需要分析一组包含0和正整数的向量

Question

我正在使用 Matlab，并且有一个 1x200 的数字向量。

我需要按照以下规则为这组数字分配一个“分数”：

1. 如果有 2 个或 3 个或 4 个连续正数，则 0.5 分
2. 如果有 5 个或更多连续正数，则 1.0 分
3. 如果没有t 任何连续的正数，例如：0 0 0 6 0 0，则为 0.0 点。 （忽略它，将正数视为零）
4. 如果一串正整数中间只有一个零，则忽略该零（将其视为正整数）
5. 如果有两个或多个连续的零，这打破了连续正数的运行。

示例： 30 43 54 0 0 0 41 54 14 10 1 0 0 0 0 32 41 98 12 0 0 0 （总共 2.0 分）

最后，应该有一个分数统计。

对于此类问题有什么有用的函数吗？

Answer 1

这是基于我对这个问题的理解，正如我在上面的问题中所指出的。我已经“取消抑制”所有输出，因此您可以看到发生了什么。

%Rules:
%1. If there are 2 or 3 or 4 consecutive positive numbers, then 0.5 point
%2. If there are five or more consecutive positive numbers, then 1.0 point
%3.  And if there isn't any consecutive positive number, for example: 
%   0 0 0 6 0 0, then 0.0 point. (ignore it, consider that positive 
%   number as zero)
%4. if there is only one zero in the middle of positive integers = ignore 
%   that zero (consider it as a positive integer)
%5. If there are two or more consecutive 0, THEN no point.

%testData = [0 30 43 54 0 0 0 41 54 14 10 1 0 0 0 0 32 41 98 12 0 0 0 1 2 0 1 2 0 ];
testData = [30 43 54 0 0 0 41 54 14 10 1 0 0 0 0 32 41 98 12 0 0 0 ];
posa = testData>0;
%add 0s at each end so that the diffs at the ends work.
diffa = diff([0 posa 0])
starts = find(diffa ==1)
ends = find(diffa==-1)

% Rule 4 if any end (-1) is immediately followed by a start, that means that there 
%   is a 0 in the middle of a run.  substitute a 1 in the position and recalc.
midZeroLengths = starts(2:end) - ends(1:(end-1));
%pad to account for the fact that we only compared part.
midZeroLengths = [midZeroLengths 0];
if any(midZeroLengths == 1);
    testData(ends(midZeroLengths==1)) = 1;
    posa = testData>0;
    %add 0s at each end so that the diffs at the ends work.
    diffa = diff([0 posa 0])
    starts = find(diffa ==1)
    ends = find(diffa==-1)  
end

runs = ends-starts
halfs = (runs > 1) & (runs < 5)
wholes = (runs > 4)
final = sum(halfs)*0.5 + sum(wholes)

Answer 2

怎么样：

str = repmat('a', 1, numel(testData));
str(testData > 0) = 'b';
m = regexp(str, 'b+(ab+)*', 'match');
n = cellfun(@numel, m);
score = 0.5 * sum(n >= 2 & n <= 4) + 1.0 * sum(n >= 5);

请注意，我还没有运行过这个，所以可能会有错误。