R help - Aug 2021 - Dynamic Creation and Use of Object Names

Hi,

I have a function PICTURE() and do something like

	COUNTRY = 'Namibia'
	NAKURVE = PICTURE(COUNTRY)
	NAKURVE
	ggsave(paste0(tolower(COUNTRY),".png"), width = 16, height = 9)

	COUNTRY = ('Germany')
	DEKURVE = PICTURE(COUNTRY)
	DEKURVE
	ggsave(paste0(tolower(COUNTRY),".png"), width = 16, height = 9)

	COUNTRY = ('Netherlands')
	NLKURVE = PICTURE(COUNTRY)
	NLKURVE
	ggsave(paste0(tolower(COUNTRY),".png"), width = 16, height = 9)

[...]

	COUNTRYGRID=grid.arrange(NAKURVE, DEKURVE, NLKURVE,
				 ncol=3)
	COUNTRYGRID
	ggsave(paste0("R7.incidence.", Sys.Date(), ".png"),
COUNTRYGRID)

I am not able to figure out (and/or find on Google) how to do this in a
loop (of sorts), ie create the variables dynamically and add them to to
the grid (dynamically, ie adding more countries)

Any ideas?

greetings, el
-- 
To email me replace 'nospam' with 'el'

On Mon, 23 Aug 2021 08:37:54 +0200
Dr Eberhard Lisse <nospam at lisse.NA> wrote:
> create the variables dynamically and add them to to
> the grid (dynamically, ie adding more countries)
In my opinion, creating variables in the global environment
programmatically may lead to code that is hard to understand and debug
[*]. A key-value data structure (a named list or a separate
environment) would avoid the potential problems from variable name
collision. How about the following:

1. Put the countries in a vector: c('Namibia', 'Germany', ...)

2. Use lapply() to get a list of objects returned from your PICTURE
   function

3. To save the pictures into individual files, loop over the list. You
   can use setNames on the step 1 or 2 to make it a named list and keep
   the country names together with their pictures:
   
   for (n in names(pictures)) {
     dev.new()
     print(pictures[[n]])
     ggsave(paste0(n, '.png'), ...)
     dev.off()
   }

   (You can also use the png() device and plot straight to the file,
   avoiding the need to draw the plot in the window for a fraction of a
   second and for ggsave().)

4. Use the grobs= argument of grid.arrange() to pass the list of
   objects to arrange instead of passing individual objects via ...

-- 
Best regards,
Ivan

[*] For example, there's this FAQ for a different language:
https://perldoc.perl.org/perlfaq7#How-can-I-use-a-variable-as-a-variable-name?