Dear Kai
When you ask again it is best to tell us what your input is and what
output you were hoping for and what you actually got. If you can make a
small data-set which shows all that then your post will be much more
likely to get a helpful response. If you want to transfer the data-set
to
Hi Jim,
Sorry to post "same" question, because
1. I was asking to use plain text format. I have to post my question again. But
I don't know if it is working.
2. I'm a beginner for R (< 2 month). It may not easy for me to ask a "clear" R
question. My current work is to transfer my SAS code into
Hi Kai,
You seem to be asking the same question again and again. This does not
give us the warm feeling that you know what you want.
testdf<-data.frame(a=c("Negative","Positive","Neutral","Random","VUS"),
b=c("No","Yes","No","Maybe","Yes"),
c=c("Off","On","Off","Off","On"),
d=c("Bad","Good","Ba
Hello,
You don't need a loop, the R way is a vectorized solution and it's also
clearer.
Create a logical index (note only one &) and assign b, c, d where it's TRUE.
i <- try$a != "Positive" & try$a != "VUS"
try <- within(try, {
b[i] <- ''
c[i] <- ''
d[i] <- ''
})
Hope this helps,
Rui
Kai,
You have made a simple mistake. And now you cannot see it. I believe this is
not uncommon among programmers. It has happened to me more times than I want
to recall.
> On May 30, 2021, at 9:28 AM, Kai Yang via R-help wrote:
>
> Hello List,I have a data frame which having the character c
Can you make R code that creates an actual sample data frame that looks like
you want the answer to look? say, just using the data.frame function and
literal strings. Oh, and read the Posting Guide... you need to send your email
using plain text format or it may get garbled when the list strips
Hello List,I have a data frame which having the character columns:
| a1 | b1 | c1 | d1 |
| a2 | b2 | c2 | d2 |
| a3 | b3 | c3 | d3 |
| a4 | b4 | c4 | d4 |
| a5 | b5 | c5 | d5 |
I need to do: if a1 not = "Positive" and not = "VUS" then values of b1, c1 and
d1 will be zero out. And do the same
Use ifelse function, not if. If is only good for one logical value at a time,
but you are working with long vectors of values simultaneously. I have no
interest in doing all of your workfor you, but the concept is
cpl_or_sngl <- dat1$b %in% c( "couple", "single" )
a_pvt <- "private" == dat1$a
da
You appear to be confusing && with & and || with | ; (the first of each
pair take a logical expression, the second of each a logical vector) ...
as well as if ... else with ifelse (the first is a flow control statement
taking a logical expression; the second is a function taking a logical
vector
Hi all,
I am trying to use the if else statement and create two new columns
based on the existing two columns. Below please find my sample data,
dat1 <-read.table(text="ID a b c d
A private couple 25 35
B private single 24 38
C none single28 32
E none none 20 36 ",header=TRUE,stringsAs
You have really tied yourself up in a knot here. Last I checked, when A==B,
then B==A. You also need to study the difference between ?if and ?ifelse, since
you are not giving the if function the scalar it expects. For example, i$ID is
a vector of three (identical, due to your use of split) value
Dear all,
I have a data.frame xy that contains numeric data and data_qual which contains
qualitative data which I want to include in a for loop with an if statement
(commented out in the code below).
The if statement should be applied if the ID in data_qual$ID is the same than
in xy$ID.
I am t
Hello, "gncl dzgn",
your problem has a flavor of homework which is usually not delt with on
this list. However, a few comments:
0. The description of your problem is rather vague, in particular, the
meaning of "input" in the description of your "conditions" is unclear! (By
the way, your main
Hi everyone,
you might find my question elementary but I am a beginner and unfortunately I
can't fix the problem.
So, I simulate this following algorithm and some values of c are NA. Therefore,
I should add these following two if-statements but I don't know how I should do
it in a for-loop.
On Jun 28, 2012, at 09:42 , Rui Barradas wrote:
> Hello,
>
> Another way is to use index vectors:
>
>
> v1.factor <- c("S","S","D","D","D",NA)
> v2.factor <- c("D","D","S","S","S","S")
>
> td2 <- test.data <- data.frame(v1.factor,v2.factor)
>
> for (i in 1:nrow(test.data) ) {
>
>[... et
On Thu, Jun 28, 2012 at 8:47 PM, James Holland wrote:
> With the multiple if statements I need to check for, I though for statements
> with the if/else if conditional statement was better than nested ifelse
> functions.
for () gives you a lot of flexibility at the expense of being verbose
& slow,
Yeah, the reason I didn't use ifelse is because I've got multiple variables
to manipulate based on the if statement, some factors and some numeric. I
have to look at the factor variables, and based on that, either use one
series of variables or another.
With the multiple if statements I need to c
Hello,
Another way is to use index vectors:
v1.factor <- c("S","S","D","D","D",NA)
v2.factor <- c("D","D","S","S","S","S")
td2 <- test.data <- data.frame(v1.factor,v2.factor)
for (i in 1:nrow(test.data) ) {
[... etc ...]
} #End FOR
# Create index vectors
na1 <- is.na(v1.factor)
na2 <-
Hi James,
On Thu, Jun 28, 2012 at 12:33 AM, James Holland wrote:
> I need to look through a dataset with two factor variables, and depending
> on certain criteria, create a new variable containing the data from one of
> those other variables.
>
> The problem is, R keeps making my new variable an
I need to look through a dataset with two factor variables, and depending
on certain criteria, create a new variable containing the data from one of
those other variables.
The problem is, R keeps making my new variable an integer and saving the
data as a 1 or 2 (I believe the levels of the factor)
Now I got results as I wanted.
Thank you all.
On Wed, Mar 7, 2012 at 2:51 PM, AAsk wrote:
> > x <- -1:4
> > x<0 # returns TRUE (1) or FALSE (0)
> [1] TRUE FALSE FALSE FALSE FALSE FALSE
> > x+as.numeric(x<0)
> [1] 0 0 1 2 3 4
>
> __
> R-help@r-project.
> x <- -1:4
> x<0 # returns TRUE (1) or FALSE (0)
[1] TRUE FALSE FALSE FALSE FALSE FALSE
> x+as.numeric(x<0)
[1] 0 0 1 2 3 4
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-
The simplest method would be:
x[x<0] <- x[x<0]+1
x <- -1:4
x
# [1] -1 0 1 2 3 4
x[x<0] <- x[x<0]+1
x
# [1] 0 0 1 2 3 4
I think where Val got confused is in thinking that
if(x<0)
is applied separately to each element of x, one at a time.
What actually happens, of course, i
Try
> ifelse( x < 0, x + 1, x)
[1] 0 0 1 2 3 4
See also ?ifelse.
HTH,
Jorge.-
On Wed, Mar 7, 2012 at 2:12 PM, Val <> wrote:
> Hi All,
>
> I have one difficulty in using the conditional if statement
>
> Assume ,
>
> x <- -1:4
> x
> [1] -1 0 1 2 3 4
>
> if x is lees than want I want to ad
You need ifelse() instead of if().
On Wed, Mar 7, 2012 at 2:12 PM, Val wrote:
> Hi All,
>
> I have one difficulty in using the conditional if statement
>
> Assume ,
>
> x <- -1:4
> x
> [1] -1 0 1 2 3 4
>
> if x is lees than want I want to add 1 and I used the following command
> if(x<0) {x
Hi All,
I have one difficulty in using the conditional if statement
Assume ,
x <- -1:4
x
[1] -1 0 1 2 3 4
if x is lees than want I want to add 1 and I used the following command
if(x<0) {x=x+1}
Warning message:
In if (x < 0) { :
the condition has length > 1 and only the first element
Hello again.
I don't understand what didn't work.
First, it seems better to use 'nrow', the result is the same
stopifnot(length(x[,1]) == nrow(x))
Then your multiple OR condition.
#if((x[i,1] || x[i,2] || x[i,3] || x[i,4]) < 5)
x <- matrix(1:24, ncol=4)
for(i in 1:nrow(x))
if(any(x
On 24.12.2011 12:03, reena wrote:
It didn't work. :(
What did not work???
Please do not misuse the R-help mailing list! Its posting guide clearly
asks you to cite the thread and specify reproducible examples that make
other able to help.
Best,
Uwe Ligges
--
View this message in cont
It didn't work. :(
--
View this message in context:
http://r.789695.n4.nabble.com/if-statement-problem-tp4230026p4230933.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/lis
This is almost Circle 8.1.7 of
'The R Inferno':
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf
but is making the mistake in the
other direction.
On 23/12/2011 22:40, reena wrote:
Hello,
I want to do fisher test for the rows in data file which has value less than
5 otherwise chi square te
reena wrote
>
> Hello,
>
> I want to do fisher test for the rows in data file which has value less
> than 5 otherwise chi square test .The p values from both test should be
> stored in one resulted file. but there is some problem with bold if
> statement. I don't know how
> implement this line
Hello,
I want to do fisher test for the rows in data file which has value less than
5 otherwise chi square test .The p values from both test should be stored in
one resulted file. but there is some problem with bold if statement. I don't
know how
implement this line properly.
x = cbind(obs1,obs
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of dpender
> Sent: Tuesday, March 08, 2011 4:07 AM
> To: r-help@r-project.org
> Subject: [R] If Statement
>
> Hi,
>
> I am having some problems us
On Tue, Mar 08, 2011 at 04:07:03AM -0800, dpender wrote:
> Hi,
>
> I am having some problems using the if statement correctly. I have used it
> many times previously so I dona't know what is different with this case.
> Here is my problem:
>
> I have a 1X10 matrix of values as follows:
>
> > H.
On Mar 8, 2011, at 7:07 AM, dpender wrote:
Hi,
I am having some problems using the if statement correctly. I have
used it
many times previously so I dona't know what is different with this
case.
Here is my problem:
I have a 1X10 matrix of values as follows:
H.MC
[,1]
[1,] 4.
Hi,
matrix has a 2 dimensions.
Is this work:
a<-matrix(rep(c(1,2),c(5,5)),ncol=1)
dim(a)
for (i in 1:10)
{
ifelse(a[i,]==1, a[i,]<-runif(1,3,4.5), a[i,])
}
a
Andrija
On Tue, Mar 8, 2011 at 1:07 PM, dpender wrote:
> Hi,
>
> I am having some problems using the if statement correctly. I have u
Hi,
I am having some problems using the if statement correctly. I have used it
many times previously so I dona't know what is different with this case.
Here is my problem:
I have a 1X10 matrix of values as follows:
> H.MC
[,1]
[1,] 4.257669
[2,] 7.023242
[3,] 4.949857
[4,] 5.1070
lly Luo
Sent: maandag 25 oktober 2010 2:01
To: r-help@r-project.org
Subject: [R] if statement and truncated distribution
Hi R helpers,
I am trying to use the if statement to generate a truncated random variable
as follows:
if (y[i]==0) { v[i] ~ rnorm(1,0,1) | (-inf ,0) }
if (y[i]==1) { v[i] ~ rnorm
Hi R helpers,
I am trying to use the if statement to generate a truncated random variable
as follows:
if (y[i]==0) { v[i] ~ rnorm(1,0,1) | (-inf ,0) }
if (y[i]==1) { v[i] ~ rnorm(1,0,1) | (0, inf) }
I guess I cannot use " | ( , ) " to restrict the range of a variable in R.
Could you let me kno
On Oct 23, 2010, at 7:44 PM, Jason Kwok wrote:
Thanks Jorge. It works.
Is there a way to keep the actual price in the price column instead of
TRUE/FALSE but filtering on when price>100?
Huh? When I use subset I get what you ask for:
> subset(x, Price > 100)
Price
2010-10-12 10
Thanks Jorge. It works.
Is there a way to keep the actual price in the price column instead of
TRUE/FALSE but filtering on when price>100?
Thanks,
Jay
On Sat, Oct 23, 2010 at 10:37 PM, Jorge Ivan Velez wrote:
> Hi Jay,
>
> If "x" is your data, you could use subset() to do what you want:
>
>
Thanks for the help Jim.
As a new user and member of this mailing list, I'm very impressed with all
the support!
Jay
On Sat, Oct 23, 2010 at 10:21 PM, jim holtman wrote:
> Need to understand how 'indexing' is done in R:
>
> > x <- read.table(textConnection(" Price
> + 2010-10
Hi Jay,
If "x" is your data, you could use subset() to do what you want:
subset(x, Price > 100)
See ?subset for more information.
HTH,
Jorge
On Sat, Oct 23, 2010 at 9:56 PM, Jason Kwok <> wrote:
>Price
> 2010-10-11 99
> 2010-10-12101
> 2010-10-13102
> 2010-10-
Need to understand how 'indexing' is done in R:
> x <- read.table(textConnection(" Price
+ 2010-10-11 99
+ 2010-10-12101
+ 2010-10-13102
+ 2010-10-14103
+ 2010-10-15 99
+ 2010-10-18 98
+ 2010-10-19 97
+ 2010-10-20101
+ 2010-10-21101
+ 2010-10-2
Price
2010-10-11 99
2010-10-12101
2010-10-13102
2010-10-14103
2010-10-15 99
2010-10-18 98
2010-10-19 97
2010-10-20101
2010-10-21101
2010-10-22101
I have this dataset and I only want to return instances when the Price is >
100.
If I use t
"Santosh Srinivas" writes:
> I'm unable to find the OR operator like other language .. any suggestions?
>
> I want to do If (condition1 OR condition 2){ do something }
>
if((condition1) | (condition2)){ do something }
--
aleblanc
__
R-help@r-project
Joshua Wiley wrote:
Hi Santosh,
I believe you are looking for |. For example:
if(3 > 5 | 3 < 4) {print(TRUE)}
In an if () statement you use || more often. | is a vector operator
that always evaluates both arguments; || is a scalar operator that quits
if the left hand argument determines t
Hi Santosh,
I believe you are looking for |. For example:
if(3 > 5 | 3 < 4) {print(TRUE)}
Cheers,
Josh
On Fri, Oct 22, 2010 at 12:51 AM, Santosh Srinivas
wrote:
> I'm unable to find the OR operator like other language .. any suggestions?
>
> I want to do If (condition1 OR condition 2){ do s
I'm unable to find the OR operator like other language .. any suggestions?
I want to do If (condition1 OR condition 2){ do something }
Thanks for answering this elementary question.
[[alternative HTML version deleted]]
__
R-help@r-proj
Thomas Lumley wrote:
> On Tue, 24 Mar 2009, Wacek Kusnierczyk wrote:
>
>>
>> thanks for the explanation. for the sake of rvalues, please do not fix
>> this bug. i find it useful to avoid messing with = and <-.
>>
>
> If you insist on calling it a bug, the chances of its getting fixed
> are higher
On Tue, 24 Mar 2009, Wacek Kusnierczyk wrote:
thanks for the explanation. for the sake of rvalues, please do not fix
this bug. i find it useful to avoid messing with = and <-.
If you insist on calling it a bug, the chances of its getting fixed are higher.
-thomas
Thomas Lumley
Thomas Lumley wrote:
> On Tue, 24 Mar 2009, Wacek Kusnierczyk wrote:
>
>> thanks. so it seems to be intentionally parsable, though i wouldn't say
>> that this gives a meaning to ':=' -- the operator has a syntactic
>> category, but no semantics. the syntactic category does not imply any
>> semant
On Tue, 24 Mar 2009, Wacek Kusnierczyk wrote:
thanks. so it seems to be intentionally parsable, though i wouldn't say
that this gives a meaning to ':=' -- the operator has a syntactic
category, but no semantics. the syntactic category does not imply any
semantics, as in
'<-' = function(a,
Romain Francois wrote:
> This is probably due to that in the gram.y file :
>
> case ':':
>if (nextchar(':')) {
>if (nextchar(':')) {
>yylval = install(":::");
>return NS_GET_INT;
>}
>else {
>yylval = install("::");
>return NS_GET;
>
Wacek Kusnierczyk wrote:
Berwin A Turlach wrote:
G'day Carl,
On Mon, 23 Mar 2009 20:11:19 -0400
Carl Witthoft wrote:
But seriously: can someone explain to me what's going on in the
rvalues.r code? I tried a simple experiment, replacing ":=" with a
"colec" in the code, and of c
Berwin A Turlach wrote:
> G'day Carl,
>
> On Mon, 23 Mar 2009 20:11:19 -0400
> Carl Witthoft wrote:
>
>
>
>> But seriously: can someone explain to me what's going on in the
>> rvalues.r code? I tried a simple experiment, replacing ":=" with a
>> "colec" in the code, and of course the line
>
Berwin A Turlach wrote:
> G'day Carl,
>
> On Mon, 23 Mar 2009 20:11:19 -0400
> Carl Witthoft wrote:
>
>
>> >From: Wacek Kusnierczyk
>> >Date: Sun, 22 Mar 2009 22:58:49 +0100
>>
>>
>> >just for fun, you could do this with multiassignment, e.g., using
>> >the (highly experimental and prematu
Berwin A Turlach wrote:
> G'day Carl,
>
> On Mon, 23 Mar 2009 20:11:19 -0400
> Carl Witthoft wrote:
>
>
>> >From: Wacek Kusnierczyk
>> >Date: Sun, 22 Mar 2009 22:58:49 +0100
>>
>>
>> >just for fun, you could do this with multiassignment, e.g., using
>> >the (highly experimental and prematu
G'day Carl,
On Mon, 23 Mar 2009 20:11:19 -0400
Carl Witthoft wrote:
> >From: Wacek Kusnierczyk
> >Date: Sun, 22 Mar 2009 22:58:49 +0100
>
>
> >just for fun, you could do this with multiassignment, e.g., using
> >the (highly experimental and premature!) rvalues:
>
> >source('http://m
>From: Wacek Kusnierczyk
>Date: Sun, 22 Mar 2009 22:58:49 +0100
>just for fun, you could do this with multiassignment, e.g., using the
>(highly experimental and premature!) rvalues:
>source('http://miscell.googlecode.com/svn/rvalues/rvalues.r')
>if (TRUE)
> c(df1, df2) := list(4
jimdare wrote:
> Hi,
>
> I tried to create the following if / else statement but I keep getting the
> error "Error: unexpected '}' in "size="large",center="none")
> }" (I have highlighted the } in bold where the error is occuring). I can't
> seem to find a reason for this, does anyone know how I
You appear to have a number of unbalanced parentheses in your
statements. Here is one that seems to work:
if (nostocks<=3)
{data1<-test[,data1stocks];
tex1<-latex(data1, file=paste(i$Species[1], "1.tex", sep=""),
rowname = NULL,
Hi,
I tried to create the following if / else statement but I keep getting the
error "Error: unexpected '}' in "size="large",center="none")
}" (I have highlighted the } in bold where the error is occuring). I can't
seem to find a reason for this, does anyone know how I can fix it?
Thanks,
Jame
jimdare wrote:
> Thanks very much
>
>
you're welcome.
just for fun, you could do this with multiassignment, e.g., using the
(highly experimental and premature!) rvalues:
source('http://miscell.googlecode.com/svn/rvalues/rvalues.r')
if (TRUE)
c(df1, df2) := list(4:8, 9:13)
Thanks very much
Wacek Kusnierczyk wrote:
>
> jimdare wrote:
>> Hi,
>>
>> How do I tell an if statement to generate two seperate outputs.
>>
>> E.g If X>5 I want to create df1 and df2:
>>
>> if (X>5) {df1<-c(4,5,6,7,8) AND df2<-c(9,10,11,12,13)}
>>
> almost there:
>
> if (X>5) {df1<-
if (X>5) {
df1<-c(4,5,6,7,8)
df2<-c(9,10,11,12,13)
}
hth,
Daniel Moreira, MD
jimdare
Sent by: r-help-boun...@r-project.org
03/22/2009 05:27 PM
To
r-help@r-project.org
cc
Subject
[R] If statement generates two outputs
Hi,
How do I tell an if statement to generate two
jimdare wrote:
> Hi,
>
> How do I tell an if statement to generate two seperate outputs.
>
> E.g If X>5 I want to create df1 and df2:
>
> if (X>5) {df1<-c(4,5,6,7,8) AND df2<-c(9,10,11,12,13)}
>
almost there:
if (X>5) {df1<-c(4,5,6,7,8); df2<-c(9,10,11,12,13)}
vQ
Hi,
How do I tell an if statement to generate two seperate outputs.
E.g If X>5 I want to create df1 and df2:
if (X>5) {df1<-c(4,5,6,7,8) AND df2<-c(9,10,11,12,13)}
Thanks,
James
--
View this message in context:
http://www.nabble.com/If-statement-generates-two-outputs-tp22650844p22650844.htm
Hi there,
Try this:
x<-your.vector
x[x>10]
HTH,
Jorge
On Mon, Feb 23, 2009 at 12:31 PM, bioinformatics_guy
wrote:
>
> right now I have a vector of about 1000 points. I'd like to iterate through
> each of these points and and test if it is greater than a certain value and
> if not, throw it o
right now I have a vector of about 1000 points. I'd like to iterate through
each of these points and and test if it is greater than a certain value and
if not, throw it out.
x=vector
y=empty vector
j=0
for i (0..length[vector])
if x[i] > 10
y[j] = x[i]
j++
Thats basical
Duncan Murdoch wrote:
> Shruthi Jayaram wrote:
>> Hi,
>>
>> How do I check for two conditions in an if loop? I want to check if a
>> value
>> lies between 2 other values.
>>
> "if" isn't normally a loop, but what you want is the vectorized
> version, the ifelse() function.
>> For example,
>> A <
Hello,
If you do
C <- A
C[A > X & A < Y] <- 0
you get what it seems you want.
Best regards,
Carlos J. Gil Bellosta
http://www.datanalytics.com
On Mon, 2009-01-05 at 03:41 -0800, Shruthi Jayaram wrote:
> A <- ts(rnorm(120), freq=12, start=c(1992,8))
> X <- 0.5
> Y <- 0.8
_
Shruthi Jayaram wrote:
Hi,
How do I check for two conditions in an if loop? I want to check if a value
lies between 2 other values.
"if" isn't normally a loop, but what you want is the vectorized version,
the ifelse() function.
For example,
A <- ts(rnorm(120), freq=12, start=c(1992,8))
X
Hi,
How do I check for two conditions in an if loop? I want to check if a value
lies between 2 other values.
For example,
A <- ts(rnorm(120), freq=12, start=c(1992,8))
X <- 0.5
Y <- 0.8
I would like to create a new vector C for which C[i] is 0 if A[i] lies in
between X and Y.
Would be grate
Try:
das$danger <- with(das, (age > 65) * (bmi > 30))
On Jan 1, 2008 5:03 PM, Gerard Smits <[EMAIL PROTECTED]> wrote:
> Hi All,
>
> I have a small dataset named das (43 cases) in which I am trying to
> create a binary outcome (1/0) based on the following code:
>
> if (das$age>65 && das$bmi>30) {
Hi Domenico,
I was incorrectly assuming it would use a vector of equal length to
my data. frame. Thanks for the clarification.
Also, thanks for the many alternate programming approaches provided by others.
Gerard
At 02:25 PM 1/1/2008, Domenico Vistocco wrote:
>You should look for your answer
08, Christos Hatzis wrote:
> >You need to use '&' instead of '&&':
> >
> >A shorter version of your code using ifelse:
> >
> >das$danger <- with(das, ifelse(age>65 & bmi>30, 1, 0))
> >
> >HTH
> >
> >-Chr
You should look for your answer using the help for the if statement (?"if").
The cond argument should be a scalar (otherwise only the first element
is used).
?"if"
.
cond: A length-one logical vector that is not 'NA'. Conditions of
length greater than one are accepted with a warnin
mp; bmi>30, 1, 0))
>
>HTH
>
>-Christos
>
> > -Original Message-
> > From: [EMAIL PROTECTED]
> > [mailto:[EMAIL PROTECTED] On Behalf Of Gerard Smits
> > Sent: Tuesday, January 01, 2008 5:04 PM
> > To: r-help@r-project.org
> > Subject: [R] i
On Behalf Of Gerard Smits
> Sent: Tuesday, January 01, 2008 5:04 PM
> To: r-help@r-project.org
> Subject: [R] if statement problem
>
> Hi All,
>
> I have a small dataset named das (43 cases) in which I am
> trying to create a binary outcome (1/0) based on the following code
Hi All,
I have a small dataset named das (43 cases) in which I am trying to
create a binary outcome (1/0) based on the following code:
if (das$age>65 && das$bmi>30) {das$danger<-1} else das$danger<-0
I am setting a flag called 'danger' to 1 of the subject is over 65
and has a BMI > 30.
I fin
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