r - Error in if function -


i have run long script decide model should use forecast. after doing accuracy tests on in , out samples of data created large if function find model best results of either "arima", "arima.wgt", "addhw", "multhw", "addhwwgt" , "multhwwgt". during script have got forecasts each of these models , want use if function view them have written

if(maxmod<-"arima")   modelf<-arimaaltfa else     if(maxmod<-"arima.wgt")       modelf<-arimaaltfb else         if(maxmod<-"addhw")           modelf<-hwabfc else             if(maxmod<-"multhw")               modelf<-hwmbfd else                 if(maxmod<-"addhwwgt")                   modelf<-hwaaltfe else                     modelf<-hwmaltff

but keep getting error

error in if (maxmod <- "arima") modelf <- arimaaltfa else if (maxmod <- "arima.wgt") modelf <- arimaaltfb else if (maxmod <- "addhw") modelf <- hwabfc else if (maxmod <- "multhw") modelf <- hwmbfd else if (maxmod <- "addhwwgt") modelf <- hwaaltfe else modelf <- hwmaltff :    argument not interpretable logical

this has happened many different things have tried eg instead of modelf<-"" tried view("",title="") , modelf<-view("",title="") still saya isn't logical... there error in way have written or there problem?

extra detail , code available if needed

you need switch function.

modelf <- switch(   maxmod,   arima     = arimaaltfa,   arima.wgt = arimaaltfb,   addhw     = hwabfc,   multhw    = hwmbfd,   addhwwgt  = hwaaltfe,   hwmaltff )

your specific problem trying assign values maxmod instead of comparing equality. although switch statement preferable, try replacing

if(maxmod<-"arima")

with

if(maxmod == "arima")

maxmod == "arima" returns true or false (a logical value).

maxmod<-"arima" assigns value "arima" variable named maxmod (and invisibly returns string).


Comments

Post a Comment

Popular posts from this blog

Python Kivy ListView: How to delete selected ListItemButton? -

Image
i'm trying learn kivy building simple todo-list app suggested dusty phillips, author of book "creating apps in kivy". this code far: from kivy.app import app kivy.uix.boxlayout import boxlayout kivy.properties import objectproperty kivy.uix.listview import listitembutton class taskbutton(listitembutton): pass class todoroot(boxlayout): task_input = objectproperty() task_list = objectproperty() def add_task(self): self.task_list.adapter.data.extend([self.task_input.text]) self.task_list._trigger_reset_populate() def del_task(self): pass class todoapp(app): def build(self): return todoroot() if __name__ == '__main__': todoapp().run() and kv file: #: import main todo #: import listadapter kivy.adapters.listadapter.listadapter #: import listitembutton kivy.uix.listview.listitembutton todoroot: <todoroot>: orientation: "vertical" task_input: task_input_view t...
Read more

ruby - How do I merge two hashes into a hash of arrays? -

http://codepad.org/wmyzqlis i'd merge these 2 arrays: a = { :a => 'a', :b => 'b', :d => 'd' } b = { :a => '1', :b => 'b', :c => 'c' } into hash looks this: {:a => ["a", "1"], :b => ["b", "b"], :c => [nil, "c"], :d => ["d", nil] } this doesn't work: p a.merge(b) { |k, v1, v2| [v1, v2] } # {:c=>"c", :a=>["a", "1"], :b=>["b", "b"], :d=>"d"} it's because hash#merge call block duplicate keys. a = { :a => 'a', :b => 'b', :d => 'd' } b = { :a => '1', :b => 'b', :c => 'c' } hash[(a.keys|b.keys).map { |key| [key, [a.fetch(key, nil), b.fetch(key, nil)]] }] # => {:a=>["a", "1"], :b=>["b", "b"], :d=>["d", nil], :c=>...
Read more