How to quote this in R | 易学教程

问题

I wrote a function as:

gener1 <- function(du){
  nth <- paste0(paste0("dum", 1:du, " * ", "X", 1:du), collapse = " + ")
  return(nth)
  }

it returns a sequence as:

"dum1 * X1 + dum2 * X2 + dum3 * X3"

Now I want to use this sequence in a next function. Simply I can just copy that and paste:

S = quote(dum1 * X1 + dum2 * X2 + dum3 * X3)

Result:

dum1 * X1 + dum2 * X2 + dum3 * X3

It works. I wonder if it is possible to automatize that I do not need to use a "copy paste approach".

The final product I want to achieve is to use S in the following situation:

fo <- substitute(BH100 ~ S * ((1 - exp(-b2*TIME))/(1-exp(-b2*100)))^b3, list(S = S))

nls(fo, data = dane, start = list(b2 = 0.01, b3 = 1.1, X1 = 20, X2 = 20, X3 = 20))

Here is an example of the data structure I have

year    T   BH100   TIME    dum1    dum2    dum3    dum4
1987    25  12.6    25  1   0   0   0
1990    28  14.9    28  1   0   0   0
1994    32  18.8    32  1   0   0   0
1983    21  13.4    21  0   1   0   0
1986    24  16.1    24  0   1   0   0
1990    28  19.6    28  0   1   0   0
1998    36  26.7    36  0   1   0   0
2002    40  27.8    40  0   1   0   0
1994    32  17.2    32  0   0   1   0
1998    36  19.4    36  0   0   1   0
2002    40  23.5    40  0   0   1   0
2008    46  26.3    46  0   0   1   0
2013    51  28.7    51  0   0   1   0
1985    23  14.6    23  0   0   0   1
1989    27  18.5    27  0   0   0   1
1990    28  19.2    28  0   0   0   1

回答1:

I'm not sure I understand your question fully, but I think you're looking for the eval function. Your gener1 function right now is generating a string: "dum1 * X1 + dum2 * X2 + dum3 * X3". You can evaluate this string using eval with parse:

expression = gener1(3)
> expression
[1] "dum1 * X1 + dum2 * X2 + dum3 * X3"
evaluated_expression = eval(parse(text=expression))
> evaluated_expression
[1] 68

So, for your example, you can generate an expression string using your gener1 function, and then use the evaluated output in your fo variable:

S = eval(parse(text = gener1(3)))
fo <- substitute(BH100 ~ S * ((1 - exp(-b2*TIME))/(1-exp(-b2*100)))^b3, list(S = S))
nls(fo, data = dane, start = list(b2 = 0.01, b3 = 1.1, X1 = 20, X2 = 20, X3 = 20))

来源：https://stackoverflow.com/questions/30728437/how-to-quote-this-in-r

标签

quote