output the first month that working time over 250 hours

[MMTG]

I have data about the working time in12 months of 1000 workers,

id m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12;

1  30  50............................................................30

2  .........................................................................

.

.

1000....................................................................

I want to get the first month that total working time over 250, if after 6 month , the first worker have 610 hours ,then the first monthis 6

I need to create a  new variable(First_month) for this  value.I have tried retain and do until ,but always have problem.

do you give me some help, thanks very much!

[Reeza · Accepted solution]

You don't need retain, you need an array and a do loop. Your requirements are unclear as you mention 250 hours and 610. Please clarify.

In the mean time try something like the following (untested)

data want;
set have;
array mths(12) m1-m12;
total_time=0;
do i=1 to 12 while(sum<250);
total_time=sum(total_time, mths(i));
if total_time>250 then first_month=i;
end;
run;

[MMTG]

sorry for the mistake, it is 250, not 610,

and i have do the same as you do ,

but it doesn't work ,because, if the 6th month have overed 250, then the 7th will also over 250, then all the first_month will be 12.

[MMTG]

I have fixed it  by combining "do until" with your procedures , you help me so much, thank you sincerely .

[ballardw]

It would help understand details of your problem if you included several complete rows of data and what the final outcome should look like for that example data.

[MMTG]

ok,

just like 

id m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12

1  20  30  40  34  60  70  80  30 60  70     90    40

2 10   24   40  20  40  30  60 20 50  30    30     30

and more obs

mu require result

id m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m11 m12  First_month_over250

1  20  30  40  34  60  70  80  30 60  70     90    40      6

2 10   24   40  20  40  30  60 20 50  30    30     30      8

and more  obs