How to derive the logit intercept value from %

I use this a lot to convert a logit coefficient (from a logistic regression model) to a probability.

p=exp(X) / (1+ exp(X))

This time I wanted to convert a probability (from the control group) into a logit.

I put X on the left side.

X= log( -p / (p-1)

This means that if I know the probability of a successful outcome occurance of the comparison group, I can get the logit intercept for the logistic regression model where only the treatment status was a predictor.  (I can use this to create a graph).

 

I tested this using a dataset.  Based on a data, I know the probability of a failure was 0.6214286.

data x;
per= 0.6214286;
intercept_derived=log((-1*per)/(per-1));
run;

I tested if I get the value for the intercept using the logistic regression.  I did.

proc logistic data=temp descending;
model Y1_to_Y2_persistence=treat_original;
run;

R question

Someone gave me am example of how to write a simple function. This was great.

addition = function(num1,num2){
answer = num1+num2
return(answer)
}
addition(10,9)
addition(5,4)

However, when I tried to use it, I realized I want to use a text instead of a number in this way.

my_matching = function(dataname){
m.out1<-matchit(data=dataname,control~gpa,method="nearest",ratio=1,
m.order="random", caliper=0.25)
}
my_matching(ABC_data1x)
my_matching(ABC_data2x)

Could you advise how I can run this without an error?

Also, how about a case like this? I used $ and . to indicate which part has to be replaced but this is not R way (This would be in SAS).

my_matching = function(data_N){
m.out1<-
matchit(data=ABC_data&data_n.x,control~gpa,method="nearest",ratio=1,
m.order="random", caliper=0.25)
}
my_matching(1)
my_matching(2)

Thank you!

R, basic function

addition = function(num1,num2){
answer = num1+num2
return(answer)
}
addition(10,9)
addition(5,4)

Thanks Isaac!

 

Use "" if dealing with a string.

addition = function(x1){
x1 = 2+3
return(x1)
}
addition(”kaz”)
addition(”john”)