Pandas - replace multiple characters from rows in a datframe

Question

I have addresses stored in "address" column in a store dataframe, I would like to create a new column with the following corrections on existing addresses:

{"ST": "STREET",
  "RD": "ROAD",
  "AVE": "AVENUE",
  "N": "NORTH",
  "W": "WEST",
  "S": "SOUTH",
  "E": "EAST",
  "STE": "SUITE",
  "HWY": "HIGHWAY",
  "DR": "DRIVE",
  "NW": "NORTH WEST",
  "NE": "NORTH EAST",
  "SW": "SOUTH WEST",
  "SE": "SOUTH EAST",
  "LN": "LANE",
  "WAY": "WAY"}

How should I move forward this?

Expected output:

101 ST LN -> 101 STREET LANE

Here is the R code to the same:

terms <- c("W","WEST","E","EAST","N","NORTH","S","SOUTH")

terms <- split(terms,rep(1:2,times = length(terms) / 2))
terms[[1]] <- paste0("\\b",terms[[1]],"(\\.|\\b|\\,)")
terms[[1]]

stri_replace_all_regex(data$address,pattern = terms[[1]], replacement = terms[[2]],vectorize_all = FALSE)

Answer 1

Late to the party, I was using the same mapping dictionary and used the following function. I do not like regex since it is difficult to make business folks understand.

add_abv is the dictionary you mentioned above

def add_abv_rep(string):
str_ex = []
for i in string_ex.split(' '):
    if i in add_abv:
        str_ex.append(add_abv[i])
    else:
        str_ex.append(i)
return ' '.join(y.upper() for y in str_ex)

Finally, apply this function to the series.

df['Address'] = df['Address'].apply(lambda x: add_abv_rep(x))