
%//检查是否小于val检查下一个是否大于val
如果是（idx）val，
结果=a（idx+1）；
打破
else%//变大
idx=idx+最大值（[楼层（N/div），1]）；
结束
结束
如果a（idx）>val，%
idx=idx-楼层（N/div）；
结束
结束%结束while循环
二进制搜索算法的toc%结束计时
%% ------------------------
%%与MATLAB的比较
matlab查找的tic%开始计时
j=查找（a>val，1）；
res2=a（j）；
matlab查找的toc%结束计时
%//基准
>>[res1，res2]=二进制搜索示例（0.556）
运行时间为0.000093秒。
运行时间为0.327183秒。
res1=
0.5560
res2=
0.5560

这是我的实现。这并不是我想要的答案，但现在，我必须假设我所追求的不是在MATLAB中实现的

关于指数的注记所有MATLAB索引都是错误的，因为它们从1开始，而不是0。我尽管如此，索引仍然从0开始。因此，贯穿始终，您将看到如下所示的索引：

数组（1+i）

访问元素i，其中i是在[0，N]中。此外，所有MATLAB范围都是错误的。它们的约定是 [a，b]，而不是[a，b]。因此您将看到如下所示的范围贯穿：0:N-1是数字的范围（通常是 N维数组）从0到N。当数组使用范围索引时，必须同时进行两项更正。1添加到顶部从顶部减去底部边界和1。结果如下：数组（1+a:b）访问[a，b]中的元素，其中a和b在[0，N]中 b>a。我真的应该用python和scipy来代替，但现在已经太迟了。下一个项目

binary_search.m:在我看来，它比@ljk07的实现要整洁得多，但它们当然还是得到了接受。谢谢，@ljk07

function i = binary_search(v, x)
%binary_search finds the first element in v greater than x
% v is a vector and x is a double. Returns the index of the desired element
% as an int64 or -1 if it doesn't exist.

% We'll call the first element of v greater than x v_f.

% Is v_f the zeroth element? This is technically covered by the algorithm,
% but is such a common case that it should be addressed immediately. It
% would otherwise take the same amount of time as the rest of them. This
% will add a check to each of the others, though, so it's a toss-up to an
% extent.
if v(1+0) > x
    i = 0;
    return;
end

% MATLAB foolishly returns the number of elements as a floating point
% constant. Thank you very much, MATLAB.
b = int64(numel(v));

% If v_f doesn't exist, return -1. This is also needed to ensure the
% algorithm later on terminates, which makes sense.
if v(1+b-1) <= x
    i = -1;
    return;
end

a = int64(0);

% There is now guaranteed to be more than one element, since if there
% wasn't, one of the above would have matched. So we split the [a, b) range
% at the top of the loop.

% The number of elements in the interval. Calculated once per loop. It is
% recalculated at the bottom of the loop, so it needs to be calculated just
% once before the loop can begin.
n = b;
while true
    % MATLAB's / operator foolishly rounds to nearest instead of flooring
    % when both inputs are integers. Thank you very much, MATLAB.
    p = a + idivide(n, int64(2));

    % Is v_f in [a, p) or [p, b)?
    if v(1+p-1) > x
        % v_f is in [a, p).
        b = p;
    else
        % v_f is in [p, b).
        a = p;
    end

    n = b - a;
    if n == 1
        i = a;
        return;
    end
end
end

二进制搜索测试.m的输出（在我的计算机上）：

当然有一个加速。在我的电脑上你可以看到加速达到大约一百万个元素。因此，除非用C实现二元搜索或者你有一个大约一百万个元素的向量，否则find速度更快，尽管它使用了一个愚蠢的算法。我原以为阈值会比这个低。我的ss是因为find主要在C中内部实现。不公平：（但无论如何，对于我的特定应用程序，我的向量大小只有1000左右，所以毕竟，find对我来说真的更快。至少直到有一天我用mex文件在C中实现二进制搜索或切换到scipy，以先发生的为准。我有点厌倦了MATLAB的一些不方便的切换。你可以通过rea在我的代码中添加注释。
为什么你把比较放在
查找的循环中而循环中？为什么tic toc 循环中？是的，我确实认为这是一个选项，看起来这是我最后要做的，因为我的时间不多了，但这不是我想要的。我希望会有做一个MATLAB函数吧。现在，我还得留下你的（尽管很好）回答不被接受，因为这不是我真正想要的。我将编辑这个问题作为回答。第一次测量的tic 在哪里？@eit这里的tic toc s有些可疑……我会对这些测量持保留态度。tic toc放置错误，但二进制搜索确实更快而不是使用蛮力O（n）查找（即使是使用while循环实现的）。我自己刚刚测试过。您使用了find ？您知道'first' 属性？find（[0,3,3,4,7]>1,1，'first'））正是你想做的。@thewaywewalk我认为你的答案只是返回索引，需要用索引进一步获取值。方法本身足够优雅。我是这样做的，但你可以用这个变量来处理初始向量：A=[0,3,3,4,7] 和A（find（A>1,1，'first'））=3@thewaywewalk我刚才在OP中对find（）说了些什么？@Ray哎哟。我的意思是它返回索引。索引更有用。现在编辑OP。 function i = binary_search(v, x) %binary_search finds the first element in v greater than x % v is a vector and x is a double. Returns the index of the desired element % as an int64 or -1 if it doesn't exist. % We'll call the first element of v greater than x v_f. % Is v_f the zeroth element? This is technically covered by the algorithm, % but is such a common case that it should be addressed immediately. It % would otherwise take the same amount of time as the rest of them. This % will add a check to each of the others, though, so it's a toss-up to an % extent. if v(1+0) > x i = 0; return; end % MATLAB foolishly returns the number of elements as a floating point % constant. Thank you very much, MATLAB. b = int64(numel(v)); % If v_f doesn't exist, return -1. This is also needed to ensure the % algorithm later on terminates, which makes sense. if v(1+b-1) <= x i = -1; return; end a = int64(0); % There is now guaranteed to be more than one element, since if there % wasn't, one of the above would have matched. So we split the [a, b) range % at the top of the loop. % The number of elements in the interval. Calculated once per loop. It is % recalculated at the bottom of the loop, so it needs to be calculated just % once before the loop can begin. n = b; while true % MATLAB's / operator foolishly rounds to nearest instead of flooring % when both inputs are integers. Thank you very much, MATLAB. p = a + idivide(n, int64(2)); % Is v_f in [a, p) or [p, b)? if v(1+p-1) > x % v_f is in [a, p). b = p; else % v_f is in [p, b). a = p; end n = b - a; if n == 1 i = a; return; end end end % Some simple tests. These had better pass... assert(binary_search([0], 0) == -1); assert(binary_search([0], -1) == 0); assert(binary_search([0 1], 0.5) == 1); assert(binary_search([0 1 1], 0.5) == 1); assert(binary_search([0 1 2], 0.5) == 1); assert(binary_search([0 1 2], 1.5) == 2); % Compare the algorithm to internal find. for n = [1 1:8] n v = sort(rand(10^n, 1)); x = 0.5; %% tic; ifind = find(v > x, 1,'first') - 1; toc; % repeat. The second time is faster usually. Some kind of JIT % optimisation... tic; ifind = find(v > x, 1,'first') - 1; toc; tic; ibs = binary_search(v, x); toc; tic; ibs = binary_search(v, x); toc; assert(ifind == ibs); end n = 1 Elapsed time is 0.000054 seconds. Elapsed time is 0.000021 seconds. Elapsed time is 0.001273 seconds. Elapsed time is 0.001135 seconds. n = 2 Elapsed time is 0.000050 seconds. Elapsed time is 0.000018 seconds. Elapsed time is 0.001571 seconds. Elapsed time is 0.001494 seconds. n = 3 Elapsed time is 0.000034 seconds. Elapsed time is 0.000025 seconds. Elapsed time is 0.002344 seconds. Elapsed time is 0.002193 seconds. n = 4 Elapsed time is 0.000057 seconds. Elapsed time is 0.000044 seconds. Elapsed time is 0.003131 seconds. Elapsed time is 0.003031 seconds. n = 5 Elapsed time is 0.000473 seconds. Elapsed time is 0.000333 seconds. Elapsed time is 0.003620 seconds. Elapsed time is 0.003161 seconds. n = 6 Elapsed time is 0.003984 seconds. Elapsed time is 0.003635 seconds. Elapsed time is 0.004209 seconds. Elapsed time is 0.003825 seconds. n = 7 Elapsed time is 0.034811 seconds. Elapsed time is 0.039106 seconds. Elapsed time is 0.005089 seconds. Elapsed time is 0.004867 seconds. n = 8 Elapsed time is 0.322853 seconds. Elapsed time is 0.323777 seconds. Elapsed time is 0.005969 seconds. Elapsed time is 0.005487 seconds.