Faster way of replacing strings in large pandas dataframe with regex

Question

I want to replace all instances of a location with just the location keyword itself, eg "Foxclore Road" with "road", "Chesture Avenue" with "avenue". The file is several GBs, with many million rows. Current working code, three methods:

startTime = time.time()
mergedAllCrimes['crime_location_approx'] = mergedAllCrimes.crime_location_approx.str.replace(r'.*(?i)road$', 'road')
endTime = time.time()
print(endTime - startTime)

startTime = time.time()
mergedAllCrimes.crime_location_approx = mergedAllCrimes.crime_location_approx.apply(lambda x: 'road' if 'road' in str.lower(x) else x)
endTime = time.time()
print(endTime - startTime)

startTime = time.time()
allCrimes.loc[allCrimes['crime_location_approx'].str.contains('Road', case=False), 'crime_location_approx'] = 'road'
endTime = time.time()
print(endTime - startTime)

my times are:

14.287408590316772
1.9554557800292969
5.129802942276001

respectively

Problem is, the second two methods (while faster), replace "Broadway" with "road", hence the need for a regex to search at the end of a string.

Is there any way to make the regex conditional method much faster? If I have a large list of replacements, it could end up taking a long time.

Answer 1

There is not much to say about your code then, Regex is slow.

A non-regex solution could be to use Python's endswidth, this works the same as r"road$"

mergedAllCrimes.crime_location_approx = mergedAllCrimes.crime_location_approx.apply(lambda x: 'road' if x.lower().endswith('road') else x)

I'm assuming all the conditional words are at the end of the string