Re: [R] simple question on data frames assignment
Hello,
You're right, sorry, I missed the parenthesis:
colordata$response <- (colordata$color == 'blue') + 0
Rui Barradas
Quoting Michael Artz :
> Fyi, This statement returned the following error
> 'Error in "Yes" + 0 : non-numeric argument to binary operator'
>
>
> On Thu, Apr 7, 2016 at 8:43 AM, wrote:
>>
>> Hello,
>>
>>Or even simpler, without ifelse,
>>
>>colordata$response <- colordata$color == 'blue' + 0
>>
>>Hope this helps,
>>
>>Rui Barradas
>>
>>
>> Citando David Barron :
>>>
>>> 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
>>> 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$response[i] <-
ifelse(colordata[i,"color"] == "blue", 1, 0)
}
which works, but I don't want to use the for loop I
want to "vecotrize"
this. How would this be implemented?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
reproducible code.
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>
>>>PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.htmland provide commented,
>>> minimal, self-contained, reproducible code.
>>
>>
>>
>>
>>
>
>
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple question on data frames assignment
Hi
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael
> Artz
> Sent: Friday, April 8, 2016 3:50 AM
> To: Hadley Wickham
> Cc: r-help@r-project.org
> Subject: Re: [R] simple question on data frames assignment
>
> Why am I better off with true and false?
I believe that because logical values can be transfered to numbers easily and
you can use them directly to subset your object without any issue with
numerical precision.
Cheers
Petr
>
> On Thu, Apr 7, 2016 at 8:41 AM, Hadley Wickham
> wrote:
>
> > == is also vectorised, and you're better off with TRUE and FALSE
> > rather than 1 and 0, so I'd recommend:
> >
> > colordata$response <- colordata$color == 'blue'
> >
> > Hadley
> >
> > On Thu, Apr 7, 2016 at 6:52 AM, David Barron
> wrote:
> > > ifelse is vectorised, so just use that without the loop.
> > >
> > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
> > >
> > > David
> > >
> > > On 7 April 2016 at 12:41, Michael Artz wrote:
> > >
> > >> Hi I'm not sure how to ask this, but its a very easy question to
> > >> answer
> > for
> > >> an R person.
> > >>
> > >> What is an easy way to check for a column value and then assigne a
> > >> new column a value based on that old column value?
> > >>
> > >> For example, Im doing
> > >> colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue",
> > >> "red", "green", "blue", "orange")) for (i in 1:nrow(colordata)){
> > >>colordata$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 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.
> > >>
> > >
> > > [[alternative HTML version deleted]]
> > >
> > > __
> > > R-help@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.
> >
> >
> >
> > --
> > http://hadley.nz
> >
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.
Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny
pouze jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně
jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze
svého systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email
jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či
zpožděním přenosu e-mailu.
V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a
to z jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout;
Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce
s dodatkem či odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným
dosažením shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost
žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně
pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně
osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi
č
Re: [R] simple question on data frames assignment
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 2red
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 2red 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$result*1
[1] 1 0 0 1 0
But why do you want to use lapply or sapply is a mystery to me.
Do you have several columns and you want to use same function to those columns?
This is probably the only case I can think of using lapply with such function.
Or you want to change each colour name to some corresponding number? If column
colour is a factor it already consists from unique numbers (as I mentioned in
previous response).
Maybe you shall use dput(yourdata) output together with desired result to help
us better understand your task.
Cheers
Petr
From: Michael Artz [mailto:michaelea...@gmail.com]
Sent: Thursday, April 7, 2016 4:17 PM
To: PIKAL Petr
Subject: Re: [R] simple question on data frames assignment
what about sapply? I guess I am not sure how to iterate with sapply() and a
function like the following
>>The below function does not work, I want to have a function that I can use
>>for sapply() later
f <- function(x) {
response <- ifelse(x[,"Churn"] == "blue", 1, 0)
}
sapply(colordata$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
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 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$color)
[1] 1 4 2 1 3
Cheers
Petr
>
> On Thu, Apr 7, 2016 at 6:52 AM, David Barron
> mailto:dnbar...@gmail.com>> wrote:
>
> > ifelse is vectorised, so just use that without the loop.
> >
> > colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
> >
> > David
> >
> > On 7 April 2016 at 12:41, Michael Artz
> > mailto:michaelea...@gmail.com>> wrote:
> >
> >> Hi I'm not sure how to ask this, but its a very easy question to
> >> answer for an R person.
> >>
> >> What is an easy way to check for a column value and then assigne a
> >> new column a value based on that old column value?
> >>
> >> For example, Im doing
> >> colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red",
> >> "green", "blue", "orange")) for (i in 1:nrow(colordata)){
> >>colordata$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
> >> 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.
> >>
> >
> >
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org<mailto:R-help@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.
Tento e-mail a jakék
Re: [R] simple question on data frames assignment
Why am I better off with true and false?
On Thu, Apr 7, 2016 at 8:41 AM, Hadley Wickham wrote:
> == is also vectorised, and you're better off with TRUE and FALSE
> rather than 1 and 0, so I'd recommend:
>
> colordata$response <- colordata$color == 'blue'
>
> Hadley
>
> On Thu, Apr 7, 2016 at 6:52 AM, David Barron 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 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$response[i] <- ifelse(colordata[i,"color"] == "blue", 1, 0)
> >> }
> >>
> >> which works, but I don't want to use the for loop I want to "vecotrize"
> >> this. How would this be implemented?
> >>
> >> [[alternative HTML version deleted]]
> >>
> >> __
> >> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> >>
> >
> > [[alternative HTML version deleted]]
> >
> > __
> > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> http://hadley.nz
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple question on data frames assignment
Fyi, This statement returned the following error
'Error in "Yes" + 0 : non-numeric argument to binary operator'
On Thu, Apr 7, 2016 at 8:43 AM, wrote:
> Hello,
>
> Or even simpler, without ifelse,
>
> colordata$response <- colordata$color == 'blue' + 0
>
> Hope this helps,
>
> Rui Barradas
>
>
> Citando David Barron :
>
> ifelse is vectorised, so just use that without the loop.
>
> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>
> David
>
> On 7 April 2016 at 12:41, Michael Artz wrote:
>
> Hi I'm not sure how to ask this, but its a very easy question to answer for
> an R person.
>
> What is an easy way to check for a column value and then assigne a new
> column a value based on that old column value?
>
> For example, Im doing
> colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red",
> "green", "blue", "orange"))
> for (i in 1:nrow(colordata)){
>colordata$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 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.
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.htmland provide commented,
> minimal, self-contained, reproducible code.
>
>
>
>
[[alternative HTML version deleted]]
__
R-help@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.
Re: [R] simple question on data frames assignment
It all makes so much sense now
On Thu, Apr 7, 2016 at 10:04 AM, Jeff Newmiller
wrote:
> lapply(colordata2[ -1 ], f )
>
> When you put the parentheses on, you are calling the function yourself
> before lapply gets a chance. The error pops up because you are giving a
> vector of numbers (the answer f gave you) to the second argument of lapply
> instead of a function.
> --
> Sent from my phone. Please excuse my brevity.
>
> On April 7, 2016 7:31:18 AM PDT, Michael Artz
> wrote:
>>
>> If you are not using an anonymous function and say you had written the
>> function out
>>
>> The below gives me the error > 'f(colordata2$color1)' is not a function,
>> character or symbol' But then how is the anonymous function working?
>>
>>
>> f <- function(col){
>> ifelse(col == 'blue', 1, 0)
>> }
>> responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )
>>
>> 'f(colordata2$color1)' is not a function, character or symbol'
>>
>> then how could you then use this fuction in lapply if not for the
>> anonymous function?
>>
>> On Thu, Apr 7, 2016 at 8:17 AM, Jeff Newmiller
>> wrote:
>>
>>> Lapply is not a vectorized function. It is compact to read, but it would
>>> not be worth using for this calculation.
>>>
>>> However, if your data frame had multiple color columns in your data
>>> frame that you wanted to make responses for then you might want to use
>>> lapply as a more compact version of a for loop to repeat this operation.
>>>
>>> colordata2 <- data.frame(id = c(1,2,3,4,5), color1 = c("blue", "red",
>>> "green", "blue", "orange"), color2 = c("orange", "green",
>>> "blue", "red", "red"))
>>> responses <- lapply( colordata2[ -1 ], function(col) { ifelse(col ==
>>> 'blue', 1, 0) } )
>>> names(responses) <- names( colordata2 )[-1]
>>>
>>> where each of the columns other than the first is handed in turn to the
>>> anonymous function that does the response calculation. The result is a data
>>> frame (list of columns) with no column names, so I give the new columns
>>> names based on the old column names. You could choose different names, e.g.
>>>
>>> names(responses) <- paste0( "response", 1:2 )
>>>
>>> but you have to be careful to fix that code whenever you change the
>>> colordata2 data frame to have more columns.
>>> --
>>> Sent from my phone. Please excuse my brevity.
>>>
>>> On April 7, 2016 4:57:04 AM PDT, Michael Artz
>>> wrote:
Thaks so much! And how would you incorporate lapply() here?
On Thu, Apr 7, 2016 at 6:52 AM, David Barron wrote:
ifelse is vectorised, so just use that without the loop.
>
> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>
> David
>
> On 7 April 2016 at 12:41, Michael Artz wrote:
>
> Hi I'm not sure how to ask this, but its a very easy question to answer
>> for
>> an R person.
>>
>> What is an easy way to check for a column value and
>> then assigne a new
>> column a value based on that old column value?
>>
>> For example, Im doing
>> colordata <- data.frame(id = c(1,2,3,4,5),
>> color = c("blue", "red",
>> "green", "blue", "orange"))
>> for (i in 1:nrow(colordata)){
>> colordata$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 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.
>
>
>
[[alternative HTML version deleted]]
--
R-help@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.
>>
[[alternative HTML version deleted]]
__
R-help@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.
Re: [R] simple question on data frames assignment
lapply(colordata2[ -1 ], f )
When you put the parentheses on, you are calling the function yourself before
lapply gets a chance. The error pops up because you are giving a vector of
numbers (the answer f gave you) to the second argument of lapply instead of a
function.
--
Sent from my phone. Please excuse my brevity.
On April 7, 2016 7:31:18 AM PDT, Michael Artz wrote:
>If you are not using an anonymous function and say you had written the
>function out
>
>The below gives me the error > 'f(colordata2$color1)' is not a
>function,
>character or symbol' But then how is the anonymous function working?
>
>
>f <- function(col){
> ifelse(col == 'blue', 1, 0)
>}
>responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )
>
>'f(colordata2$color1)' is not a function, character or symbol'
>
>then how could you then use this fuction in lapply if not for the
>anonymous
>function?
>
>On Thu, Apr 7, 2016 at 8:17 AM, Jeff Newmiller
>
>wrote:
>
>> Lapply is not a vectorized function. It is compact to read, but it
>would
>> not be worth using for this calculation.
>>
>> However, if your data frame had multiple color columns in your data
>frame
>> that you wanted to make responses for then you might want to use
>lapply as
>> a more compact version of a for loop to repeat this operation.
>>
>> colordata2 <- data.frame(id = c(1,2,3,4,5), color1 = c("blue", "red",
>> "green", "blue", "orange"), color2 = c("orange", "green",
>> "blue", "red", "red"))
>> responses <- lapply( colordata2[ -1 ], function(col) { ifelse(col ==
>> 'blue', 1, 0) } )
>> names(responses) <- names( colordata2 )[-1]
>>
>> where each of the columns other than the first is handed in turn to
>the
>> anonymous function that does the response calculation. The result is
>a data
>> frame (list of columns) with no column names, so I give the new
>columns
>> names based on the old column names. You could choose different
>names, e.g.
>>
>> names(responses) <- paste0( "response", 1:2 )
>>
>> but you have to be careful to fix that code whenever you change the
>> colordata2 data frame to have more columns.
>> --
>> Sent from my phone. Please excuse my brevity.
>>
>> On April 7, 2016 4:57:04 AM PDT, Michael Artz
>
>> wrote:
>>>
>>> Thaks so much! And how would you incorporate lapply() here?
>>>
>>> On Thu, Apr 7, 2016 at 6:52 AM, David Barron
>wrote:
>>>
>>> ifelse is vectorised, so just use that without the loop.
colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
David
On 7 April 2016 at 12:41, Michael Artz
>wrote:
Hi I'm not sure how to ask this, but its a very easy question to
>answer
> for
> an R person.
>
> What is an easy way to check for a column value and then assigne
>a new
> column a value based on that old column value?
>
> For example, Im doing
> colordata <- data.frame(id = c(1,2,3,4,5),
> color = c("blue", "red",
> "green", "blue", "orange"))
> for (i in 1:nrow(colordata)){
> colordata$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 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.
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> --
>>>
>>> R-help@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.
>>>
>>>
[[alternative HTML version deleted]]
__
R-help@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.
Re: [R] simple question on data frames assignment
If you are not using an anonymous function and say you had written the
function out
The below gives me the error > 'f(colordata2$color1)' is not a function,
character or symbol' But then how is the anonymous function working?
f <- function(col){
ifelse(col == 'blue', 1, 0)
}
responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )
'f(colordata2$color1)' is not a function, character or symbol'
then how could you then use this fuction in lapply if not for the anonymous
function?
On Thu, Apr 7, 2016 at 8:17 AM, Jeff Newmiller
wrote:
> Lapply is not a vectorized function. It is compact to read, but it would
> not be worth using for this calculation.
>
> However, if your data frame had multiple color columns in your data frame
> that you wanted to make responses for then you might want to use lapply as
> a more compact version of a for loop to repeat this operation.
>
> colordata2 <- data.frame(id = c(1,2,3,4,5), color1 = c("blue", "red",
> "green", "blue", "orange"), color2 = c("orange", "green",
> "blue", "red", "red"))
> responses <- lapply( colordata2[ -1 ], function(col) { ifelse(col ==
> 'blue', 1, 0) } )
> names(responses) <- names( colordata2 )[-1]
>
> where each of the columns other than the first is handed in turn to the
> anonymous function that does the response calculation. The result is a data
> frame (list of columns) with no column names, so I give the new columns
> names based on the old column names. You could choose different names, e.g.
>
> names(responses) <- paste0( "response", 1:2 )
>
> but you have to be careful to fix that code whenever you change the
> colordata2 data frame to have more columns.
> --
> Sent from my phone. Please excuse my brevity.
>
> On April 7, 2016 4:57:04 AM PDT, Michael Artz
> wrote:
>>
>> Thaks so much! And how would you incorporate lapply() here?
>>
>> On Thu, Apr 7, 2016 at 6:52 AM, David Barron wrote:
>>
>> ifelse is vectorised, so just use that without the loop.
>>>
>>> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>>>
>>> David
>>>
>>> On 7 April 2016 at 12:41, Michael Artz wrote:
>>>
>>> Hi I'm not sure how to ask this, but its a very easy question to answer
for
an R person.
What is an easy way to check for a column value and then assigne a new
column a value based on that old column value?
For example, Im doing
colordata <- data.frame(id = c(1,2,3,4,5),
color = c("blue", "red",
"green", "blue", "orange"))
for (i in 1:nrow(colordata)){
colordata$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 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.
>>>
>>>
>>>
>>
>> [[alternative HTML version deleted]]
>>
>> --
>>
>> R-help@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.
>>
>>
[[alternative HTML version deleted]]
__
R-help@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.
Re: [R] simple question on data frames assignment
Hello,
Or even simpler, without ifelse,
colordata$response <- colordata$color == 'blue' + 0
Hope this helps,
Rui Barradas
Citando David Barron :
> 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 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$response[i] <- ifelse(colordata[i,"color"] == "blue", 1, 0)
>> }
>>
>> which works, but I don't want to use the for loop I want to "vecotrize"
>> this. How would this be implemented?
>>
>> [[alternative HTML version deleted]]
>>
>> __
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.htmland provide commented,
> minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple question on data frames assignment
== is also vectorised, and you're better off with TRUE and FALSE
rather than 1 and 0, so I'd recommend:
colordata$response <- colordata$color == 'blue'
Hadley
On Thu, Apr 7, 2016 at 6:52 AM, David Barron wrote:
> ifelse is vectorised, so just use that without the loop.
>
> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>
> David
>
> On 7 April 2016 at 12:41, Michael Artz wrote:
>
>> Hi I'm not sure how to ask this, but its a very easy question to answer for
>> an R person.
>>
>> What is an easy way to check for a column value and then assigne a new
>> column a value based on that old column value?
>>
>> For example, Im doing
>> colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red",
>> "green", "blue", "orange"))
>> for (i in 1:nrow(colordata)){
>>colordata$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 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.
>>
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.
--
http://hadley.nz
__
R-help@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.
Re: [R] simple question on data frames assignment
Lapply is not a vectorized function. It is compact to read, but it would not be
worth using for this calculation.
However, if your data frame had multiple color columns in your data frame that
you wanted to make responses for then you might want to use lapply as a more
compact version of a for loop to repeat this operation.
colordata2 <- data.frame(id = c(1,2,3,4,5), color1 = c("blue", "red",
"green", "blue", "orange"), color2 = c("orange", "green",
"blue", "red", "red"))
responses <- lapply( colordata2[ -1 ], function(col) { ifelse(col == 'blue', 1,
0) } )
names(responses) <- names( colordata2 )[-1]
where each of the columns other than the first is handed in turn to the
anonymous function that does the response calculation. The result is a data
frame (list of columns) with no column names, so I give the new columns names
based on the old column names. You could choose different names, e.g.
names(responses) <- paste0( "response", 1:2 )
but you have to be careful to fix that code whenever you change the colordata2
data frame to have more columns.
--
Sent from my phone. Please excuse my brevity.
On April 7, 2016 4:57:04 AM PDT, Michael Artz wrote:
>Thaks so much! And how would you incorporate lapply() here?
>
>On Thu, Apr 7, 2016 at 6:52 AM, David Barron
>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
>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$response[i] <- ifelse(colordata[i,"color"] == "blue",
>1, 0)
>>> }
>>>
>>> which works, but I don't want to use the for loop I want to
>"vecotrize"
>>> this. How would this be implemented?
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> __
>>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>
> [[alternative HTML version deleted]]
>
>__
>R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] simple question on data frames assignment
Hi
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Michael
> Artz
> Sent: Thursday, April 7, 2016 1:57 PM
> To: David Barron
> Cc: r-help@r-project.org
> Subject: Re: [R] simple question on data frames assignment
>
> Thaks so much! And how would you incorporate lapply() here?
Why do you want to use lapply? What is a result you want to achieve?
Actually color is factor and it has a numeric value "inside".
> as.numeric(colordata$color)
[1] 1 4 2 1 3
Cheers
Petr
>
> On Thu, Apr 7, 2016 at 6:52 AM, David Barron 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 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 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.
> >>
> >
> >
>
> [[alternative HTML version deleted]]
>
> __
> R-help@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.
Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny
pouze jeho adresátům.
Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně
jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze
svého systému.
Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email
jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat.
Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či
zpožděním přenosu e-mailu.
V případě, že je tento e-mail součástí obchodního jednání:
- vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a
to z jakéhokoliv důvodu i bez uvedení důvodu.
- a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout;
Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce
s dodatkem či odchylkou.
- trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným
dosažením shody na všech jejích náležitostech.
- odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost
žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně
pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně
osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi
či osobě jím zastoupené známá.
This e-mail and any documents attached to it may be confidential and are
intended only for its intended recipients.
If you received this e-mail by mistake, please immediately inform its sender.
Delete the contents of this e-mail with all attachments and its copies from
your system.
If you are not the intended recipient of this e-mail, you are not authorized to
use, disseminate, copy or disclose this e-mail in any manner.
The sender of this e-mail shall not be liable for any possible damage caused by
modifications of the e-mail or by delay with transfer of the email.
In case that this e-mail forms part of business dealings:
- the sender reserves the right to end negotiations about entering into a
contract in any time, for any reason, and without stating any reasoning.
- if the e-mail contains an offer, the recipient is entitled to immediately
accept such offer; The sender of this e-mail (offer) excludes any acceptance of
the offer on the part of the recipient conta
Re: [R] simple question on data frames assignment
Thaks so much! And how would you incorporate lapply() here?
On Thu, Apr 7, 2016 at 6:52 AM, David Barron wrote:
> ifelse is vectorised, so just use that without the loop.
>
> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>
> David
>
> On 7 April 2016 at 12:41, Michael Artz wrote:
>
>> Hi I'm not sure how to ask this, but its a very easy question to answer
>> for
>> an R person.
>>
>> What is an easy way to check for a column value and then assigne a new
>> column a value based on that old column value?
>>
>> For example, Im doing
>> colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red",
>> "green", "blue", "orange"))
>> for (i in 1:nrow(colordata)){
>>colordata$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 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.
>>
>
>
[[alternative HTML version deleted]]
__
R-help@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.
Re: [R] simple question on data frames assignment
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 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$response[i] <- ifelse(colordata[i,"color"] == "blue", 1, 0)
> }
>
> which works, but I don't want to use the for loop I want to "vecotrize"
> this. How would this be implemented?
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] simple question on data frames assignment
Hi I'm not sure how to ask this, but its a very easy question to answer for
an R person.
What is an easy way to check for a column value and then assigne a new
column a value based on that old column value?
For example, Im doing
colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue", "red",
"green", "blue", "orange"))
for (i in 1:nrow(colordata)){
colordata$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 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.
