Question

假设我有一个数据集：

Jday = datenum('2009-01-01 00:00','yyyy-mm-dd HH:MM'):1/24:...
    datenum('2009-01-05 23:00','yyyy-mm-dd HH:MM');
DateV = datevec(Jday);
DateV(4,:) = [];
DateV(15,:) = [];
DateV(95,:) = [];

Dat = rand(length(Jday),1)

如何删除少于24次测量的所有日子。例如，在第一天只有23次测量，因此我需要移除整整一天，我怎么能为所有阵列重复这个？

Answer 1

相当长的答案，但我认为它应该是有用的。我会使用containers.Map来做到这一点。可能有更快的方式，但也许现在这个会很好。

Jday = datenum('2009-01-01 00:00','yyyy-mm-dd HH:MM'):1/24:...
    datenum('2009-01-05 23:00','yyyy-mm-dd HH:MM');

DateV = datevec(Jday);
DateV(4,:) = [];
DateV(15,:) = [];
DateV(95,:) = [];


% create a map
dateMap = containers.Map();



% count measurements in each date (i.e. first three columns of DateV)
for rowi = 1:1:size(DateV,1)

    dateRow = DateV(rowi, :);
    dateStr = num2str(dateRow(1:3));

    if ~isKey(dateMap, dateStr)
        % initialize Map for a given date with 1 measurement (i.e. our
        % counter of measuremnts
        dateMap(dateStr)  = 1;
        continue;
    end
    % increment measurement counter for given date
    dateMap(dateStr)  = dateMap(dateStr) + 1;
end


% get the dates
dateStrSet = keys(dateMap);




for keyi = 1:numel(dateStrSet)

    dateStrCell = dateStrSet(keyi);  
    dateStr = dateStrCell{1};

    % get number of measurements in a given date
    numOfmeasurements = dateMap(dateStr);

    % if less then 24 do something about it, e.g. save the date
    % for later removal from DateV
    if numOfmeasurements < 24
        fprintf(1, 'This date has less than 24 measurement: %s\n', dateStr);
    end
end

结果是：

This date has less than 24 measurement: 2009     1     1
This date has less than 24 measurement: 2009     1     5

Answer 2

快速解决方案是按年份，月份，日期对unique()进行分组，然后使用accumarray()计算每天的观察次数，并将逻辑索引的两个步骤排除在少于24个障碍物的情况下：

% Count observations per day [unDate,~,subs] = unique(DateV(:,1:3),'rows'); counts = [unDate accumarray(subs,1)] counts = 2009 1 1 22 2009 1 2 24 2009 1 3 24 2009 1 4 24 2009 1 5 23

然后，将标准应用于计数并检索逻辑索引

% index only those that meet criteria idxC = counts(:,end) == 24 idxC = 0 1 1 1 0 % keep those which meet criteria (optional, for visual inspection) counts(idxC,:) ans = 2009 1 2 24 2009 1 3 24 2009 1 4 24

最后，通过Dat找到属于第二轮逻辑indexinf的counts成员{/ 1}}：

ismember()

matlab - 只保留24个值存在的日子

2 个答案: