标题: Question: How to assign time to different time period [打印本页] 作者: shiyiming 时间: 2009-7-31 22:11 标题: Question: How to assign time to different time period Dear friends,
I have a data set about surgery. I have patient in time and patient out time. I need to calculate how many minutes are used in each time period
(from 7am to 3pm, 3pm to 5pm, 5pm to 7pm and 7pm to 11pm). The data like this:
Could anyone help this out? Many thanks.作者: shiyiming 时间: 2009-8-1 12:53 标题: Re: Question: How to assign time to different time period [code:jwtajwyx]data original;
input date:ddmmyy10. indate:ddmmyy10. intime:time5. outdate:ddmmyy10. outtime:time5.;
id=_n_;
intime=dhms(indate,hour(intime),minute(intime),second(intime));
outtime=dhms(outdate,hour(outtime),minute(outtime),second(outtime));
drop indate outdate;
format date yymmdd10. intime outtime datetime13.;
datalines;
01/04/2009 01/04/2009 8:02 01/04/2009 12:52
01/04/2009 01/04/2009 20:51 02/04/2009 0:24
01/04/2009 01/04/2009 17:45 01/04/2009 20:39
01/04/2009 01/04/2009 15:27 01/04/2009 16:20
01/04/2009 01/04/2009 9:53 01/04/2009 11:48
02/04/2009 02/04/2009 11:35 03/04/2009 11:40
29/04/2009 29/04/2009 14:55 30/04/2009 15:01
;
/* 创建内部开始、结束时间 */
data temp;
set original;
inner_intime=intime-dhms(date,0,0,0);
inner_outtime=inner_intime+(outtime-intime);
format inner_intime inner_outtime datetime13.;
run;
/* 拆分跨天记录 */
data temp(drop=temp i);
set temp;
day_int=ceil(inner_outtime/86399)-1;
if day_int>0 then do;
temp=inner_outtime;
inner_outtime=86399;
output;
if day_int>1 then do;
do i=1 to day_int-1;
inner_intime=dhms(i,0,0,0);
inner_outtime=dhms(i,23,59,59);
output;
end;
end;
inner_intime=dhms(day_int,0,0,0);
inner_outtime=temp;
output;
end;
else output;
run;
/* 计算分钟数 */
data temp(drop=i start end);
set temp;
start=hour(inner_intime);
end=hour(inner_outtime);
array arr{0:23} hour0-hour23;
do i=start to end;
arr(i)=60;
if i=start then arr(i)=60-minute(inner_intime);
if i=end then do;
if second(inner_outtime)=59 then arr(i)=minute(inner_outtime)+1;
else arr(i)=minute(inner_outtime);
end;
if start=end then do;
arr(i)=minute(inner_outtime-inner_intime);
if arr(i)=0 then arr(i)=1;
end;
end;
temp_ut73=sum(of hour7-hour14);
temp_ut35=sum(of hour15-hour16);
temp_ut57=sum(of hour17-hour18);
temp_ut711=sum(of hour19-hour22);
run;
/* 合计分钟数 */
data final;
do _n_=1 by 1 until(last.id);
set temp;
by id;
ut73=sum(temp_ut73,ut73);
ut35=sum(temp_ut35,ut35);
ut57=sum(temp_ut57,ut57);
ut711=sum(temp_ut711,ut711);
end;
keep date intime outtime ut:;
run;[/code:jwtajwyx]作者: shiyiming 时间: 2009-8-2 14:26 标题: Re: Question: How to assign time to different time period 死猪头作者: shiyiming 时间: 2009-8-2 17:31 标题: Re: Question: How to assign time to different time period <!-- s:( --><img src="{SMILIES_PATH}/icon_sad.gif" alt=":(" title="Sad" /><!-- s:( --> 我写了80行,死猪头就写了一行...
丢死人了...
学习了一下午刚明白了个大概,要是领会的不对一定告诉我昂.
[code:3rh0sysv]假设格式如下：
#表示根据公式计算的min()，max()的值
LE RE
I # I
INxxI 123456789012345678901234567890 Ixxxxxxx
xxxxxxI 123456789012345678901234567890 IxxOUT
I # I
intime<=outtime时：
DATE LE RE
I # I =(3-1)*30+(25-5)=80
1-1 I IN6789012345678901234567890 I xxxxxxx -->30(多加了5)
1-2 xxxxxx I 123456789012345678901234567890 I xxxxxxx -->30
1-3 xxxxxx I 1234567890123456789012345OUT I -->25-5(已减去第1行多加的5)
I # I
DATE LE RE
I # I =(3-1)*30+(30-0)=90
1-1 INxx I 123456789012345678901234567890 I xxxxxxx -->30
1-2 xxxxxx I 123456789012345678901234567890 I xxxxxxx -->30
1-3 xxxxxx I 123456789012345678901234567890 I xxOUT -->30-0
I # I
intime>outtime时：
DATE LE RE
I # I =(3-1)*30-(25-5)=40
1-1 I 12345 IN67890 I xxxxxxx -->30(多加了20,已把第3行的5加这里)
1-2 xxxxxx I 123456789012345678901234567890 I xxxxxxx -->30
1-3 xxxxxx I ^^^^^OUT I -->-(25-5)(减去第1行多加的20)
I # I
DATE LE RE
I I =(3-1)*30-(30-0)=30
1-1 I # I INxxx -->30(多加了30)
1-2 xxxxxx I 123456789012345678901234567890 I xxxxxxx -->30
1-3 xxOUT I # I -->-(30-0)(减去多加的30)
I I
[/code:3rh0sysv]