[R] simple question on data frames assignment

[Jeff Newmiller]

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?
>>>
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>>>
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>>
>>
>
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>
>______________________________________________
>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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