difference of # of days from 1st row to 2nd row and 2nd row to 3rd row so on by ID

[mahossain]

i.e first # of days (difference)  will be missing and the 2nd will be difference of Jan 22, 2020-Jan 01, 2020 and so on by ID
ID       Date                   # of days (difference)
1       Jan 01, 2020           
1       Jan 22, 2020
1       Feb 02, 2020
2      Jan 01, 2020
2      Jan 24, 2020
2      Feb 12, 2020

[Reeza · Accepted solution]

data want;
set have;

by id;

num_days = dif(date);

if first.id then call missing(num_days);

run;

Use the DIF() function. 

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@mahossain wrote:

i.e first # of days (difference)  will be missing and the 2nd will be difference of Jan 22, 2020-Jan 01, 2020 and so on by ID
ID       Date                   # of days (difference)
1       Jan 01, 2020           
1       Jan 22, 2020
1       Feb 02, 2020
2      Jan 01, 2020
2      Jan 24, 2020
2      Feb 12, 2020

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[mahossain]

Thanks Reeza!