topic Re: IF ELSE STATEMENT is not working as expected in New SAS User

IF ELSE STATEMENT is not working as expected

Deanna_Payne — Sun, 13 Dec 2020 00:10:29 GMT

Hello All,

I am having troubles with my IF ELSE statement. It is not assigning the values properly for some reason.

I am trying to code a new variables Code' to be:

1 if Case is "I25.YY"

2 if Case is "150.YY"

3 if Case is "163.YY" and

4 for any other value

and I want a new variable 'Death'

to be 1 if Code = 1,2,3

and 0 if Code = 4

So I created a substring of Case named "Placeholder" to only contain the first three characters to make the new variable CODE easier to recognize. But when I input my IF ELSE statements, it doesn't assign the values correctly and I don't know why. Please help:

Dataset I have that isn't correct

obs	ssn	date	Case	placeholder	Code	Death
101	252-81-5641	2016-12-30	I22.13	I22	4	0

102

112-32-1992

2016-06-27

I24.23

I24

103

498-73-1990

2016-10-16

I25.33

I25

104

372-39-2093

2016-09-05

I25.42

I25

105

127-41-0067

2016-10-13

I25.19

I25

106

478-37-8847

2015-10-30

I25.91

I25

107

528-08-1992

2015-05-26

I25.21

I25

108

574-46-0715

2015-09-08

I25.31

I25

109

068-44-0495

2015-03-19

I25.92

I25

110

494-72-4043

2016-10-13

I39.41

I39

111

018-93-3430

2016-07-03

I41.21

I41

112

538-92-4707

2016-11-20

I49.13

I49

113

798-08-2998

2016-10-22

I50.32

I50

114

424-92-9797

2016-07-04

I50.21

I50

115

418-15-3055

2015-08-30

I50.91

I50

116

434-75-5248

2016-10-07

I50.11

I50

117

423-12-7794

2016-05-29

I50.42

I50

118

326-80-9737

2015-04-17

I50.91

I50

119

524-43-5860

2016-12-12

I50.13

I50

120

986-08-2045

2015-05-25

I50.13

I50

121

013-78-8406

2015-08-16

I62.92

I62

122

484-63-7323

2015-05-23

I63.42

I63

Dataset I want

obs	ssn	date	Case	Death
101	252-81-5641	2016-12-30	I22.13	4

102

112-32-1992

2016-06-27

I24.23

103

498-73-1990

2016-10-16

I25.33

104

372-39-2093

2016-09-05

I25.42

105

127-41-0067

2016-10-13

I25.19

106

478-37-8847

2015-10-30

I25.91

107

528-08-1992

2015-05-26

I25.21

108

574-46-0715

2015-09-08

I25.31

109

068-44-0495

2015-03-19

I25.92

110

494-72-4043

2016-10-13

I39.41

111

018-93-3430

2016-07-03

I41.21

112

538-92-4707

2016-11-20

I49.13

113

798-08-2998

2016-10-22

I50.32

114

424-92-9797

2016-07-04

I50.21

115

418-15-3055

2015-08-30

I50.91

116

434-75-5248

2016-10-07

I50.11

117

423-12-7794

2016-05-29

I50.42

118

326-80-9737

2015-04-17

I50.91

119

524-43-5860

2016-12-12

I50.13

120

986-08-2045

2015-05-25

I50.13

121

013-78-8406

2015-08-16

I62.92

122

484-63-7323

2015-05-23

I63.42

Code I have that is not currently working

DATA Want.NDI;
SET Have.NDI;

PLACEHOLDER = SUBSTR(ICD10, 1, 3.);

IF PLACEHOLDER = 'I25' THEN Code = 1;
IF PLACEHOLDER = 'I50' THEN Code = 2;
IF PLACEHOLDER = 'I63' THEN Code = 3;
ELSE Code = '4';

IF Code = 4 THEN death = 0;
ELSE death = 1;

DROP PLACEHOLDER;

RUN;

other information that might be useful:

SAS studio~~ Release: 3.8 (Basic Edition)

no errors or warnings in log

**Original Post has been altered to avoid being googlesearched for future classes*;

Re: IF ELSE STATEMENT is not working as expected

Kurt_Bremser — Mon, 09 Nov 2020 04:48:52 GMT

IF PLACEHOLDER = 'I63' THEN CODCD = 3;
ELSE CODCD = '4';

This statement means that ALL values other than I63 will result in codcd =4; you need to make the branches mutually exclusive, and the best way to do that is the data step SELECT:

select (placeholder);
  when ('I25') codcd = 1;
  when ('I50') codcd = 2;
  when ('I63') codcd = 3;
  otherwise codcd = 4;
end;

(with your original code, you would need to insert ELSE branches:

IF PLACEHOLDER = 'I25' THEN CODCD = 1;
else IF PLACEHOLDER = 'I50' THEN CODCD = 2;
else IF PLACEHOLDER = 'I63' THEN CODCD = 3;
ELSE CODCD = '4';

Re: IF ELSE STATEMENT is not working as expected

DeJanae07 — Thu, 10 Dec 2020 15:12:48 GMT

Hi Deanna,

I happen to be in Lala's Programming course as well and for this portion of the project. I did the same exact thing for my IF THEN statements that you did

However for next portion of your code, I used different numbers:

IF CODCD = (1, 2, 3) THEN RELDEATHIND = 1;
ELSE RELDEATHIND = 0;

Lastly, I did not DROP the PLACEHOLDER in my code either.

Try these changes and see if you have success with creating the dataset then.

De'Janae'

Re: IF ELSE STATEMENT is not working as expected

ballardw — Mon, 09 Nov 2020 16:36:22 GMT

@DeJanae07 wrote:

Hi Deanna,

I happen to be in Lala's Programming course as well and for this portion of the project. I did the same exact thing for my IF THEN statements that you did below:

PLACEHOLDER = SUBSTR(ICD10, 1, 3.);

IF PLACEHOLDER = 'I25' THEN CODCD = 1;
IF PLACEHOLDER = 'I50' THEN CODCD = 2;
IF PLACEHOLDER = 'I63' THEN CODCD = 3;
ELSE CODCD = '4';

However for next portion of your code, I used different numbers:

IF CODCD = (1, 2, 3) THEN HYPRELDEATHIND = 1;
ELSE HYPRELDEATHIND = 0;

Lastly, I did not DROP the PLACEHOLDER in my code either.

Try these changes and see if you have success with creating the dataset then.

De'Janae'

You really want to check the value of your CODCD variable for ICD "I25" and "I50".

If you didn't drop your place holder variable then you may want to try this code (a moderately common way to check recoding)

Proc freq data=have;
   tables placeholder *codcd/list missing;
run;

This sort of code will generate one line for each combination of Placeholder and Codcd, including missing values if any. This is a relatively easy check on single variable recoding as you are doing.

Re: IF ELSE STATEMENT is not working as expected

sjbothwell — Mon, 09 Nov 2020 17:28:17 GMT

Hi Deanna,

It looks like you specify CODCD with quotations in your first IF statement: CODCD = '4';

But then in your second statement you have if without quotations: CODCD = 4

If you put quotations around the 4 in the second statement does that solve the problem?

Samantha

Re: IF ELSE STATEMENT is not working as expected

bshiferaw27 — Thu, 10 Dec 2020 15:15:54 GMT

Hello,

Try using this code below, the issue might be that you would have to format them again as shown below.

I hope it helps!

ICD10a = SUBSTR(ICD10,1,3);
IF ICD10a = 'I25' THEN CODCd = 1;
ELSE IF ICD10a = 'I50' THEN CODCd = 2;
ELSE IF ICD10a = 'I63' THEN CODCd = 3;
ELSE CODCd = 4;

IF CODCd IN(1, 2, 3) THEN HypRelDeathInd = 1;
ELSE HypRelDeathInd = 0;

DROP ICD10a;
RUN;

To get the formats as to how you mentioned in the second table, try using this:

PROC FORMAT;
VALUE CODCd
1 = "I25.XX"
2 = "I50.XX"
3 = "I63.XX"
4 = "Other COD";

RUN;