Trying to return more than just the top result from sklearn NearestNeighbors

Question

I'm trying to compare a list of names (duplicated into a clean file and a messy file). I then compare the files against each other. My problem is that it returns only the top 1 result for each, which is itself (the identical record in each file). What I am trying to capture is the second result, which would be the closest match, not being itself.

names =  pd.read_csv('C:/Temp/messynames.txt', sep='\t')
org_names = names['VariationName'].unique()
vectorizer = TfidfVectorizer(min_df=1, analyzer=ngrams)
tf_idf_matrix = vectorizer.fit_transform(org_names)

clean_org_names = pd.read_csv('C:/Temp/cleannames.txt', sep='\t')
org_name_clean = clean_org_names['StandardName'].unique()
vectorizer = TfidfVectorizer(min_df=1, analyzer=ngrams, lowercase=False)
tfidf = vectorizer.fit_transform(org_name_clean)

nbrs = NearestNeighbors(n_neighbors=3, n_jobs=-1).fit(tfidf) 
unique_org = set(names['VariationName'].values)

def getNearestN(query):
  queryTFIDF_ = vectorizer.transform(query)
  distances, indices = nbrs.kneighbors(queryTFIDF_)
  return distances, indices
distances, indices = getNearestN(unique_org)

unique_org = list(unique_org) #need to convert back to a list

matches = []
for i,j in enumerate(indices):
  temp = [round(distances[i][0],2), clean_org_names.values[j][0][0],unique_org[i]]
  matches.append(temp)

matches = pd.DataFrame(matches, columns=['Match confidence (lower is better)','Matched name','Original name'])
matches.to_csv('C:/Temp/matchednames.txt', sep='\t', encoding='utf-8', index=False, quoting=3)

For a file with the following four names:

NOKIA
NOKIAA
NOKIA LMD
NOKIA LTD

The results looks like this:

Match confidence                         Matched name Original name
0                                 0.0    NOKIA LMD     NOKIA LMD
1                                 0.0       NOKIAA        NOKIAA
2                                 0.0        NOKIA         NOKIA
3                                 0.0    NOKIA LTD     NOKIA LTD

I'm trying to get to something more like:

Match confidence                         Matched name Original name
0                                 0.1    NOKIA LTD     NOKIA LMD
1                                 0.1        NOKIA        NOKIAA

Answer 1

My guess would be that in the following piece of code:

for i,j in enumerate(indices):
  temp = [round(distances[i][0],2), clean_org_names.values[j][0][0],unique_org[i]]
  matches.append(temp)

The variable (array element):

distances[i][0]

contains the top element match for the $i^{th}$ line. Replace the second array index so it becomes:

distances[i][1]

You may want to increase one of the two zero indices in the following line as well to have a distance like value:

clean_org_names.values[j][0][0]