A have a list that has the following structure.
mylist=list(y~ A,
y ~ A+B,
y ~ A+B+C)
I want to replace (recode) the “y “ with a “z”. my goal is
mylist=list(z~ A,
z ~ A+B,
z ~ A+B+C)
Q: How to replace (recode) values in a list?
I have tried this:
for i in range(len(mylist)):
mylist[i] = mylist[i].replace('y','z')
is not working
The update function is useful for formulas.
Just include a . to indicate any formula side to retain. So, for your problem the following is a quick one-liner.
lapply(mylist, update, new = z~.)
update has an update.formula methodI would alternatively suggest to use R built in formulas manipulation functionality. This allows us to operate on different terms of a fromula separately without using regex
lapply(mylist, function(x) reformulate(as.character(terms(x))[3], "z"))
# [[1]]
# z ~ A
# <environment: 0x59c6040>
#
# [[2]]
# z ~ A + B
# <environment: 0x59c0308>
#
# [[3]]
# z ~ A + B + C
# <environment: 0x59bb7b8>
Since you have a list of formulas as a start, you can convert the formulas to characters, use gsub to do the replacement and convert it back to formula. Use env parameter to specify the environment of the formulas to make sure they are the same as original list:
lapply(mylist, function(f) formula(gsub("y", "z", format(f)), env = .GlobalEnv))
# [[1]]
# z ~ A
# [[2]]
# z ~ A + B
# [[3]]
# z ~ A + B + C
To take care of the concern of @David Arenburg, so that the replacement of y will always happen on the left side of the formula, we can use a more restricted regular expression:
lapply(mylist, function(f) formula(gsub("y(\\s)?(?=~)", "z", format(f), perl = T), env = .GlobalEnv))
# [[1]]
# z ~ A
# [[2]]
# z ~ A + B
# [[3]]
# z ~ A + B + C
y ~ city + country? Or how will you handle different names in the response variable? This regex isn't targeting just the response variable.
env = .GlobalEnv assign the environment of an object. Since the formula is generated in a function, by default, it will have an environment local to the function which is not the case in your original list. Assign the global environment to these formulas will make sure they are visible in the global R session.Just like expressions, formulas have replaceable parts. So you could use [[<- to replace parts of the formula. The y value is the second in the expression list, as ~ is a function and hence the first.
lapply(mylist, "[[<-", 2, as.name("z"))
# [[1]]
# z ~ A
#
# [[2]]
# z ~ A + B
#
# [[3]]
# z ~ A + B + C
lapply(mylist, function(x) reformulate(as.character(terms(x))[3], "z"))maybe in R