每分钟SAS_Stan.Dev

我想问一个简短的问题。我想我可以用简单的例子更好地解释。 因此，我有以下数据：

 Time          Value
 13:45          0.2
 13:45          0.4
 13:45          0.3
 13:46          0.1
 13:46          0.2
 13:46          0.3
 13:46          0.5
 13:46          0.4

  Time          Value        St.D
  13:45          0.2           0.1  (it is the standard deviation of 0.2,0.4 and 0.3 - so st.dev for 13:45)
  13:45          0.4           0.1
  13:45          0.3           0.1
  13:46          0.1           0.1528  (it is the standard deviation of 0.1,0.2,0.3,0.5 and 0.6 - so st.dev for 13:46)
  13:46          0.2           0.1528 
  13:46          0.3           0.1528 
  13:46          0.5           0.1528 
  13:46          0.6           0.1528 

我想再添加一列。此列中的值应为每分钟的标准偏差。所以，我想得到以下数据：

 Time          Value
 13:45          0.2
 13:45          0.4
 13:45          0.3
 13:46          0.1
 13:46          0.2
 13:46          0.3
 13:46          0.5
 13:46          0.4

  Time          Value        St.D
  13:45          0.2           0.1  (it is the standard deviation of 0.2,0.4 and 0.3 - so st.dev for 13:45)
  13:45          0.4           0.1
  13:45          0.3           0.1
  13:46          0.1           0.1528  (it is the standard deviation of 0.1,0.2,0.3,0.5 and 0.6 - so st.dev for 13:46)
  13:46          0.2           0.1528 
  13:46          0.3           0.1528 
  13:46          0.5           0.1528 
  13:46          0.6           0.1528 

非常感谢您的帮助。

准备数据：

data a;
  time ="13:45";
  value=0.2;
  output;
  time ="13:45";
  value=0.4;
  output;
  time ="13:45";
  value=0.3;
  output;
  time ="13:46";
  value=0.1;
  output;
  time ="13:46";
  value=0.2;
  output;
  time ="13:46";
  value=0.3;
  output;
  time ="13:46";
  value=0.5;
  output;
  time ="13:46";
  value=0.6;
  output;
run;

计算STDEV：

proc summary data=a stddev nonobs noprint nway;
  by time;
  var value;
  output out=b(drop=_type_ _freq_) stddev()=;
run;
proc sql noprint;
  CREATE TABLE res AS
  SELECT a.*
      ,b.value as stddev
  FROM a
  LEFT JOIN b
  ON a.time=b.time
  ;
quit;

然而，13:46的标准差分与您预期的不同。此外，13:46的示例数据中有一点输入错误（[0.1,0.2,0.3,0.4,0.5]，[0.1,0.2,0.3,0.5,0.6]）。

准备数据：

data a;
  time ="13:45";
  value=0.2;
  output;
  time ="13:45";
  value=0.4;
  output;
  time ="13:45";
  value=0.3;
  output;
  time ="13:46";
  value=0.1;
  output;
  time ="13:46";
  value=0.2;
  output;
  time ="13:46";
  value=0.3;
  output;
  time ="13:46";
  value=0.5;
  output;
  time ="13:46";
  value=0.6;
  output;
run;

计算STDEV：

proc summary data=a stddev nonobs noprint nway;
  by time;
  var value;
  output out=b(drop=_type_ _freq_) stddev()=;
run;
proc sql noprint;
  CREATE TABLE res AS
  SELECT a.*
      ,b.value as stddev
  FROM a
  LEFT JOIN b
  ON a.time=b.time
  ;
quit;

然而，13:46的标准差分与您预期的不同。此外，13:46的示例数据中有一点输入错误（[0.1,0.2,0.3,0.4,0.5]，[0.1,0.2,0.3,0.5,0.6]）