[R] simple question on data frames assignment

[Michael Artz]

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.
>>
>>

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