3

I have a data frame like this, 

df
col1             col2
 A        'the value is zero'
 B        'this is a cat'
 C        'the value is one'
 D        'nothing is here'
 E        'the colour is blue'
 F        'this is dog'
 G        'empty sequence'
 H        'the colour is red'
 I        'the colour is green'         1

Now I want the similar kind of strings as flagged as 1 and others as zero, so the final data frame should look like, 

col1             col2                 col1
 A        'the value is zero'           1
 B        'this is a cat'               1
 C        'the value is one'            1
 D        'nothing is here'             0
 E        'the colour is blue'          1
 F        'this is dog'                 1
 G        'empty sequence'              0
 H        'the colour is red'           1
 I        'the colour is green'         1

The 0 and 1 can be obtained using SequenceMatcher(SequenceMatcher(None, s1, s2).ratio()) function and with some threshold value we can make it to zero or one.

But if I use for loops to find the similarity between each other then it will take longer time to execute. Looking for some pandas shortcuts/pythonic way to do this efficiently.

yatu
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Kallol
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1 Answers1

9

Similarly to is it possible to do fuzzy match merge with python pandas?, we can use difflib and check if we find more than 1 similar string (to exclude its own) by looking at the length of the list returned by difflib.get_close_matches:

import difflib

df['col1'] = [(len(difflib.get_close_matches(x, df['col2'], cutoff=0.7))>1)*1 
              for x in df['col2']]

print(df)

   col1                            col2
0     1             'the value is zero'
1     1                 'this is a cat'
2     1              'the value is one'
3     0               'nothing is here'
4     1            'the colour is blue'
5     1                   'this is dog'
6     0                'empty sequence'
7     1             'the colour is red'
8     1           'the colour is green'

Similarity matrix based on fuzzy matching

One could also be interested in obtaining a similarity matrix setting all values in a pivoted column to 1 if the strings are similar. For this we could proceed similarly as above, but keeping the entire list, exploding it and pivoting the resulting dataframe with pd.crosstab:

df['sim'] = [difflib.get_close_matches(x, df['col2'], cutoff=0.7)  for x in df['col2']]
sim_df = df.explode('sim')
pd.crosstab(sim_df.col2, sim_df.sim)

sim             empty sequence  nothing is here  the colour is blue... the value is zero  this is a cat  this is dog
col2
empty sequence      1                0                     0         ...        0                   0            0
nothing is here     0                1                     0         ...        0                   0            0
the colour is blue  0                0                     1         ...        0                   0            0
the colour is green 0                0                     1         ...        0                   0            0
the colour is red   0                0                     1         ...        0                   0            0
the value is one    0                0                     0         ...        1                   0            0
the value is zero   0                0                     0         ...        1                   0            0
this is a cat       0                0                     0         ...        0                   1            1
this is dog         0                0                     0         ...        0                   1            1
yatu
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