Question

说我有以下数据库：

Min Rank Qty
2   1    100
2   2    90
2   3    80 
2   4    70
5   1    110
5   2    100
5   3    90
5   4    80
5   5    70
7   1    120
7   2    110
7   3    100
7   4    90

我需要让数据库具有连续的分钟值，如下所示：

Min Rank Qty
2   1    100
2   2    90
2   3    80 
2   4    70
3   1    100
3   2    90
3   3    80 
3   4    70
4   1    100
4   2    90
4   3    80 
4   4    70
5   1    110
5   2    100
5   3    90
5   4    80
5   5    70
6   1    110
6   2    100
6   3    90
6   4    80
6   5    70
7   1    120
7   2    110
7   3    100
7   4    90

我如何在SAS中执行此操作？我只需复制前一分钟。每分钟的观察次数各不相同......可以是4或5或更多。

Answer 1

想象代码会难以实现，问题是它很快就会开始变得混乱。

如果您的数据集不是太大，您可以考虑采用以下方法：

/* We find all gaps. the output dataset is a mapping: the data of which minute (reference_minute) do we need to create each minute of data*/
data  MINUTE_MAPPING (keep=current_minute reference_minute);
    set YOUR_DATA;
    by min;
    retain last_minute 2; *set to the first minute you have;

    if _N_ NE 1 and first.min then do;
        /* Find gaps, map them to the last minute of data we have*/
        if last_minute+1 < min then do;
            do current_minute=last_minute+1 to min-1;
                reference_minute=last_minute;
                output;
            end;
        end;

        /* For the available data, we map the minute to itself*/
        reference_minute=min;
        current_minute=min;
        output;

        *update;
        last_minute=min;
    end;
run;

/* Now we apply our mapping to the data */
*you must use proc sql because it is a many-to-many join, data step merge would give a different outcome;
proc sql;
    create table RESULT as
    select YD.current_minute as min, YD.rank, YD.qty
    MINUTE_MAPPING as MM
    join YOUR_DATA as YD
    on (MM.reference_minute=YD.min)
    ;
quit;

更高效的方法将涉及数组欺骗。但是我发现这种方法更具吸引力（免责声明：最初的想法），之后对其他人更快掌握（免责声明：imho）。

为了更好的衡量，阵列方法：

data  RESULT (keep=min rank qty);
    set YOUR_DATA;
    by min;
    retain last_minute; *assume that first record really is first minute;
    array last_data{5} _TEMPORARY_;

    if _N_ NE 1 and first.min and last_minute+1 < min then do; *gap found;
        do current_min=last_minute+1 to min-1;
            *store data of current record;
            curr_min=min;
            curr_rank=rank;
            curr_qty=qty;

            *produce records from array with last available data;
            do iter=1 to 5;
                min = current_minute;
                rank = iter;
                qty = last_data{iter};
                if qty NE . then output; *to prevent output of 5th element where there are only 4;
            end;

            *put back values of actual current record before proceeding;
            min=curr_min;
            rank=curr_rank;
            qty=curr_qty;
        end;

        *update;
        last_minute=min;
    end;

    *insert data for use on later missing minutes;
    last_data{rank}=qty;
    if last.min and rank<5 then last_data{5}=.;

    output; *output actual current data point;
run;

希望它有所帮助。注意，目前我无法访问SAS客户端。所以未经测试的代码可能包含几个拼写错误。

Answer 2

除非你有一些荒谬的观察，否则我认为调换会使这很容易。

此刻我无法访问sas所以请耐心等待（如果你不能让它工作，我明天可以测试一下）。

proc transpose data=data out=data_wide prefix=obs_;
  by minute;
  id rank;
  var qty;
run;

*sort backwards so you can use lag() to fill in the next minute;
proc sort data=data_wide;
  by descending minute;
run;

data data_wide; set data_wide;
  nextminute = lag(minute);
run;

proc sort data=data_wide;
  by minute;
run;

*output until you get to the next minute;
data data_wide; set data_wide;
  *ensure that the last observation is output;
  if nextminute = . then output;
  do until (minute ge nextminute);
    output;
    minute+1;
  end;
run;

*then you probably want to reverse the transpose;
proc transpose data=data_wide(drop=nextminute)
               out=data_narrow(rename=(col1=qty));
  by minute;
  var _numeric_;
run;

*clean up the observation number;
data data_narrow(drop=_NAME_); set data_narrow;
  rank = substr(_NAME_,5)*1;
run;

同样，我现在无法测试，但它应该有用。

其他人可能有一个聪明的解决方案，因此您不必反向排序/滞后/转发排序。我觉得我之前已经解决了这个问题，但现在对我来说明显的解决办法是让它按照你以前做过的任何事情排序（你可以用降序进行转置没问题）来节省额外的排序。 / p>

SAS：按数据块填充缺失值

2 个答案: