Finding exact duplicates texts is quite straightforward and fast in Python. This can be useful for removing duplicate entries in a dataset. I tried this on the Adzuna Job Salary Predictions Kaggle Competition job ad texts and found it worked well.

Naively finding exact duplicates by comparing every pair would be O(N^2), but if we sort the input, which is O(N log(N)), then duplicate items are adjacent. This scales really well to big datasets, and then the duplicate entries can be handled efficiently with itertools groupby to do something like uniq. In this case we can get the indices of the exact duplicates from a list (adding them with enumerate).

from itertools import groupby

def second(x):
return x[1]

def exact_duplicate_indices(items)
exact_duplicates = []
for _key, group in groupby(sorted(enumerate(items), key=second), key=second):
group = list(group)
if len(group) > 1:
exact_duplicates.append([item[0] for item in group])
return exact_duplicates

If memory was an issue because the items are really large we could even hash them to make the sorted possible. With the 400,000 ad descriptions, on average 1600 characters long, this took 1.3s on my laptop.

I could then look at the size of the duplicate clusters and investigate them:

from collections import counter
Counter(len(cluster) for cluster in duplicates)

This gave a table of frequencies; most clusters only have one duplicate, but there are a few with many duplicates.

Cluster Size Frequency
2 5836
3 293
4 71
5 26
6 7
7 8
8 5
9 7
10 1
12 1
13 1
14 2
15 1
24 1

I then inspected the largest clusters. In my case I kept them in the same order as a dataframe they came from and could index into them with iloc.

# Get the 20 largest clusters
megaclusters = sorted(exact_duplicates, key=len, reverse=True)[:20]

# Look it up in the original dataframe
df.iloc[megaclusters[0]]

This let me discover that in most of these cases it's due to being posted in multiple locations. I verified this with a quick analysis that the size of the cluster was almost the number of unique locations.

[(len(megacluster), len(df.iloc[megacluster].LocationRaw.unique())) for megacluster in megaclusters]

This is useful for exact duplicat, but what about slight variations? In this case the largest two clusters were in fact the same ad posted to different sites with a slight variation in the footer text. I'll look at this in the next part of the series.