如何基于R中另一列中的值替换列值？

Question

I'm trying to develop my R skills after a few years working with Pandas, and have a problem that's got me stumped.

I've split a column of data in a dataframe called df that broadly takes the following form:

"MN - place1 - time"
...
"ST - place2 - time"

I've used the separate function to split the data into three columns and aimed to isolate the middle column as the updated column:

cleaning_df <- separate(data = data, col = location, into = c("type", 'location', "time_data"), sep = "-")

It takes the form:

type    location    time_data
MN      place1     time
ST      place2      time

Unfortunately, there are typos which mean that hyphens aren't used to separate the first two fields.

For instance:

"STPlace2 - time"

Which separate can't handle - or I couldn't work out how.

Luckily, there aren't too many mistakes so I'd created a simple lookup table, location_lookup, which I was hoping to use as a dataframe to correct the data.

It's of the form:

Broken_type     Correct_middle
STPlace2        Place2
...             ...

With Pandas, I could write a straightforward, if un-pythonic and un-Pandas, apply function to go line-by-line through the newly-generated 'type' and 'place' columns.

It would then update values in 'place' where the value in 'type' matched in the look-up.

Is there a neater way of doing this? I've not been able to work through a solution using joins which would clearly be more efficient.

UPDATE:

The output from the separate function from my example, along with the error would be:

type     place     time
MN       place1    time
ST       place2    time
STPlace2 time      NA

I want to be able to create a function or join to use the look-up table

Broken_type     Correct_middle
STPlace2        Place2
...             ...

to identify that the third row in the above left column is wrong, and replace the value 'time' with 'Place2.

The eventual output column would then be:

place
place1
place2
Place2

Answer 1

We can pass regex on extract

extract(data, location, into = c("type", "location", "time_data"),
           "(.{2})[^[:alnum:]]*([[:alnum:]]+)\\s+-\\s+(.*)")
#   type location time_data
#1   MN   place1      time
#2   ST   place2      time
#3   ST   Place2      time

data

data <- structure(list(location = c("MN - place1 - time", "ST - place2 - time", 
"STPlace2 - time")), .Names = "location", class = "data.frame", row.names = c(NA, 
-3L))

Answer 2

It's not exactly elegant, but...

df$location <- sapply(1:length(df$type), function(x){
  if (df$type[x] %in% location_lookup$Broken_type){
    location_lookup$Correct_middle[match(df$type[x], location_lookup$Broken_type)]
  } else {

    df$place[x]
  }

})