Correct typos inside a column using word distance

Question

if have a column inside a pandas df containing a bunch of names:

NAME
-------
robert
robert
robrt
marie
ann

I'd like to merge similar ones in order to correct/uniform typos, resulting in:

NAME
-------
robert
robert
robert
marie
ann

I would like to use Levenshtein distance in order to search for similar records. Also, solutions using other metrics are much appreciated.

Thanks a lot in advance

All examples on Stackoverflow seem to compare multiple columns, so I have not been able to find a nice solution to my problem.

Answer 1

One possible approach is the following:

import pandas as pd
from sklearn.cluster import AgglomerativeClustering
from Levenshtein import distance
import numpy as np

df = pd.DataFrame({'NAME': ['robert', 'robert', 'robrt', 'marie', 'ann']})

def merge_similar_names(df, column):
    unique_names = df[column].str.lower().str.strip().unique()
    distances = np.zeros((len(unique_names), len(unique_names)))
    for i in range(len(unique_names)):
        for j in range(i, len(unique_names)):
            d = distance(unique_names[i], unique_names[j])
            distances[i, j] = d
            distances[j, i] = d
    clusterer = AgglomerativeClustering(n_clusters=None, distance_threshold=2, linkage='complete', affinity='precomputed')
    clusters = clusterer.fit_predict(distances)
    name_clusters = pd.DataFrame({'NAME': unique_names, 'CLUSTER': clusters})
    df = pd.merge(df, name_clusters, on='NAME')
    most_common_names = df.groupby('CLUSTER')[column].apply(lambda x: x.value_counts().index[0]).reset_index()
    df = pd.merge(df, most_common_names, on='CLUSTER')
    df.rename(columns={column+'_y': column}, inplace=True)
    return df

df = merge_similar_names(df, 'NAME')

print(df)

which will give you

   NAME_x  CLUSTER    NAME
0  robert        0  robert
1  robert        0  robert
2   robrt        0  robert
3   marie        2   marie
4     ann        1     ann