search for: wsapply

...mtcars$cyl), summary(lm(mpg ~ wt, .))$r.squared) "Not a big saving in typing" you can say but multiplied by the number of [lsv]apply usage and a neater look, I think, the idea merits to be considered. To illustrate a possible implementation, I propose a wrapper example for sapply(): wsapply=function(l, fun, ...) { ??? s=substitute(fun) ??? if (is.name(s) || is.call(s) && s[[1]]==as.name("function")) { ??????? sapply(l, fun, ...) # legacy call ??? } else { ??????? sapply(l, function(d) eval(s, list(.=d)), ...) ??? } } Now, we can do: wsapply(split(mtcars, mtca...

...6, 2020 at 12:03 PM > > Subject: Re: [Rd] suggestion: "." in [lsv]apply() > > To: William Dunlap <wdunlap at tibco.com> > > Cc: r-devel <r-devel at r-project.org> > > > > > > Thanks Bill, > > > > Clearly, my first proposition for wsapply() is quick and dirty one. > > However, if "." becomes a reserved variable with this new syntax, > > wsapply() can be fixed (at least for your example and alike) as: > > > > wsapply=function(l, fun, ...) { > > .=substitute(fun) > > if (is.name(...

...39;s not in any way "neater", not only is it less readable, it's just > plain wrong. What if the expression returned a function? do you mean like in l=sapply(1:3, function(i) function(x) i+x) l[[1]](3) # 4 l[[2]](3) # 5 This is indeed a corner case but a pair of () or {} can keep wsapply() in course: l=wsapply(1:3, (function(x) .+x)) l[[1]](3) # 4 l[[2]](3) # 5 > How do you know that you don't want to apply the result of the call? A small example (if it is significantly different from the one above) would be very helpful for me to understand this point. > For the s...

...sa-toulouse.fr> > Date: Thu, Apr 16, 2020 at 12:03 PM > Subject: Re: [Rd] suggestion: "." in [lsv]apply() > To: William Dunlap <wdunlap at tibco.com> > Cc: r-devel <r-devel at r-project.org> > > > Thanks Bill, > > Clearly, my first proposition for wsapply() is quick and dirty one. > However, if "." becomes a reserved variable with this new syntax, > wsapply() can be fixed (at least for your example and alike) as: > > wsapply=function(l, fun, ...) { > .=substitute(fun) > if (is.name(.) || is.call(.) && .[...

...;, not only is it less readable, it's >>> just plain wrong. What if the expression returned a function? >> do you mean like in l=sapply(1:3, function(i) function(x) i+x) >> l[[1]](3) # 4 l[[2]](3) # 5 This is indeed a corner case but a pair >> of () or {} can keep wsapply() in course: l=wsapply(1:3, (function(x) >> .+x)) l[[1]](3) # 4 l[[2]](3) # 5 >>> How do you know that you don't want to apply the result of the call? >> A small example (if it is significantly different from the one above) >> would be very helpful for me to unders...

...it's >>>>> just plain wrong. What if the expression returned a function? >>>> do you mean like in l=sapply(1:3, function(i) function(x) i+x) >>>> l[[1]](3) # 4 l[[2]](3) # 5 This is indeed a corner case but a pair >>>> of () or {} can keep wsapply() in course: l=wsapply(1:3, >>>> (function(x) .+x)) l[[1]](3) # 4 l[[2]](3) # 5 >>>>> How do you know that you don't want to apply the result of the call? >>>> A small example (if it is significantly different from the one >>>> above) would...

...uot;, not only is it less readable, it's just plain wrong. What if the expression returned a function? > do you mean like in > l=sapply(1:3, function(i) function(x) i+x) > l[[1]](3) > # 4 > l[[2]](3) > # 5 > > This is indeed a corner case but a pair of () or {} can keep wsapply() in course: > l=wsapply(1:3, (function(x) .+x)) > > l[[1]](3) > > # 4 > > l[[2]](3) > > # 5 >> How do you know that you don't want to apply the result of the call? > A small example (if it is significantly different from the one above) would be very help...

Passing in a function passes not only an argument list but also an environment from which to get free variables. Since your function doesn't pay attention to the environment you get things like the following. > wsapply(list(1,2:3), paste(., ":", deparse(s))) [[1]] [1] "1 : paste(., \":\", deparse(s))" [[2]] [1] "2 : paste(., \":\", deparse(s))" "3 : paste(., \":\", deparse(s))" Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Apr 16, 20...

...t less readable, it's > >>> just plain wrong. What if the expression returned a function? > >> do you mean like in l=sapply(1:3, function(i) function(x) i+x) > >> l[[1]](3) # 4 l[[2]](3) # 5 This is indeed a corner case but a pair > >> of () or {} can keep wsapply() in course: l=wsapply(1:3, (function(x) > >> .+x)) l[[1]](3) # 4 l[[2]](3) # 5 > >>> How do you know that you don't want to apply the result of the call? > >> A small example (if it is significantly different from the one above) > >> would be very helpful...

...that uses a different syntax (and I'm sure someone has already done that in the package space), but what you propose is incompatible with *apply in R (and very much not R syntax). Cheers, Simon > To illustrate a possible implementation, I propose a wrapper example for sapply(): > > wsapply=function(l, fun, ...) { > s=substitute(fun) > if (is.name(s) || is.call(s) && s[[1]]==as.name("function")) { > sapply(l, fun, ...) # legacy call > } else { > sapply(l, function(d) eval(s, list(.=d)), ...) > } > } > > Now,...