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Jyuen204
Obsidian | Level 7

I have a table where I have an AGENT_ID, AGENT_NAME, TEAM_LEAD_NM, DATE

 

I am trying to use Proc SQL to generate a rank because the ID and NAME are duplicated.

 

proc sql;
select AGENT_ID, AGENT_NAME, TEAM_LEAD_NM, rank() over (Partition by AGENT_LAN_ID
                                                                                                             order by sys_load_dtm desc) as RANK
from TABLE
where AGENT_NAME = 'Smith, John';
run;

 

The error i get is:

____
22
76
ERROR 22-322: Syntax error, expecting one of the following: !, !!, &, *, **, +, ',', -, /, <, <=, <>, =, >, >=, ?, AND, BETWEEN,
CONTAINS, EQ, EQT, GE, GET, GT, GTT, LE, LET, LIKE, LT, LTT, NE, NET, OR, ^=, |, ||, ~=.

ERROR 76-322: Syntax error, statement will be ignored.

 

 

what is the proper syntax in PROC SQL to rank my data? 

 

Thank you!

21 REPLIES 21
Jyuen204
Obsidian | Level 7

I am trying to build a table of of this select statement.

PeterClemmensen
Tourmaline | Level 20

Rank over is not available in PROC SQL.

Jyuen204
Obsidian | Level 7

Is there any other method in doing this within SAS?

PaigeMiller
Diamond | Level 26

@Jyuen204 wrote:

Is there any other method in doing this within SAS?


Yes I mentioned a method in my earlier reply and provided a link to the documentation (which includes examples of code)

--
Paige Miller
ballardw
Super User

@Jyuen204 wrote:

Is there any other method in doing this within SAS?


Almost certainly more than one.

 

It might help to show some starting data and what you expect the result to be when finished.

 

 

Jyuen204
Obsidian | Level 7

ID          Name               Manager           DATE

XXX1     Smith, John     Doe, Jane         15NOV2019

XXX1     Smith, John     Doe, Jane         13NOV2019

XXX1     Smith, John     Crane, Frasier   3NOV2019

 

Want to get the following out put in a another table

 

ID          Name               Manager           DATE              RANK

XXX1     Smith, John     Doe, Jane         15NOV2019    1

XXX1     Smith, John     Doe, Jane         13NOV2019    2

XXX1     Smith, John     Crane, Frasier   3NOV2019      3

PaigeMiller
Diamond | Level 26

Hi

 

I'm afraid this doesn't explain anything. How are these rankings computed? 

Or are they simply sequence numbers, not true rankings?

--
Paige Miller
Jyuen204
Obsidian | Level 7
That's the data within my table, I need to rank the data an place into another table (in the ranked format above.
I am ranking by date descending
PaigeMiller
Diamond | Level 26

So you are essentially asking for sequence numbers? (assuming you have sorted the data by descending date).

--
Paige Miller
Jyuen204
Obsidian | Level 7
Well my data isnt as clean as I am going to have multiple names and multiple dates
PaigeMiller
Diamond | Level 26

@Jyuen204 wrote:
Well my data isnt as clean as I am going to have multiple names and multiple dates

This doesn't explain anything, and it doesn't answer my question.

 

Give us a small example of the un-clean data, and show us and explain to us what you want to do with it.

--
Paige Miller
Tom
Super User Tom
Super User

@Jyuen204 wrote:

ID          Name               Manager           DATE

XXX1     Smith, John     Doe, Jane         15NOV2019

XXX1     Smith, John     Doe, Jane         13NOV2019

XXX1     Smith, John     Crane, Frasier   3NOV2019

 

Want to get the following out put in a another table

 

ID          Name               Manager           DATE              RANK

XXX1     Smith, John     Doe, Jane         15NOV2019    1

XXX1     Smith, John     Doe, Jane         13NOV2019    2

XXX1     Smith, John     Crane, Frasier   3NOV2019      3


What is your definition of RANK?  Are you just counting observations?

data want;
  set have;
  by id descending date;
  if first.id then rank=0;
  rank+1;
run;

Counting distinct dates?

data want;
  set have;
  by id descending date;
  if first.id then rank=0;
  rank+first.date;
run;
Jyuen204
Obsidian | Level 7

 

TABLE1

 

ID          Name               Manager           DATE

XXX1     Smith, John     Doe, Jane         15NOV2019

XXX1     Smith, John     Doe, Jane         13NOV2019

XXX1     Smith, John     Crane, Frasier   3NOV2019

 

I am needing to rank a table of data like here. so that the most recent date is ranked the highest.

I need to create a secondary table of this data where RANK = 1

 

so of the data examples above, I want to rank the data in the table, and create a table of the rank 1 rows.

So I am only left with :
XXX1   Smith, John    Doe, Jane

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