if else statement

Hello,

I have a data frame like this:
head(b)
FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a
PHENO =2 if b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana
"I tried this but I am not sure if this is correct:"

Does it provide the expected result for all possible combinations of 1/2/NA
for both variables?

On Mon, May 4, 2020 at 1:16 PM Ana Marija <sokovic.anamarija at gmail.com>
wrote:
Hello,

I have a data frame like this:

head(b)
       FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a
PHENO =2 if b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana

        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Patrick S. Malone, Ph.D., Malone Quantitative
NEW Service Models: http://malonequantitative.com

He/Him/His

	[[alternative HTML version deleted]]
To expand on Patrick's response...

You can use the expand.grid function to generate a test table containing all combinations. However, we would not be in a position to verify that the results you get when you apply your logic to the test table are what you want... you know the requirements much better than we do. Nor is that kind of service what this mailing list is for, so please focus on showing what you cannot figure out how to accomplish rather than asking us to do or check your work for you.
"I tried this but I am not sure if this is correct:"

Does it provide the expected result for all possible combinations of
1/2/NA
for both variables?

On Mon, May 4, 2020 at 1:16 PM Ana Marija <sokovic.anamarija at gmail.com>
wrote:

Hello,

I have a data frame like this:

head(b)
       FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a
PHENO =2 if b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana

        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Sent from my phone. Please excuse my brevity.
Thank you for the tip about table function, it seems correct:
table(b$FLASER, b$PLASER, exclude = NULL)
1   2 <NA>
  1    836 691    6
  2     14  70    8
  <NA>   0  45   28
table(b$pheno,exclude = NULL)
1    2 <NA>
 836  828   34

On Mon, May 4, 2020 at 12:45 PM Jeff Newmiller <jdnewmil at dcn.davis.ca.us>
wrote:
To expand on Patrick's response...

You can use the expand.grid function to generate a test table containing
all combinations. However, we would not be in a position to verify that the
results you get when you apply your logic to the test table are what you
want... you know the requirements much better than we do. Nor is that kind
of service what this mailing list is for, so please focus on showing what
you cannot figure out how to accomplish rather than asking us to do or
check your work for you.

On May 4, 2020 10:33:12 AM PDT, "Patrick (Malone Quantitative)" <
malone at malonequantitative.com> wrote:
"I tried this but I am not sure if this is correct:"

Does it provide the expected result for all possible combinations of
1/2/NA
for both variables?

On Mon, May 4, 2020 at 1:16 PM Ana Marija <sokovic.anamarija at gmail.com>
wrote:

Hello,

I have a data frame like this:

head(b)
       FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a
PHENO =2 if b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana

        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

--
Sent from my phone. Please excuse my brevity.

Hello,

Here is a way, using logical indices.

b$pheno <- NA
b$pheno[b$FLASER == 1 & b$PLASER == 1] <- 1
b$pheno[b$FLASER == 2 | b$PLASER == 2] <- 2

Hope this helps,

Rui Barradas

?s 18:15 de 04/05/20, Ana Marija escreveu:
Hello,

I have a data frame like this:

head(b)
        FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a
PHENO =2 if b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana

	[[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Your ifelse expression looks fine.  What goes wrong with it?
Hello,

I have a data frame like this:

head(b)
       FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a
PHENO =2 if b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana

        [[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hi

another possible version 

b$pheno <- ((b$FLASER==2) | (b$PLASER==2))+1

Cheers
Petr
-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of Rui Barradas
Sent: Monday, May 4, 2020 8:32 PM
To: sokovic.anamarija at gmail.com; r-help <r-help at r-project.org>
Subject: Re: [R] if else statement

Hello,

Here is a way, using logical indices.

b$pheno <- NA
b$pheno[b$FLASER == 1 & b$PLASER == 1] <- 1 b$pheno[b$FLASER == 2 |
b$PLASER == 2] <- 2

Hope this helps,

Rui Barradas

?s 18:15 de 04/05/20, Ana Marija escreveu:
Hello,

I have a data frame like this:

head(b)
        FID   IID FLASER PLASER
1: fam1000 G1000      1      1
2: fam1001 G1001      1      1
3: fam1003 G1003      1      2
4: fam1005 G1005      1      1
5: fam1009 G1009      NA      2
6: fam1052 G1052      1      1
...
unique(b$PLASER)
[1]  1  2 NA
unique(b$FLASER)
[1]  1  2 NA

how can I do if else statement so that I am creating a PHENO =2 if
b$FLASER=2 or b$PLASER=2
PHENO=1 if b$FLASER=1 and b$PLASER=1
otherwise PHENO=NA

I tried this but I am not sure if this is correct:
b$pheno=ifelse(b$PLASER==1 & b$FLASER==1,1,ifelse(b$PLASER==2 |
b$FLASER==2,2,NA))

Thanks
Ana

	[[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

______________________________________________
R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.