SAS中的状态持续时间

我有一个关于SAS和变量特定状态持续时间分析的问题。我想知道我的数据集中的每个人在状态a中持续停留多长时间，直到状态b出现。如果状态c发生在状态a之后，则持续时间应设置为零。请注意，如果pre_period处于状态a，我也会将持续时间设置为零，但如果我在之后获得另一个状态a，则应该将其计算在内

数据看起来有点像这样：

    pre_period    week1 week2 week3 week4 week5 week6 week7 ...
id1 b             b     a     a     a     b     c     c     ...
id2 a             a     a     a     b     a     b     b     ...
id3 b             b     a     a     b     a     a     b     ...
id4 c             c     c     a     a     a     a     a     ...
id5 a             b     a     b     b     a     a     b     ...
id6 b             a     a     a     a     a     a     a     ...

    dur1 dur2 dur3 dur4 ...
id1 3    .    .    .    ...
id2 1    .    .    .    ...
id3 3    1    .    .    ...
id4 5    .    .    .    ...
id5 1    2    .    .    ...
id6 7    .    .    .    ...

sas代码中的示例集：

data work.sample_data;
input id $ pre_period $  (week1-week7) ($);
datalines;
id1 b b a a a b c c
id2 a a a a b a b b
id3 b b a a b a a b
id4 c c c a a a a a
id5 a b a b b a a b
id6 b a a a a a a a
;

所以对于id1，它应该给我一个3的持续时间，对于ID21，对于ID33和1，对于ID45，对于ID51和2，对于ID67

因此，输出应该有点像这样：

    pre_period    week1 week2 week3 week4 week5 week6 week7 ...
id1 b             b     a     a     a     b     c     c     ...
id2 a             a     a     a     b     a     b     b     ...
id3 b             b     a     a     b     a     a     b     ...
id4 c             c     c     a     a     a     a     a     ...
id5 a             b     a     b     b     a     a     b     ...
id6 b             a     a     a     a     a     a     a     ...

    dur1 dur2 dur3 dur4 ...
id1 3    .    .    .    ...
id2 1    .    .    .    ...
id3 3    1    .    .    ...
id4 5    .    .    .    ...
id5 1    2    .    .    ...
id6 7    .    .    .    ...

我是SAS的初学者，没有找到解决这个问题的方法。请注意，数据集包含数千行和大约1000列，因此对于一个人，我可能有几个状态a的间隔，我都希望捕获这些间隔（因此在输出中有几个持续时间变量）

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我很感激你的建议。谢谢

在这些情况下，用有限状态机来思考是明智的。通过这种方式，如果您的需求发生变化，那么在以后扩展状态机是非常容易的

持续时间在三种情况下有效（包括结果集中给出的不合法情况）：

• 如果出现以下情况，则应计算状态

  a

  的连续持续时间
  • 它以状态

    b

    结束
  • 当数据集结束时，它仍处于状态

    a

  • 只要它不是在第一周开始的，当前期状态是

    a

首先，我们必须考虑前期需求，我们可以将此状态称为

前期锁定状态

：

    do week = 1 to last_week;
        if current_state = pre_period_locked_state then do;
            if 'a' not = pre_period or 'a' not = week_state then do;
            current_state = duration_state;
        end;

        if current_state = no_duration_state then do;
            if 'a' = week_state then do;
                 current_state = duration_state;
            end;
        end;

        if current_state = dispatch_state then do;
            if 'b' = week_state or 'a' = week_state and week = last_week then do;
                duration{duration_index} = duration_count;
                duration_index = duration_index + 1;
            end;
            duration_count = 0; 
            current_state = no_duration_state;
        end;

下一个问题是当状态不是

a

时，这里称为

no\u duration\u state

：

    do week = 1 to last_week;
        if current_state = pre_period_locked_state then do;
            if 'a' not = pre_period or 'a' not = week_state then do;
            current_state = duration_state;
        end;

        if current_state = no_duration_state then do;
            if 'a' = week_state then do;
                 current_state = duration_state;
            end;
        end;

        if current_state = dispatch_state then do;
            if 'b' = week_state or 'a' = week_state and week = last_week then do;
                duration{duration_index} = duration_count;
                duration_index = duration_index + 1;
            end;
            duration_count = 0; 
            current_state = no_duration_state;
        end;

这是我们的空闲状态，只有在新的持续时间开始时才会改变。下一个状态名为

duration\u state

，定义为：

        if current_state = duration_state then do;
            if 'a' = week_state then do;
                duration_count = duration_count + 1;
            end;
            if ('a' not = week_state or week = last_week) and 0 < duration_count then do;
                current_state = dispatch_state;
             end;
        end;

这将负责对输出表进行索引，并确保只存储有效的持续时间

我在下面添加了

id7

，因为样本数据没有任何以b以外的状态结束的持续时间

data work.sample_data;
input id $ pre_period $  (week1-week7) ($);
datalines;
id1 b b a a a b c c
id2 a a a a b a b b
id3 b b a a b a a b
id4 c c c a a a a a
id5 a b a b b a a b
id6 b a a a a a a a
id7 b a a c a a a a
;

完整sas代码状态机：

 data work.duration_fsm;
    set work.sample_data;
    array weeks{*} week1-week7;
    array duration{*} dur1-dur7;

    *states;
    initial_reset_state = 'initial_reset_state';
    pre_period_locked_state = 'pre_period_locked_state';
    duration_state = 'duration_state';
    no_duration_state = 'no_duration_state';
    dispatch_state = 'dispatch_state';
    length current_state $ 50;

    *initial values;
    current_state = initial_reset_state;
    last_week = dim(weeks);

    keep id dur1-dur7;

    do week = 1 to last_week;
        if current_state = initial_reset_state then do;
            duration_count = 0;
            duration_index = 1;
        current_state = pre_period_locked_state;
        end;
        week_state = weeks{week};
        if current_state = pre_period_locked_state then do;
            if 'a' not = pre_period and 'a' = week_state then do;
                    current_state = duration_state;
                end;
            else if 'a' = pre_period and 'a' not = week_state then do;
                current_state = no_duration_state;
            end;
        end;
        if current_state = no_duration_state then do;
            if 'a' = week_state then do;
                 current_state = duration_state;
            end;
        end;
        if current_state = duration_state then do;
            if 'a' = week_state then do;
                duration_count = duration_count + 1;
            end;
            if ('a' not = week_state or week = last_week) and 0 < duration_count then do;
                current_state = dispatch_state;
             end;
        end;
        if current_state = dispatch_state then do;
            if 'b' = week_state or  'a' = week_state and week = last_week then do;
                duration{duration_index} = duration_count;
                duration_index = duration_index + 1;
            end;
            duration_count = 0; 
            current_state = no_duration_state;
        end;
    end;
    run;

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您期望的结果似乎与您的要求不一致。第6周id 1出现状态c-这不意味着id 1的持续时间为0吗？对于id 2，pre_period有状态a，那么id 2不也应该有持续时间0吗？也许我有点不清楚：如果pre_period是a，那么下面的a不应该被计算在内，但是如果个人离开状态，它应该像其他人一样被处理。如果c紧跟在a之后，则c的情况应该是相关的。谢谢你对我描述的提示！目前还不清楚，如果存在由b或c状态分隔的状态a的多次运行，那么您需要哪种类型的输出。您希望每个id有一行多列来保存每次运行的长度，还是多行？试着输入你希望输出数据集与输入数据集的格式相同的内容。我编辑了这篇文章，希望它现在更清晰。非常感谢。您应该花一些时间来创建将构建一些示例数据（数据步骤/数据线）的代码。喜欢这个问题吗