Hi

The question does not make much sense so as your code. Maybe you shall spend 
some time with R tutorials.


1.       lapply or sapply is basically hidden cycle

2.       function shall return something, yours does not

So if you want some binary outcome from a vector you can use e.g.

f <- function(vector, token) {
response <- vector==token
response
}

So with your colordata

> colordata
  id  color
1  1   blue
2  2    red
3  3  green
4  4   blue
5  5 orange

You can get this

> colordata$result <- f(colordata$color, "blue")
> colordata
  id  color result
1  1   blue   TRUE
2  2    red  FALSE
3  3  green  FALSE
4  4   blue   TRUE
5  5 orange  FALSE

or if you insist on numbers you can transfer it easily.

> colordata$result*1
[1] 1 0 0 1 0

But why do you want to use lapply or sapply is a mystery to me.

Do you have several columns and you want to use same function to those columns? 
This is probably the only case I can think of using lapply with such function.

Or you want to change each colour name to some corresponding number? If column 
colour is a factor it already consists from unique numbers (as I mentioned in 
previous response).

Maybe you shall use dput(yourdata) output together with desired result to help 
us better understand your task.

Cheers
Petr


From: Michael Artz [mailto:michaelea...@gmail.com]
Sent: Thursday, April 7, 2016 4:17 PM
To: PIKAL Petr <petr.pi...@precheza.cz>
Subject: Re: [R] simple question on data frames assignment

what about sapply?   I guess I am not sure how to iterate with sapply() and a 
function like the following

>>The below function does not work, I want to have a function that  I can use 
>>for sapply() later
f <- function(x) {
  response <- ifelse(x[,"Churn"] == "blue", 1, 0)
}

sapply(colordata$color, f(x))

does that question make sense?  I just want to have a function that I can pass 
to sapply()

On Thu, Apr 7, 2016 at 7:44 AM, PIKAL Petr 
<petr.pi...@precheza.cz<mailto:petr.pi...@precheza.cz>> wrote:
Hi

> -----Original Message-----
> From: R-help 
> [mailto:r-help-boun...@r-project.org<mailto:r-help-boun...@r-project.org>] On 
> Behalf Of Michael
> Artz
> Sent: Thursday, April 7, 2016 1:57 PM
> To: David Barron <dnbar...@gmail.com<mailto:dnbar...@gmail.com>>
> Cc: r-help@r-project.org<mailto:r-help@r-project.org>
> Subject: Re: [R] simple question on data frames assignment
>
> Thaks so much!  And how would you incorporate lapply() here?

Why do you want to use lapply? What is a result you want to achieve?

Actually color is factor and it has a numeric value "inside".

> as.numeric(colordata$color)
[1] 1 4 2 1 3

Cheers
Petr

>
> On Thu, Apr 7, 2016 at 6:52 AM, David Barron 
> <dnbar...@gmail.com<mailto:dnbar...@gmail.com>> wrote:
>
> > ifelse is vectorised, so just use that without the loop.
> >
> > colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
> >
> > David
> >
> > On 7 April 2016 at 12:41, Michael Artz 
> > <michaelea...@gmail.com<mailto:michaelea...@gmail.com>> wrote:
> >
> >> Hi I'm not sure how to ask this, but its a very easy question to
> >> answer for an R person.
> >>
> >> What is an easy way to check for a column value and then assigne a
> >> new column a value based on that old column value?
> >>
> >> For example, Im doing
> >>  colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red",
> >> "green", "blue", "orange"))  for (i in 1:nrow(colordata)){
> >>    colordata$response[i] <- ifelse(colordata[i,"color"] == "blue", 1,
> >> 0)  }
> >>
> >> which works,  but I don't want to use the for loop I want to "vecotrize"
> >> this.  How would this be implemented?
> >>
> >>         [[alternative HTML version deleted]]
> >>
> >> ______________________________________________
> >> R-help@r-project.org<mailto:R-help@r-project.org> mailing list -- To 
> >> UNSUBSCRIBE and more, see
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> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> >
>
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>
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- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; 
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