R：grepl在字符串上选择第一个字符

Question

I apologize in advance, this might be a repeat question. However, I just spent the two last hours over stackoverflow, and can't seem to find a solution.

I want to use grepl to detect rows that begin with a digit, that's what I tried to use but It didn't give me the rigt answer:

   grep.numeric=as.data.frame(grepl("^[:digit:]",df_mod$name))

I guess that the problem is from the regular expression "^[:digit:]" , but I couldn't figure it out.

UPDATE

My dataframe looks like this, It's huge, but below is an example:

  ID       mark         name
   1       whatever     name product
   2       whatever     10 product
   3       whatever     250 product
   4       another_mark other product

I want to detect products which their names begin with a number.

UPDATE 2

applying grep.numeric=grepl("^[[:digit:]]",df_mod$name) on the example below give me the right answer which is:

    grep.numeric
   [1] FALSE  TRUE  TRUE FALSE

But, what drive me crazy is when I pply this fuction to my real dataframe:

   grep.numeric=grepl("^[[:digit:]]",df_mod[217,]$nom)

give me this result:

   grep.numeric
   [1] FALSE

But actually, what I have is this :

   df_mod[217,]$nom
   [1]  100 lipo 30 gélules

Please help me.

Answer 1

Apparently, some of your values have leading spaces, so you could either modify your regex to (or something similar)

grepl("^\\s*[[:digit:]]", df_mod$name)

Or use the built in trimws function

grepl("^[[:digit:]]", trimws(df_mod$name))