Use string comparisons to split a column in R - regex

Question

To the best of my search this question hasn't been asked before.
I have a dataframe column called Product. This column has the company name as well as product model in just one column.
product.df <- data.frame("Product" = c("Company1 123M UG", "Company1 234M-I", "Company2 763-87-U","Company2 777-87", "Company3 Name1 87M", "Company3 Name1 O77M", "Company3 Name1 765-U MP"))
I want to split out the company names and product model number from this single column into two columns. I need a function that can find similar words between rows and classify them as Company names and the rest of the letters as product model number. No two rows as far as i can tell have same model numbers. So in the case above. I would get this answer
new.product.df <- data.frame("CompanyName" = c("Company1", "Company1", "Company2","Company2", "Company3 Name1", "Company3 Name1", "Company3 Name1"), "Model" = c("123M UG", "234M-I", "763-87-U", "777-87", "87M", "O77M", "765-U MP"))
I need a function that can compare two strings and return me similar continuous letters and dissimilar letters.

Answer 1

If you're guaranteed the first word is always a company name, then simply do a fixed split on the first space with max 2 output:
require(stringi)
stri_split_fixed(product.pd[,1], ' ', n=2)
or:
apply(product.df, 2, function(...) { stri_split_fixed(..., ' ', n=2) } )
[1] "Company1" "123M UG"
[1] "Company1" "234M-I"
[1] "Company2" "763-87-U"
[1] "Company2" "777-87"
[1] "Company3" "Name1 87M"
[1] "Company3" "Name1 O77M"
[1] "Company3" "Name1 765-U MP"

Answer 2

Try this
new.product.df <- data.frame(company=
unlist(lapply(strsplit(as.character(product.df$Product), split=" .[0-9]"), function(x) x[1])),
name =
unlist(lapply(strsplit(as.character(product.df$Product), split="[1|2] "), function(x) x[2]))
)

Answer 3

according to your data the separator between company and product is the first space character , so the first step we need to convert this first space character to something else , in this example to __ , later I'll tell you why we need to do this .
this is your actual data
Product
1 Company1 123M UG
2 Company1 234M-I
3 Company2 763-87-U
4 Company2 777-87
5 Company3 Name1 87M
6 Company3 Name1 O77M
7 Company3 Name1 765-U MP
this code to do this kind of conversion
product.df$Product <- sub(product.df$Product , pattern = " " , replacement = "__" ,
perl = T)
the data should be something like this
Product
1 Company1__123M UG
2 Company1__234M-I
3 Company2__763-87-U
4 Company2__777-87
5 Company3__Name1 87M
6 Company3__Name1 O77M
7 Company3__Name1 765-U MP
then use the tidyr library to separate this new data frame
library("tidyr")
new.product.df <- separate( product.df , Product , c("Company" , "Model") , sep = "__")
the reason behind converting space character to __ is that company name also may include space character like companies 123M UG & Name1 87M this will cause error later so the first step in this solution is to avoid this later when separating the column.
of course it will be better if we separated on the first occurrence of space character , but I don't know how because the global modifier is turned on by default for separator regex , so any suggestions are welcome