HI Jim
thank you so much! This is amazing answer!!!
Ana
On Sat, Jun 13, 2020 at 4:09 AM Jim Lemon wrote:
>
> Right, back from shopping. Since you have fourteen rows containing NAs
> and you only want seven, we can infer that half of them must go. As
> they are neatly divided into seven rows in
On 2020-06-13 19:09 +1000, Jim Lemon wrote:
> Right, back from shopping. Since you have fourteen rows containing NAs
> and you only want seven, we can infer that half of them must go. As
> they are neatly divided into seven rows in which only one NA appears
> and seven in which two stare meaningles
Dear Ana,
pmax could also fit here.
pmax(b$FLASER, b$PLASER, na.rm = TRUE)
Bests,
Mark
> --
>
> Message: 21
> Date: Sat, 13 Jun 2020 19:09:11 +1000
> From: Jim Lemon
> To: sokovic.anamar...@gmail.com
> Cc: Rasmus Liland , r-help
>
Right, back from shopping. Since you have fourteen rows containing NAs
and you only want seven, we can infer that half of them must go. As
they are neatly divided into seven rows in which only one NA appears
and seven in which two stare meaninglessly out at us. I will assume
that the latter are the
Great idea!
Here it is:
> b[is.na(b$FLASER) | is.na(b$PLASER),]
FID IID FLASER PLASER pheno
1: fam1837 G1837 1 NA 2
2: fam2410 G2410 NA NA 2
3: fam2838 G2838 NA 2 2
4: fam3367 G3367 1 NA 2
5: fam3410 G3410 1 NA 2
6: fam
Since you have only a few troublesome NA values, if you look at them,
or even better, post them:
b[is.na(b$FLASER) | is.na(b$PLASER),]
perhaps we can work out the appropriate logic to get rid of only the
ones you don't want.
Jim
On Sat, Jun 13, 2020 at 12:50 PM Ana Marija wrote:
>
> Hi Rasmus,
Hi Rasmus,
thank you for getting back to be, the command your provided seems to
add all 11 NAs to 2s
> b$pheno <-
+ ifelse(b$PLASER==2 |
+ b$FLASER==2 |
+ is.na(b$PLASER) |
+ is.na(b$PLASER) & b$FLASER %in% 1:2 |
+ is.na
On 2020-06-13 11:30 +1000, Jim Lemon wrote:
> On Fri, Jun 12, 2020 at 8:06 PM Jim Lemon wrote:
> > On Sat, Jun 13, 2020 at 10:46 AM Ana Marija wrote:
> > >
> > > I am trying to make a new column
> > > "pheno" so that I reduce the number
> > > of NAs
> >
> > it looks like those two NA values in
>
Obviously my guess was wrong. I thought you wanted to impute the value
of "pheno" from FLASER if PLASER was missing. From just your summary
table, it's hard to guess the distribution of NA values. My guess that
the two undesirable NAs were cases where PLASER was missing and FLASER
was 2. My tactic
Hi Jim,
I tried it:
> b$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |is.na(b$PLASER) & b$FLASER ==
> 2,2,1)
> table(b$pheno,exclude = NULL)
12
859 828 11
> b$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |is.na(b$FLASER) & b$PLASER ==
> 2,2,1)
> table(b$pheno,exclude = NULL)
12
859
Hi Ana,
>From your desired result, it looks like those two NA values in PLASER
are the ones you want to drop.
If so, try this:
b$pheno<-ifelse(b$PLASER==2 | b$FLASER==2 |
is.na(b$PLASER) & b$FLASER == 2,2,1)
and if I have it the wrong way round, swap FLASER and PLASER in the
bit I have added.
J
Hi
another possible version
b$pheno <- ((b$FLASER==2) | (b$PLASER==2))+1
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Rui Barradas
> Sent: Monday, May 4, 2020 8:32 PM
> To: sokovic.anamar...@gmail.com; r-help
> Subject: Re: [R] if else statement
>
Your ifelse expression looks fine. What goes wrong with it?
On Tue, 5 May 2020 at 05:16, Ana Marija 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
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$PLASER == 2] <- 2
Hope this helps,
Rui Barradas
Às 18:15 de 04/05/20, Ana Marija escreveu:
Hello,
I have a data frame like this:
head(b)
FID IID FLA
Thank you for the tip about table function, it seems correct:
> table(b$FLASER, b$PLASER, exclude = NULL)
1 2
1836 6916
2 14 708
0 45 28
> table(b$pheno,exclude = NULL)
12
836 828 34
On Mon, May 4, 2020 at 12:45 PM Jeff Newmiller
wrote:
> T
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
"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
wrote:
> Hello,
>
> I have a data frame like this:
>
> > head(b)
>FID IID FLASER PLASER
> 1: f
Dear all,
All works well. Thank you so much for your help.
D## Function 1
wet_dry1 <- function(x,thresh=0.1)
{ for(column in 1:dim(x)[2]) x[,column] <- ifelse(x[,column]>=thresh,1,0)
return(x)
}
wet_dry1(dt)
## Function 2
wet_dry2 <- ( dt >= 0.1)*1
wet_dry2
wet_total <- colSums(wet_dry2)
p
Your f1() has an unneeded for loop in it.
f1a <- function(mat) mat > 0.1, 1, 0)
would do the same thing in a bit less time.
However, I think that a simple
mat > 0.1
would be preferable. The resulting TRUEs and FALSEs
are easier to interpret than the 1s and 0s that f1a()
produces and arithme
I'm sorry, but I have to take issue with this particular use case of
ifelse(). When the goal is to generate a logical vector, ifelse() is
very inefficient. It's better to apply a logical condition directly to
the object in question and multiply the result by 1 to make it
numeric/integer rather than
Thank you jim.
On Saturday, June 6, 2015, Jim Lemon wrote:
> Hi rosalinazairimah,
> I think the problem is that you are using "if" instead of "ifelse". Try
> this:
>
> wet_dry<-function(x,thresh=0.1) {
> for(column in 1:dim(x)[2]) x[,column]<-ifelse(x[,column]>=thresh,1,0)
> return(x)
> }
> we
Hi rosalinazairimah,
I think the problem is that you are using "if" instead of "ifelse". Try this:
wet_dry<-function(x,thresh=0.1) {
for(column in 1:dim(x)[2]) x[,column]<-ifelse(x[,column]>=thresh,1,0)
return(x)
}
wet_dry(dt)
and see what you get.
Also, why can I read your message perfectly w
Please do not post in HTML. It made your posting unreadable. R-help is a plain
text list and when it removes all the HTML tags often the result is gibberish
Have a look at
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and http://adv-r.had.co.nz/Reproduci
Hi
I will not inspect your function as it is corrupted by HTML posting.
If your data frame is named rain
newrain <- (rain>.1)*1
gives you new data frame with reqired coding.
However I am not sure, what do you want to do next. Do you want to merge those
2 data frames so as coded column is besi
e(1:3, .Label = c("samas4", "samas5", "samas6"
), class = "factor")), .Names = c("FID", "IID"), class = "data.frame",
row.names = c(NA,
-3L))
> -Original Message-
> From: Kate Ignatius [mailto:kate.ignat...@g
Ooops,
I edited the code wrong to make it more easier for interpretation and
got X and Y's mixed up. Try this:
for(i in length(1:(nrow(X{
Y$IID1new <- ifelse((as.character(Y[,2]) == as.character(X[,i]) &
Y$IID1new != ''), as.character(as.matrix(X[,(nrow(X)+i)])),'')
}
The second should
Hi
Please, be more clear in what do you want. I get many errors trying your code
and your explanation does not help much.
> for(i in length(1:(2*nrow(X{
+ Y$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
X$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
It is well explained here
http://www.burns-stat.com/pages/Tutor/R_inferno.pdf page 67.
On Sat, Jan 21, 2012 at 9:56 PM, Ery Arias-Castro wrote:
> Hello,
>
> This example seems strange to me:
>
> > if (2 > 3) print('Yes'); else print('No')
> Error: unexpected 'else' in " else"
>
> > {if (2 > 3) p
Don't use the ';'
> if (2 > 3) print('Yes'); else print('No')
Error: unexpected 'else' in " else"
No suitable frames for recover()
> if (2 > 3) print('Yes') else print('No')
[1] "No"
>
'else' is part of the 'if'. The ';' implies the start of a new
statement. You do not need the ';' when writing
On 03/10/2009 3:33 PM, Marianne Promberger wrote:
Two problems with your code:
(1) The "else" has to be on the same line as the closing "}" -- this was
confusing for me as well.
(2) "and" probably should be "&", but read up on logical operators: ?&
Better to always use && in an if.
Duncan M
Two problems with your code:
(1) The "else" has to be on the same line as the closing "}" -- this was
confusing for me as well.
(2) "and" probably should be "&", but read up on logical operators: ?&
Try:
ini=3
b=4
if (ini==1) {
a=3
} else if (ini>1 & b>2 ) {
a=3
} else {
a=6
}
HTH,
Ma
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