I have dataframe with 2 unequal columns:

And i am trying to fill the new third column with all of the many-words that contain one-word. (5 or less) So it would be like:

I actually found a way, but it is very slow.

1. Go with loop in column 'many-rows".

   1.1 Within loop create a dictionary, where key is cell from "many-words" and value is list made with split

2. Go with loop in column "one-word"

   2.1 Within loop create another loop in keys,values of dictionary in 1.1

   2.2.Within these to loops check whether list from 1.1 contains word from one-word

   2.3 If it does - concatenate corresponding cell in third column with the key of dictionary on a condition, that amount of concatenations is 5 or less.

I am actually looping through dataframe-column cells, and creating dicts and lists from it, which i read is very very bad.

I am novice in Python but i am pretty sure that my way is unholy.

There is got to be a better, faster, and cleaner way. Maybe something with vectorization?

Thank you!

CodePudding user response：

You can use iterrows to loop over your df rows and build a list of Many-Words containing One-word:

df["Many-Words with One-word"] = pd.Series([
  df[df["Many-Words"].str.lower().str.contains(row["One-word"].lower())]["Many-Words"].to_list()
    for _, row in df.iterrows()
])

Note: using lower to make the match case-insensitive.

Output:

  One-word                  Many-Words                           Many-Words with One-word
0     Bird          Bird with no blood  [Bird with no blood, Stone that killed the bir...
1    Stone  Stone that killed the bird      [Stone that killed the bird, stone and blood]
2    Blood         Bird without brains              [Bird with no blood, stone and blood]
3   <none>             stone and blood                                                 []