I have the following csv
id;price;editor
k1;10,00;ed1
k1;8,00;ed2
k3;10,00;ed1
k3;11,00;ed2
k2;10,50;ed1
k1;9,50;ed3
If I do the following
Group the data by only id and find min price for each group. Index the original dataframe based on the minimum values to include the editor column.
Note: I am assuming that the comma in price column is a typo
df.loc[df['price'] == df.groupby('id')['price'].transform('min')]
id price editor
1 k1 8.0 ed2
2 k3 10.0 ed1
4 k2 10.5 ed1
get rid of the editor part:
df_reduced= df.groupby(['id'])['price'].min()
no need to include 'transformed' as somebody else stated
Much like @Wen-Ben I choose to use sort_values
and drop_duplicates
, however, I converted the values using pd.read_csv
with the decimal
parameter.
from io import StringIO
csvfile = StringIO("""id;price;editor
k1;10,00;ed1
k1;8,00;ed2
k3;10,00;ed1
k3;11,00;ed2
k2;10,50;ed1
k1;9,50;ed3""")
df = pd.read_csv(csvfile, delimiter =';', decimal=',')
df.sort_values(['id','price']).drop_duplicates(['id'])
Output:
id price editor
1 k1 8.0 ed2
4 k2 10.5 ed1
2 k3 10.0 ed1
drop_duplicate
+ sort_values
#df['price'] = pd.to_numeric(df['price'].str.replace(",", "."))
df.sort_values('price').drop_duplicates(['id'])
Out[423]:
id price editor
1 k1 8.0 ed2
2 k3 10.0 ed1
4 k2 10.5 ed1
The instruction
df_reduced= df.groupby(['id', 'editor'])['price'].min()
will give you the min price per each unique id-editor pair, you want the min per id. However, since your price field has a string format, you first need to cast it to numeric in order to run the groupby:
df['price'] = pd.to_numeric(df1['price'].str.replace(",", "."))
df.loc[df.groupby('id')['price'].idxmin()]
Output
id price editor
1 k1 8.0 ed2
4 k2 10.5 ed1
2 k3 10.0 ed1