Text manipulation using lapply()

Question

Question

Are there other ways I may use lapply() to paste the column name for each element within each column in a data frame?

Problem

At the moment, I'm resorting to re-using lapply() for each column.

Reproducible Example

# create data
df <- data.frame(
  Chicago_Has = c("Lou Malnati's", "Wrigley Field", "CTA" )
  , Seattle_Has = c("Piroshky Piroshky", "Safeco Field", "KCMT" )
  , stringsAsFactors = FALSE
)

# paste column name into each element 
# within each column
df[ "Chicago_Has" ] <- lapply( X = df[ "Chicago_Has" ]
                               , FUN = function(i) paste( "Chicago_Has", i, sep = " " )
                               )
df[ "Seattle_Has" ] <- lapply( X = df[ "Seattle_Has" ]
                               , FUN = function(i) paste( "Seattle_Has", i, sep = " ")
)

# examine the desired result of the data frame
df

#                 Chicago_Has                   Seattle_Has
# 1 Chicago_Has Lou Malnati's Seattle_Has Piroshky Piroshky
# 2 Chicago_Has Wrigley Field      Seattle_Has Safeco Field
# 3           Chicago_Has CTA              Seattle_Has KCMT

Thoughts

At the moment, I think this method involves too much copying and pasting. If I store colnames( df ) as a character vector, I don't know how I would use that object - while only using lapply() once - to obtain my desired results. I think it involves using multiple counters in the function I'm supplying in FUN, but am unsure how to proceed.

Answer 1

df[]=Map(paste,names(df),df)
df
                Chicago_Has                   Seattle_Has
1 Chicago_Has Lou Malnati's Seattle_Has Piroshky Piroshky
2 Chicago_Has Wrigley Field      Seattle_Has Safeco Field
3           Chicago_Has CTA              Seattle_Has KCMT