How can I compare different rows of one column with Levenshtein distance metric in pandas?

Question

I have a table like this:

id name
1 gfh
2 bob
3 boby
4 hgf

etc.

I am wondering how can I use Levenshtein metric to compare different rows of my 'name' column?

I already know that I can use this to compare columns:

L.distance('Hello, Word!', 'Hallo, World!')

But how about rows?

Answer 1

Here is a way to do it with pandas and numpy:

from numpy import triu, ones
t = """id name
1 gfh
2 bob
3 boby
4 hgf"""

df = pd.read_csv(pd.core.common.StringIO(t), sep='\s{1,}').set_index('id')
print df

        name
id      
1    gfh
2    bob
3   boby
4    hgf

Create dataframe with list of strings to mesure distance:

dfs = pd.DataFrame([df.name.tolist()] * df.shape[0], index=df.index, columns=df.index)
dfs = dfs.applymap(lambda x: list([x]))
print dfs

    id      1      2       3      4
id                             
1   [gfh]  [bob]  [boby]  [hgf]
2   [gfh]  [bob]  [boby]  [hgf]
3   [gfh]  [bob]  [boby]  [hgf]
4   [gfh]  [bob]  [boby]  [hgf]

Mix lists to form a matrix with all variations and make upper right corner as NaNs:

dfd = dfs + dfs.T
dfd = dfd.mask(triu(ones(dfd.shape)).astype(bool))
print dfd

id            1            2            3    4
id                                            
1           NaN          NaN          NaN  NaN
2    [gfh, bob]          NaN          NaN  NaN
3   [gfh, boby]  [bob, boby]          NaN  NaN
4    [gfh, hgf]   [bob, hgf]  [boby, hgf]  NaN

Measure L.distance:

dfd.applymap(lambda x: L.distance(x[0], x[1]))