Cannot get groupby records based on their minimum value using pandas in python

Question

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

import pandas as pd 

df = pd.read_csv('Testing.csv', delimiter =';')
df_reduced= df.groupby(['id', 'editor'])['price'].min()

Instead of getting

k1;8,00;ed2
k2;10,50;ed1
k3;10,00;ed1

I get

k1;10,00;ed1
    8,00;ed2
    9,50;ed3
k2;10,50;ed1
k3;10,00;ed1
   11,00;ed2 

So can I get three id's with their minimum values?

Answer 1

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 

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

get rid of the editor part:

df_reduced= df.groupby(['id'])['price'].min()

no need to include 'transformed' as somebody else stated