ilto:r-help-boun...@r-project.org] On Behalf Of Kirsten
> Morehouse
> Sent: Monday, July 24, 2017 2:24 PM
> To: R-help
> Subject: [R] Ifelse statements and combining columns
>
> Hi everyone,
>
> I'm having some trouble with my ifelse statements.
>
> I'm try
That ifelse statement is a mess, it is missing brackets, comma separators
between arguments and & or | between conditions. The bracket error points
towards the invalid bracket you had in the second ifelse since it expects
and yes and a no argument alongside conditions.
ifelse(test = (dat$cond == "
On 24/07/2017 8:57 AM, Jeff Newmiller wrote:
Not a reproducible example, so a bit of guessing here, but
a) don't try to assign results to variables inside the ifelse. That is, remove all the
single-equals signs and "test" variables. If you really need to conditionally assign
variables then us
Not a reproducible example, so a bit of guessing here, but
a) don't try to assign results to variables inside the ifelse. That is, remove
all the single-equals signs and "test" variables. If you really need to
conditionally assign variables then use "if"... but chances are good you don't
need
Hello,
Your ifelse statement is a mess. I cannot make sense of it. Let me try
to explain where I've lost it.
dat$cond <- ifelse(test = dat$cond == "cond1"
This is already very bad, do you mean dat$cond == "cond1" ?
Maybe you mean that condition OR the others below, but then there's one
'|' mi
Hi everyone,
I'm having some trouble with my ifelse statements.
I'm trying to put 12 conditions within 3 groups. Here is the code I have so
far:
dat$cond <- ifelse(test = dat$cond == "cond1" | dat$cond == "cond2" |
dat$cond == "cond3" dat$cond == "cond4"
yes = "Uniform"
reas Nord
> Cc: r-help@r-project.org
> Subject: Re: [R] ifelse for creating discriminating variable based on two
> conditions
>
> Hi Andreas,
> Try this:
>
> fruit_2sds<-by(data2$molecule,data2$fruit,sd)*2
> data2$newcol<-ifelse(data2$molecule>fruit_2sds[
Hi Andreas,
Try this:
fruit_2sds<-by(data2$molecule,data2$fruit,sd)*2
data2$newcol<-ifelse(data2$molecule>fruit_2sds[data2$fruit],1,0)
or even just:
data$newcol<-as.numeric(data2$molecule>fruit_2sds[data2$fruit])
Jim
On Fri, Oct 14, 2016 at 5:17 PM, Andreas Nord wrote:
>
> Dear list,
>
> Apo
Dear list,
Apologies for a likely naive question.
I am trying to create a discriminating dummy variable using 'ifelse' based on
conditions in two variables.
Specifically, I want to assign levels in a new factor as '0' or '1' based on a
user-defined cut off. I.e. something similar to:
>da
ifelse() often has problems constructing the right type of return value.
if you want to keep the data as a factor (with its existing levels)
use x[condition] <- value instead of ifelse(condition, value, x). E.g.,
> x <- factor(c("Large","Small","Small","XLarge"),
levels=c("Small","Med","Large"
Apologies - you're right. Missed it in the pdf.
K.
On Sun, Sep 28, 2014 at 10:22 AM, Bert Gunter wrote:
> Inline.
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> Cliffo
Inline.
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
Clifford Stoll
On Sun, Sep 28, 2014 at 6:38 AM, Kate Ignatius wrote:
> Strange that,
>
> I did put everything with as.charact
Strange that,
I did put everything with as.character but all I got was the same...
class of dbpmn[,2]) = factor
class of dbpmn[,21] = factor
class of dbpmn[,20] = data.frame
This has to be a problem ???
I can put reproducible output here but not sure if this going to of
help here. I think its
I believe you are in Circle 8.2.7 of
The R Inferno.
http://www.burns-stat.com/documents/books/the-r-inferno/
Pat
On 28/09/2014 05:49, Kate Ignatius wrote:
Quick question:
I am running the following code on some variables that are factors:
dbpmn$IID1new <- ifelse(as.character(dbpmn[,2]) ==
as
On Sun, 28 Sep 2014 12:49:41 AM Kate Ignatius wrote:
> Quick question:
>
> I am running the following code on some variables that are factors:
>
> dbpmn$IID1new <- ifelse(as.character(dbpmn[,2]) ==
> as.character(dbpmn[,(21)]), dbpmn[,20], '')
>
> Instead of returning some value it gives me this
Not reproducible, ball in your court. However, in the meantime, my suggestion
is to not do that. Convert to character before you alter the factor, then
convert back when you are done.
---
Jeff Newmiller
Quick question:
I am running the following code on some variables that are factors:
dbpmn$IID1new <- ifelse(as.character(dbpmn[,2]) ==
as.character(dbpmn[,(21)]), dbpmn[,20], '')
Instead of returning some value it gives me this:
c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1))
Playing around wi
On 16/01/2014 8:46 AM, ONKELINX, Thierry wrote:
You want
y <- ifelse(x == 'a', 1, 2)
or use if, rather than ifelse, i.e.
if (x == 'a') {
y <- 1
} else {
y <- 2
}
ifelse() is mainly used when you want to work with whole vectors of
decisions, e.g.
x <- 1:10
ifelse(x > 5, 1, 0)
Duncan M
rzonden: donderdag 16 januari 2014 14:32
Aan: r
Onderwerp: [R] ifelse...
Hi,
Sorry for the newbie question! My code:
x <- 'a'
ifelse(x == 'a',y <- 1, y <- 2)
print(y)
Shouldn't this assign a value of 1? When I execute this I get:
> x <- 'a'
Hi,
Sorry for the newbie question! My code:
x <- 'a'
ifelse(x == 'a',y <- 1, y <- 2)
print(y)
Shouldn't this assign a value of 1? When I execute this I get:
> x <- 'a'
> ifelse(x == 'a',y <- 1, y <- 2)
[1] 1
> print(y)
[1] 2
Am I doing something really daft???
thanks!
> sessionInfo()
R ve
Hi Adel,
If the problem is the spacing, then
library(stringr)
1*(long_df$country_name %in% str_trim(countrydiff))
# [1] 1 0 0 1 1 0 0 0 0 0
A.K.
Dear Arun
Thanks for your reply, it made me realize that the problem was
not in the code but in the levels() of the factors. Some countries had
som
Dear Arun
Thanks for your reply, it made me realize that the problem was not in the
code but in the levels() of the factors. Some countries had some extra
spacing which made the ifelse() function not work. So if I modify your code
(added space to countrydiff), it will then look something like thi
Hi,
Suggestion 1: read
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
and bookmark it for future reference.
Suggestion 2:
set.seed(123)
countrydiff <- letters[1:5]
long_df <- data.frame(country_name = sample(letters[1:8], 20, replace=TRUE))
long_df$povdat <
Dear list-members,
I have the following problem: I have a vector (countrydiff) with length 72
and another vector (long_df$country_name) which is about 12000 long.
Basically what I want to do is to if the factor level (or string name) in
long_df$country_name appears on the countrydiff, then long_d
Hi,
Please show a reproducible example.
countrydiff <- c("Albania", "Algeria", "Belarus", "Canada", "Germany")
long_df <- data.frame(country_name = c("Algeria", "Guyana", "Hungary",
"Algeria", "Canada", "Iran", "Iran", "Norway","Uruguay", "Zimbabwe") )
ifelse(long_df$country_name %in% countrydif
On Dec 3, 2013, at 5:37 AM, Hadley Wickham wrote:
> A better solution to this problem is to use character indexing:
>
> x <- c("Tuesday", "Thursday", "Sunday")
> c(Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5,
> Saturday = 6, Sunday = 7)[x]
>
> http://adv-r.had.co.nz/Subsett
A better solution to this problem is to use character indexing:
x <- c("Tuesday", "Thursday", "Sunday")
c(Monday = 1, Tuesday = 2, Wednesday = 3, Thursday = 4, Friday = 5,
Saturday = 6, Sunday = 7)[x]
http://adv-r.had.co.nz/Subsetting.html#lookup-tables-character-subsetting
Hadley
On Mon, Dec 2
vectors.
>
> Or, you can use factors.
>> factor(x, levels=LongDayNames, labels=ShortDayNames)
>[1] Wed Mon Wed
>Levels: Mon Tue Wed
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
> > -Original Message-
> > From:
vectors.
>
> Or, you can use factors.
>> factor(x, levels=LongDayNames, labels=ShortDayNames)
>[1] Wed Mon Wed
>Levels: Mon Tue Wed
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
> > -Original Message-
> > From:
ors.
Or, you can use factors.
> factor(x, levels=LongDayNames, labels=ShortDayNames)
[1] Wed Mon Wed
Levels: Mon Tue Wed
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Beh
Hi. Thanks. Which part? I looked at the below but I don't think it clearly
addresses the nested ifelse situation.
ifelse {base}R DocumentationConditional Element Selection Description
ifelse returns a value with the same shape as test which is filled with
elements selected from either yes or no
On 13-12-02 7:49 PM, Bill wrote:
It seems so inefficient. I mean the whole first vector will be
evaluated. Then if the second if is run the whole vector will be
evaluated again. Then if the next if is run the whole vector will be
evaluted again. And so on. And this could be only to test the first
It seems so inefficient. I mean the whole first vector will be evaluated.
Then if the second if is run the whole vector will be evaluated again. Then
if the next if is run the whole vector will be evaluted again. And so on.
And this could be only to test the first element (if it is false for each
i
On 13-12-02 7:33 PM, Bill wrote:
ifelse ((day_of_week == "Monday"),1,
ifelse ((day_of_week == "Tuesday"),2,
ifelse ((day_of_week == "Wednesday"),3,
ifelse ((day_of_week == "Thursday"),4,
ifelse ((day_of_week == "Friday"),5,
ifelse ((day_of_week == "Saturday"),6,7)))
In cod
ifelse ((day_of_week == "Monday"),1,
ifelse ((day_of_week == "Tuesday"),2,
ifelse ((day_of_week == "Wednesday"),3,
ifelse ((day_of_week == "Thursday"),4,
ifelse ((day_of_week == "Friday"),5,
ifelse ((day_of_week == "Saturday"),6,7)))
In code like the above, day_of_week is a vector
Hello:
This seems like an obvious question, but I am having trouble answering it.
I am new to R, so I apologize if its too simple to be posting. I have
searched for solutions to no avail.
I have data that I am trying to set up for further analysis ("training
data"). What I need is 12 groups based
ailto:r-help-boun...@r-project.org] On
> Behalf
> Of Jonathan Greenberg
> Sent: Tuesday, September 10, 2013 12:40 PM
> To: r-help
> Subject: [R] ifelse question (I'm not sure why this is working)...
>
> R-helpers:
>
> One of my intrepid students came up with a solu
R-helpers:
One of my intrepid students came up with a solution to a problem where
they need to write a function that takes a vector x and a "scalar" d,
and return the indices of the vector x where x %% d is equal to 0 (x
is evenly divisible by d). I thought I had a good handle on the
potential so
On 09/11/2013 05:40 AM, Jonathan Greenberg wrote:
R-helpers:
One of my intrepid students came up with a solution to a problem where
they need to write a function that takes a vector x and a "scalar" d,
and return the indices of the vector x where x %% d is equal to 0 (x
is evenly divisible by d)
On Sep 10, 2013, at 12:40 PM, Jonathan Greenberg wrote:
> R-helpers:
>
> One of my intrepid students came up with a solution to a problem where
> they need to write a function that takes a vector x and a "scalar" d,
> and return the indices of the vector x where x %% d is equal to 0 (x
> is even
On Sep 10, 2013, at 23:39 , Jim Lemon wrote:
> On 09/11/2013 05:40 AM, Jonathan Greenberg wrote:
>> R-helpers:
>>
>> One of my intrepid students came up with a solution to a problem where
>> they need to write a function that takes a vector x and a "scalar" d,
>> and return the indices of the ve
roject.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of William Dunlap
> Sent: Tuesday, September 10, 2013 12:59 PM
> To: Jonathan Greenberg; r-help
> Subject: Re: [R] ifelse question (I'm not sure why this is working)...
>
> > remainderFunction<-function(
CO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Ron Michael
> Sent: Sunday, August 11, 2013 1:18 PM
> To: r-help@r-project.org
> Subject: [R] ifelse() applied on function
>
Hi,
On Sun, Aug 11, 2013 at 1:18 PM, Ron Michael wrote:
> Hi,
>
> How can I apply ifelse function to chose appropriate function?
>
> My goal is user will chose lapply() function if "ChooseFn = T" otherwise to
> chose sfLapply() from snowfall package.
>
> I am basically trying to avoid repeatatio
Hi,
How can I apply ifelse function to chose appropriate function?
My goal is user will chose lapply() function if "ChooseFn = T" otherwise to
chose sfLapply() from snowfall package.
I am basically trying to avoid repeatation to write all internal steps, if user
choses lapply or sfLapply. F
NJ 08854.
848-445-7669
http://fsrm.rutgers.edu/
- Original Message -
From: "David Carlson"
To: "Neville O'Reilly" , r-help@r-project.org
Sent: Wednesday, June 26, 2013 1:15:06 PM
Subject: RE: [R] Ifelse leading to inconsistent result
I don't see that yo
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Neville O'Reilly
Sent: Wednesday, June 26, 2013 10:41 AM
On Jun 26, 2013, at 17:40 , Neville O'Reilly wrote:
> I have used ifelse in count variables to count the number of times in a
> simulation the values of a vector of logprice fall within mutually exclusive
> ranges. However, there is a double count in the result i.e. i am getting
> output indic
On 26/06/2013 11:40 AM, Neville O'Reilly wrote:
I have used ifelse in count variables to count the number of times in a
simulation the values of a vector of logprice fall within mutually exclusive
ranges. However, there is a double count in the result i.e. i am getting output
indicating values
I have used ifelse in count variables to count the number of times in a
simulation the values of a vector of logprice fall within mutually exclusive
ranges. However, there is a double count in the result i.e. i am getting output
indicating values falling in mutually exclusive ranges. Here is the
1$group2<-cumsum(c(TRUE, v[-length(v)] != v[-1] ))# works
test1
# V1 group group2
#1 a 1 1
#2 a 1 1
#3 a 1 1
#4 b 2 2
#5 b 2 2
#6 b 2 2
#7 a 3 3
#8 a 3 3
#9 a 3 3
#10 d 5 4
#11 d
n...@r-project.org] On
> Behalf
> Of Jeff
> Sent: Thursday, January 24, 2013 4:08 PM
> To: arun
> Cc: R help; Jeffrey Fuerte
> Subject: Re: [R] ifelse to speed up loop?
>
> Thank you for the tips. I should mention that the V1 is numeric in this case
> but I am trying
&g
Group
> #1 1 1
> #2 1 1
> #3 1 1
> #4 2 2
> #5 2 2
> #6 2 2
> #7 1 3
> # 1 3
> #9 1 3
> #10 2 4
> #11 2 4
> #12 2 4
> A.K.
>
>
>
> ----- Original Message -
> From: Jeffrey Fu
, 2013 6:48 PM
Subject: RE: [R] ifelse to speed up loop?
I like
v <- test[,1]
c(TRUE, cumsum( v[-length(v)] != v[-1] ))
(R's arithmetic on logicals treats TRUE as 1 and FALSE as 0.)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-
Yes.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: arun [mailto:smartpink...@yahoo.com]
> Sent: Thursday, January 24, 2013 3:53 PM
> To: William Dunlap
> Cc: R help
> Subject: Re: [R] ifelse to speed up loop?
>
> Hi,
>
roject.org] On
> Behalf
> Of arun
> Sent: Thursday, January 24, 2013 3:37 PM
> To: Jeffrey Fuerte
> Cc: R help
> Subject: Re: [R] ifelse to speed up loop?
>
> HI,
>
> This might be better.
> test$group<-cumsum(abs(c(1,diff(test[,1]
> A.K.
>
>
>
HI,
This might be better.
test$group<-cumsum(abs(c(1,diff(test[,1]
A.K.
- Original Message -
From: Jeffrey Fuerte
To: r-help@r-project.org
Cc:
Sent: Thursday, January 24, 2013 4:20 PM
Subject: [R] ifelse to speed up loop?
Hello,
I'm not sure how to explain what I'
I should add that if your V1 column is numeric, fiddling around with
?diff or equivalent will also identify the groupings for you and may
be faster than rle().
-- Bert
On Thu, Jan 24, 2013 at 3:32 PM, Bert Gunter wrote:
> Your query is a little unclear to me, but I suspect
> ?rle
> is what you w
2 2
#7 1 3
# 1 3
#9 1 3
#10 2 4
#11 2 4
#12 2 4
A.K.
- Original Message -
From: Jeffrey Fuerte
To: r-help@r-project.org
Cc:
Sent: Thursday, January 24, 2013 4:20 PM
Subject: [R] ifelse to speed up loop?
Hello,
I'm not sure how to explain what
Your query is a little unclear to me, but I suspect
?rle
is what you want.
-- Bert
On Thu, Jan 24, 2013 at 1:20 PM, Jeffrey Fuerte wrote:
> Hello,
>
> I'm not sure how to explain what I'm looking for but essentially I have a
> test dataset that looks like this:
>
> test:
>V1
> 1 1
> 2 1
Hello,
I'm not sure how to explain what I'm looking for but essentially I have a
test dataset that looks like this:
test:
V1
1 1
2 1
3 1
4 2
5 2
6 2
7 1
8 1
9 1
10 2
11 2
12 2
And what I want to be able to do is create another column that captures a
"grouping" variable t
Thanks, it worked without quotation marks.
--
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__
R-help@r-project.org mailing list
https://stat
Try leaving the quote marks off the NA:
data[,4] <- ifelse(data[,3]<1,data[,1]/(1-data[,3]),NA)
You are putting a character value with the "NA" which converts
everything else to a character. If you had examined the dataframe
with 'str(data)' you would probably have seen the problem.
On Wed, Nov
Hi,
I have a data frame and I need to add another column.
I am using: data[,4] <- ifelse(data[,3]<1,data[,1]/(1-data[,3]),"NA")
This is returning values but are in quotation marks. The error I am getting
is "non-numeric argument to binary operator"
I need column four in data to be numeric. Trie
)
#[1] 1 0 1 1 1 1 0 1 1 1 1 1 0 1 1
A.K.
- Original Message -
From: brunosm
To: r-help@r-project.org
Cc:
Sent: Thursday, October 25, 2012 12:55 PM
Subject: Re: [R] ifelse reformulation
Arun, thank you very much... but a new problem arose...
What if... in the vari
1 1 1 1 0 1 1
as.vector(apply(test1,1,function(x)
ifelse(any(x[4:6]==40|x[4:6]==50|x[4:6]==60|x[4:6]==70),0,1)))
#[1] 1 0 1 1 1 1 1 1 1 1 1 1 0 1 1
A.K.
From: Bruno Marques
To: arun
Sent: Wednesday, October 24, 2012 12:47 PM
Subject: Re: [R] ifelse reformula
brunosm
> Sent: Friday 12 October 2012 11:51
> To: r-help@r-project.org
> Subject: [R] ifelse reformulation
>
> Hi, i'm trying to simplify some R code but i got stucked in this:
>
> test<-data.frame(cbind(id,x1,x2,x3,x4,x5,x6,x7))
> test
>
> > test
>id
Hi arun kirshna
just let me ask you something else.
Imagin that i only want to "search" in the variables x3,x4 and x5. How can i
do this?
Regards,
Bruno
--
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Sent from the R help mailing li
p@r-project.org
Cc:
Sent: Friday, October 12, 2012 10:18 AM
Subject: Re: [R] ifelse reformulation
That was exacly what i was looking for! Thanks a lot!
Cheers!
--
View this message in context:
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Sent from the R help mailing l
That was exacly what i was looking for! Thanks a lot!
Cheers!
--
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
Hi, i'm trying to simplify some R code but i got stucked in this:
test<-data.frame(cbind(id,x1,x2,x3,x4,x5,x6,x7))
test
> test
id x1 x2 x3 x4 x5 x6 x7
1 1 36 26 21 32 31 27 31
2 2 45 21 46 50 22 36 29
3 3 49 47 35 44 33 31 46
4 4 42 32 38 28 39 45 32
5 5 29 42 39 48 25 35 34
6 6 39
Ifelse is a vector function and is absolutely inappropriate for that use.
Use "if" instead.
Also, read the help for Sys.sleep... you need to tell it how long you want to
sleep. You should compute how long that is from now and sleep that long.
---
Hi Everyone,
I am trying to run a time based query and need some of your help.
Not much of data or packages.
Just a simple one.
Query I am trying to execute.
ifelse ((as.numeric(as.POSIXct("2012-10-01 20:38:00"))),
(rnorm(1,2,1)),(Sys.sleep()))
Note. Why I am using as.numeric is as I have a li
Hah - I guess I didn't mean I understood it in full as I expect I will
run into it again without anticipating it.
But, now that I know the "old adage" I will look there first when I
run into a problem.
Also, I used the square root of machine precision instead - thanks for
that, too.
Thank you, t
On Fri, Aug 24, 2012 at 7:29 PM, Jennifer Sabatier
wrote:
> AHHH I GOT IT!!
>
> And I *think* I understand about floating point arithmetic..
Well then you're doing much better than the rest of us: it's quite a
difficult subject and only gets trickier as you think about it more.
(N
AHHH I GOT IT!!
And I *think* I understand about floating point arithmetic..
In this case vn$PM.DIST.TOT is the sum of proportions. So, it should
be anywhere 0 and 1.
In our case, if it's anything other than 1 when vn$PM.EXP is greater
than 0 then it means something is wrong wi
I see that you got other responses while I was composing an answer.
Your 'example.csv' did come through for me, but I still can't
replicate your PM.DIST_flag variable. Specifically, observations
30, 33, 36 and 40 are wrong.
I agree with Rui, that there's something else going on. The data
you've s
Hi Michael,
Thanks for letting me know how to post data. I will try to upload it
that way in a second.
I can usually use code to make a reproducible dataset but this time
with the ifelse behaving strangely (perhaps, it's probably me) I
didn't think I could do it easily so I figured I would just
Oh, sorry, I first though you couldn't post data to the list, but then
I thought I remembered other people doing so, so I tried to post it.
Here is a copy.
Thanks,
Jen
On Fri, Aug 24, 2012 at 5:49 PM, Rui Barradas wrote:
> No data arrived to me.
>
> Rui Barradas
> Em 24-08-2012 22:46, Jennifer
On Fri, Aug 24, 2012 at 4:50 PM, R. Michael Weylandt
wrote:
> On Fri, Aug 24, 2012 at 4:46 PM, Jennifer Sabatier
> wrote:
>> Hi Michael,
>>
>> No, I never use attach(), exactly for the reasons you state. To do
>> due diligence I did a search of code for the function and it didn't
>> come up (I w
On Fri, Aug 24, 2012 at 4:46 PM, Jennifer Sabatier
wrote:
> Hi Michael,
>
> No, I never use attach(), exactly for the reasons you state. To do
> due diligence I did a search of code for the function and it didn't
> come up (I would have been shocked because I never us it!).
>
> Now that real data
No data arrived to me.
Rui Barradas
Em 24-08-2012 22:46, Jennifer Sabatier escreveu:
Hi Michael,
No, I never use attach(), exactly for the reasons you state. To do
due diligence I did a search of code for the function and it didn't
come up (I would have been shocked because I never us it!).
N
Hi Michael,
No, I never use attach(), exactly for the reasons you state. To do
due diligence I did a search of code for the function and it didn't
come up (I would have been shocked because I never us it!).
Now that real data is up, does your suggestion still apply? I am
reading it now.
Thanks
Off the wall / wild guess, do you use attach() frequently? Not
entirely sure how it would come up, but it tends to make weird errors
like this occur.
M
On Fri, Aug 24, 2012 at 4:36 PM, Jennifer Sabatier
wrote:
> Hi Rui,
>
> Thanks so much for responding but I think with my HTML problem the vn
>
Hi Rui,
Thanks so much for responding but I think with my HTML problem the vn
data you made must not be the same. I tried running your code on the
data (I uploaded a copy) and I got the same thing I had before.
Jen
On Fri, Aug 24, 2012 at 5:28 PM, Rui Barradas wrote:
>
> 165114 1 0 0 0 0 417
BTW - no one else who has replied to this topic was snobby or
unfriendly and I thank you very much for trying to help me.
It's just Bert is not the first to respond to my request for help as
such. As someone looking forward to becoming an advanced R programmer
in my statistical work it is discour
Hello,
Michael's standard guess, FAQ 7.31, was also mine, but is wrong. The
error is in Jennifer's flag column, not the result of her ifelse. (!)
x <- scan(what="character", text="
PM.EXP PM.DIST.TOT PM.DIST_flag 0 0 0 0 0 0 0 0 0 177502 1 0 31403 1
0 0 0 0 1100549 1 0 38762 1 0 0 0
Hi Peter,
I'm really sorry, I thought I was in plain text. I don't use any
formatting in my emails and in Gmail the HTML looks the same as plain
text.
Anyway, I've attached the data (I didn't think we could do that but I
am frequently wrong).
I say many cases because this is just a subset of >3
Bert,
I will thank you not to condescend to me, as I am too damn old (40) to
be treated that way. You didn't even offer a solution to my problem.
You only came to chastise me with regards to your assumptions about
me, which is very annoying.
While I am at the beginner level of R, I am not an idi
On 2012-08-24 13:22, Jennifer Sabatier wrote:
Hi R-Helpers,
I don't think I need to post a dataset for this question but if I do, I
can. Anyway, I am having a lot of trouble with the ifelse command.
Here is my code:
vn$PM.DIST_flag <- ifelse( (vn$PM.EXP > 0.0) & (vn$PM.DIST.TOT != 1.0), 1,
0
... and if Michael is correct, there is a lesson here: Think of how
much time and aggravation you would have saved yourself if you had
FIRST made an effort to read the docs. The FAQ's are there for a
reason. As is An Introduction to R, which also should be read before
posting on this list.
If Mich
Oops, sorry, I thought I was in plain text. I can't tell the
difference because I use so little formatting in my emails.
Try this (a truncated version since I have to hand space everything):
PM.EXP PM.DIST.TOT PM.DIST_flag
0 00
6417 1
On Fri, Aug 24, 2012 at 3:22 PM, Jennifer Sabatier
wrote:
> Hi R-Helpers,
>
> I don't think I need to post a dataset for this question but if I do, I
> can. Anyway, I am having a lot of trouble with the ifelse command.
>
You probably should have: dput() makes it super easy as well.
> Here is my
Hi R-Helpers,
I don't think I need to post a dataset for this question but if I do, I
can. Anyway, I am having a lot of trouble with the ifelse command.
Here is my code:
vn$PM.DIST_flag <- ifelse( (vn$PM.EXP > 0.0) & (vn$PM.DIST.TOT != 1.0), 1,
0 )
And here is my output that doesn't make ANY
You failed to tell us any of: the actual code you used, the data you were
working with, or the actual error you got. Your "pseudo-code" looks okay to me,
so I suggest you read the Posting Guide and try again. (Anything other than a
complete reproducible example of your problem is unlikely to eli
At 07:41 PM 7/10/2012, you wrote:
Seems to work for me:
> x <- data.frame(old1 = sample(c(1,2,8), 10, TRUE), old2 = 1:10)
> x$new <- ifelse(x$old1 == 8, 1, x$old2)
> x
Thanks Jim and Dan. The problem ended up with a missing values issue
that threw me off. When your post confirmed that my co
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jeff
> Sent: Tuesday, July 10, 2012 5:26 PM
> To: r-help
> Subject: [R] ifelse help
>
>
> I'm sure this is easy, but I'm new to R and can&
Seems to work for me:
> x <- data.frame(old1 = sample(c(1,2,8), 10, TRUE), old2 = 1:10)
> x$new <- ifelse(x$old1 == 8, 1, x$old2)
> x
old1 old2 new
1 11 1
2 22 2
3 23 3
4 84 1
5 15 5
6 86 1
7 87 1
8 28 8
9 2
I'm sure this is easy, but I'm new to R and can't find any example of
the following.
Here's what I'm trying to do in pseudo-code.
data$newvar <- ifelse(data$oldvar1 == 8, 1,data$oldvar2)
In other words, if the existing variable equals 8, then the new
variable should equal 1, otherwise the n
A generalized procedure for real life datasets might be
out<-rep(matrix(arg),ncol(ma))-ma
out[which(arg==0),]<--out[which(arg==0),]
out
Ozgur
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