Python: Expand dataframe rows with specific column value in terms of 1/8th

user_7818560 0
Viewed 297

Have got input dataframe like below:

df

Store   Item        Space
11      Grape       0.125
11      Beans       0.0
12      Mango       0.25
13      Beetroot    0.375
13      Carrot      0.5

Need to expand given df row w.r.t. 'Space' column. Values in 'Space' column will always be in 1/8th term. E.g.: 0.125(1/8) equivalent to 1 Part among 8, 0.25 - 2 Part, 0.375 - 3 Part, 0.5 - 4 Part, 0.625 - 5 Part, 0.75 - 6 Part, 0.875 - 7 Part, 1.0 - 8 Part. So, rows in df should be expand w.r.t. how many parts of 1/8 present for each 'Item' in 'Space' column. And if any 'Item' holds 0 space that Item need to be dropped.

Expected Output:

Store   Item        Space
11      Grape       0.125
12      Mango       0.125
12      Mango       0.125
13      Beetroot    0.125
13      Beetroot    0.125
13      Beetroot    0.125
13      Carrot      0.125
13      Carrot      0.125
13      Carrot      0.125
13      Carrot      0.125

Any help appreciated. Thank You!

expand pandas dataframe python-3.x python
1 Answers

You can use

eigth = 0.125
result = df['Item'].repeat(df['Space']/eigth).to_frame().assign(Space=eigth)

How to improve this line of code to include in any additional columns present in df and to include those as well? Question edited accordingly.

result = df.apply(pd.Series.repeat, repeats=df['Space']/eigth).assign(Space=eigth)
user_36200030
Related