CODEWITHSUNDEEP
pythonpandas
Charlie
Charlie

Reputation: 49

Replace value in a column and row with a given row index in dataframe

I have a dataframe of urls and verified urls and add a column with the levenstein ratio, which compares the two types of urls for each row.

Here is an example of my pandas dataframe:

                       url                  url_ok2
13          10hanover.org/                      NaN
15  111140.cevadosite.com/      aerorealestate.net/
42         18brownlow.com/      18brownlow.com:443/
57           1granary.com/    1granary.com/journal/
61             1rs.org.uk/                  1rs.io/
79   2020visionnetwork.eu/  network.crowdhelix.com/

Here is my script:

import Levenshtein as lev

to_test['lev_ratio'] = None
for i in range(to_test.shape[0]):
    to_test.iloc[i]['lev_ratio'] =  lev.ratio(str(to_test.iloc[i].url),str(to_test.iloc[i].url_ok2))

But the values are not replaced, see dataframe after running script: url url_ok2 lev_ratio 13 10hanover.org/ NaN None 15 111140.cevadosite.com/ aerorealestate.net/ None 42 18brownlow.com/ 18brownlow.com:443/ None 57 1granary.com/ 1granary.com/journal/ None 61 1rs.org.uk/ 1rs.io/ None 79 2020visionnetwork.eu/ network.crowdhelix.com/ None

But when I check lev.ratio(str(to_test.iloc[i].url),str(to_test.iloc[i].url_ok2)), it gives me the corresponding value, i.e. lev.ratio(str(to_test.iloc[0].url),str(to_test.iloc[0].url_ok2)) returns

0.45454545454545453

How can I replace the values in lev_ratio column for each row?

Upvotes: 0

Views: 32

Answers (1)

Ghanshyam Yadav
Ghanshyam Yadav

Reputation: 73

Try using .apply to the dataFrame:

df['lev_ratio'] = df.apply(lambda x: lev.ratio(str(x.url),str(x.url_ok2)), axis=1)

Upvotes: 1

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