Displaying 4 results from an estimated 4 matches for "colordata2".
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colordata
2016 Apr 07
2
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$color1)' is not a function, character or...
2016 Apr 07
0
simple question on data frames assignment
lapply(colordata2[ -1 ], f )
When you put the parentheses on, you are calling the function yourself before lapply gets a chance. The error pops up because you are giving a vector of numbers (the answer f gave you) to the second argument of lapply instead of a function.
--
Sent from my phone. Please excuse my brev...
2016 Apr 07
0
simple question on data frames assignment
...ction. It is compact to read, but it would not be worth using for this calculation.
However, if your data frame had multiple color columns in your data frame that you wanted to make responses for then you might want to use lapply as a more compact version of a for loop to repeat this operation.
colordata2 <- data.frame(id = c(1,2,3,4,5), color1 = c("blue", "red",
"green", "blue", "orange"), color2 = c("orange", "green",
"blue", "red", "red"))
responses <- lapply( colordata2[ -1 ], function(col)...
2016 Apr 07
4
simple question on data frames assignment
Thaks so much! And how would you incorporate lapply() here?
On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote:
> ifelse is vectorised, so just use that without the loop.
>
> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>
> David
>
> On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote:
>