Thaks so much! And how would you incorporate lapply() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at 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 <michaeleartz at 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$response[i] <- ifelse(colordata[i,"color"] == "blue", 1, 0) >> } >> >> which works, but I don't want to use the for loop I want to "vecotrize" >> this. How would this be implemented? >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide >> http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. >> > >[[alternative HTML version deleted]]
Hi> -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Michael > Artz > Sent: Thursday, April 7, 2016 1:57 PM > To: David Barron <dnbarron at gmail.com> > Cc: r-help at r-project.org > Subject: Re: [R] simple question on data frames assignment > > Thaks so much! And how would you incorporate lapply() here?Why do you want to use lapply? What is a result you want to achieve? Actually color is factor and it has a numeric value "inside".> as.numeric(colordata$color)[1] 1 4 2 1 3 Cheers Petr> > On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at 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 <michaeleartz at 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 at 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 at 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 jak?koliv k n?mu p?ipojen? dokumenty jsou d?v?rn? a jsou ur?eny pouze jeho adres?t?m. Jestli?e jste obdr?el(a) tento e-mail omylem, informujte laskav? neprodlen? jeho odes?latele. Obsah tohoto emailu i s p??lohami a jeho kopie vyma?te ze sv?ho syst?mu. Nejste-li zam??len?m adres?tem tohoto emailu, nejste opr?vn?ni tento email jakkoliv u??vat, roz?i?ovat, kop?rovat ?i zve?ej?ovat. Odes?latel e-mailu neodpov?d? za eventu?ln? ?kodu zp?sobenou modifikacemi ?i zpo?d?n?m p?enosu e-mailu. V p??pad?, ?e je tento e-mail sou??st? obchodn?ho jedn?n?: - vyhrazuje si odes?latel pr?vo ukon?it kdykoliv jedn?n? o uzav?en? smlouvy, a to z jak?hokoliv d?vodu i bez uveden? d?vodu. - a obsahuje-li nab?dku, je adres?t opr?vn?n nab?dku bezodkladn? p?ijmout; Odes?latel tohoto e-mailu (nab?dky) vylu?uje p?ijet? nab?dky ze strany p??jemce s dodatkem ?i odchylkou. - trv? odes?latel na tom, ?e p??slu?n? smlouva je uzav?ena teprve v?slovn?m dosa?en?m shody na v?ech jej?ch n?le?itostech. - odes?latel tohoto emailu informuje, ?e nen? opr?vn?n uzav?rat za spole?nost ??dn? smlouvy s v?jimkou p??pad?, kdy k tomu byl p?semn? zmocn?n nebo p?semn? pov??en a takov? pov??en? nebo pln? moc byly adres?tovi tohoto emailu p??padn? osob?, kterou adres?t zastupuje, p?edlo?eny nebo jejich existence je adres?tovi ?i osob? j?m zastoupen? zn?m?. 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. 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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 <michaeleartz at gmail.com>
wrote:>Thaks so much! And how would you incorporate lapply() here?
>
>On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com>
>wrote:
>
>> ifelse is vectorised, so just use that without the loop.
>>
>> colordata$response <- ifelse(colordata$color == 'blue', 1,
0)
>>
>> David
>>
>> On 7 April 2016 at 12:41, Michael Artz <michaeleartz at
gmail.com>
>wrote:
>>
>>> Hi I'm not sure how to ask this, but its a very easy question
to
>answer
>>> for
>>> an R person.
>>>
>>> What is an easy way to check for a column value and then assigne a
>new
>>> column a value based on that old column value?
>>>
>>> For example, Im doing
>>> colordata <- data.frame(id = c(1,2,3,4,5), color =
c("blue", "red",
>>> "green", "blue", "orange"))
>>> for (i in 1:nrow(colordata)){
>>> colordata$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 at 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 at 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]]
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$color1)' is not a function, character or symbol'
then how could you then use this fuction in lapply if not for the anonymous
function?
On Thu, Apr 7, 2016 at 8:17 AM, Jeff Newmiller <jdnewmil at
dcn.davis.ca.us>
wrote:
> Lapply is not a vectorized function. It is compact to read, but it would
> not be worth using for this calculation.
>
> However, if your data frame had multiple color columns in your data frame
> that you wanted to make responses for then you might want to use lapply as
> a more compact version of a for loop to repeat this operation.
>
> colordata2 <- data.frame(id = c(1,2,3,4,5), color1 = c("blue",
"red",
> "green", "blue", "orange"), color2 =
c("orange", "green",
> "blue", "red", "red"))
> responses <- lapply( colordata2[ -1 ], function(col) { ifelse(col =>
'blue', 1, 0) } )
> names(responses) <- names( colordata2 )[-1]
>
> where each of the columns other than the first is handed in turn to the
> anonymous function that does the response calculation. The result is a data
> frame (list of columns) with no column names, so I give the new columns
> names based on the old column names. You could choose different names, e.g.
>
> names(responses) <- paste0( "response", 1:2 )
>
> but you have to be careful to fix that code whenever you change the
> colordata2 data frame to have more columns.
> --
> Sent from my phone. Please excuse my brevity.
>
> On April 7, 2016 4:57:04 AM PDT, Michael Artz <michaeleartz at
gmail.com>
> wrote:
>>
>> Thaks so much! And how would you incorporate lapply() here?
>>
>> On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at
gmail.com> wrote:
>>
>> ifelse is vectorised, so just use that without the loop.
>>>
>>> colordata$response <- ifelse(colordata$color == 'blue',
1, 0)
>>>
>>> David
>>>
>>> On 7 April 2016 at 12:41, Michael Artz <michaeleartz at
gmail.com> wrote:
>>>
>>> Hi I'm not sure how to ask this, but its a very easy question
to answer
>>>> for
>>>> an R person.
>>>>
>>>> What is an easy way to check for a column value and then
assigne a new
>>>> column a value based on that old column value?
>>>>
>>>> For example, Im doing
>>>> colordata <- data.frame(id = c(1,2,3,4,5),
>>>> color = c("blue", "red",
>>>> "green", "blue", "orange"))
>>>> for (i in 1:nrow(colordata)){
>>>> colordata$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 at 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 at 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]]
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$result*1
[1] 1 0 0 1 0
But why do you want to use lapply or sapply is a mystery to me.
Do you have several columns and you want to use same function to those columns?
This is probably the only case I can think of using lapply with such function.
Or you want to change each colour name to some corresponding number? If column
colour is a factor it already consists from unique numbers (as I mentioned in
previous response).
Maybe you shall use dput(yourdata) output together with desired result to help
us better understand your task.
Cheers
Petr
From: Michael Artz [mailto:michaeleartz at gmail.com]
Sent: Thursday, April 7, 2016 4:17 PM
To: PIKAL Petr <petr.pikal at precheza.cz>
Subject: Re: [R] simple question on data frames assignment
what about sapply? I guess I am not sure how to iterate with sapply() and a
function like the following
>>The below function does not work, I want to have a function that I can
use for sapply() later
f <- function(x) {
response <- ifelse(x[,"Churn"] == "blue", 1, 0)
}
sapply(colordata$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.pikal at
precheza.cz<mailto:petr.pikal at precheza.cz>> wrote:
Hi
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at
r-project.org<mailto:r-help-bounces at r-project.org>] On Behalf Of
Michael
> Artz
> Sent: Thursday, April 7, 2016 1:57 PM
> To: David Barron <dnbarron at gmail.com<mailto:dnbarron at
gmail.com>>
> Cc: r-help at r-project.org<mailto:r-help at 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$color)
[1] 1 4 2 1 3
Cheers
Petr
>
> On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at
gmail.com<mailto:dnbarron at gmail.com>> wrote:
>
> > ifelse is vectorised, so just use that without the loop.
> >
> > colordata$response <- ifelse(colordata$color == 'blue', 1,
0)
> >
> > David
> >
> > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at
gmail.com<mailto:michaeleartz at gmail.com>> wrote:
> >
> >> Hi I'm not sure how to ask this, but its a very easy question
to
> >> answer for an R person.
> >>
> >> What is an easy way to check for a column value and then assigne a
> >> new column a value based on that old column value?
> >>
> >> For example, Im doing
> >> colordata <- data.frame(id = c(1,2,3,4,5), color =
c("blue", "red",
> >> "green", "blue", "orange")) for (i
in 1:nrow(colordata)){
> >> colordata$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 at r-project.org<mailto:R-help at 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 at r-project.org<mailto:R-help at 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.
________________________________
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[[alternative HTML version deleted]]