converting columns of data frame to factor

[nirgrahamuk]

its not clear what you are trying to say.
maybe you should restate your issue in full

[str_guru]

rather than input parameter for columns and labels like
factor_list <- data.frame (colnames=c("col1","col2"....),col_labels =c("sales","OPS"....)

in this i was giving a list of var and their labels but i want to give parameters separately
like
var <- c("col1","col2","col3","col4")
labels <- c("Sales","OPS","Management","HR")

i mean the question is same as I asked but the parameter will be like separate var and label....

but i will apply the same way like below you helped me
(ldat <- list(slice(data,1:20),
slice(data,20:39)))

purrr::map(ldat,
~mutate(.,
!!!parse_exprs(list_of_mutators)))

i am also updating in question for more understanding

[nirgrahamuk]

I provided that in my answer.
You can produce the table of parameters from the two vectors provided.

[str_guru]

actually that is based on faclist (a list of col and labels but if I want to give separate vector of col and labels)

[nirgrahamuk]

I showed how to make faclist from two vectors.
Read it again

converting columns of data frame to factor

#how to make this faclist faclist <- list( col1 = "Sales", col2 = "OPS", col3 = "Management", col4 = "HR" ) #from var <- c("col1","col2","col3","col4") labels <- c("Sales","OPS","Management","HR") #do faclist2 <- setNames(purrr::map(labels,identity), var) #check all.equal(faclist,faclist2)

[str_guru]

sorry if i am unable to make you understand....
Thanks, but my requirement is like below, I have N list of data frames like below

df1<- data
  df1$cc1<-1
  df2<- subset(df, col5 == 'Local')
  df$cc2<-ifelse(df$col5 == 'Local',1,NA)
  lst<-list(df$cc1, df$cc2)
  ldat<-list("ALL" = df, "Local" =df2)

after defining list of dataframes , it should work like ....
 
colnames=c("col1","col2"...."col4")
col_labels =c("sales","OPS"...."HR")
# so here I will be just needed to update the list of columns and their labels

conv_frac <- function(dataset,colnames,collabels){
for(i in 1:ldat)
lapply(factor,ldat(i))  # may be lapply or any thing else

}

# then  will apply function like this

conv_frac(dataset = ldat, col =colnames  , labels = col_labels)

so the input parameters for function will be ldat, colnames, col_labels

[nirgrahamuk]

I answered this already. you chose my answer as a solution.
you wanted to change a detail: - rather than desrcribe the relationship with a table , to do so with two vectors,.. i showed how to construct the table from the two vectors, thereby solving that requirement.

I'll let whoever else wants to continue this thread to continue without me

[str_guru]

should i create a new topic for that....??

[jmcvw]

You could very easily adapt this piece of code to do what you want.

conv_frac <- function(dataset,colnames,collabels){
for(i in 1:ldat)
lapply(factor,ldat(i))  # may be lapply or any thing else
}

Might I suggest another look at my answer. It is very similar to this code you have written. You are very nearly there with this. If you compare your attempt with mine, I think you will gain some understanding.

[str_guru]

I have tried to create a full function like below but doesn't work , please help if you can

###############

"sub_function"

fac_conv <- function(x) {
paste0(
"factor(", col_names[],
",labels=c('",
paste0(labels[],
collapse = "','"
), "'))"
)
}

fac_conv_lapply <- function(data,col_names,labels) {
list_of_fac <- purrr::map_chr(seq_len(length(col_names)),
fac_conv)
ldat <- map(data, ~ {
.x[col_names] <- map2(.x %>%
dplyr::select(col_names),
labels, ~ factor(.x, labels= .y))
.x} )

}

col_name <- c("col1","col2","col3","col4")
label <- c("Sales","Ops","admin","HR")

fac_conv_lapply(data = ldat,col_names = col_name, labels = label )