Calculate the minimum value of a certain column for all the groups and subtract the value from all the values of a certain column of that group

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df = pd.DataFrame([['2018-02-03',42],
                   ['2018-02-03',22],
                   ['2018-02-03',10],
                   ['2018-02-03',32],
                   ['2018-02-03',10],
                   ['2018-02-04',8],
                   ['2018-02-04',2],
                   ['2018-02-04',12],
                   ['2018-02-03',20],
                   ['2018-02-05',30],
                   ['2018-02-05',5],
                   ['2018-02-05',15]])
df.columns = ['product','date','quantity']

I want to create groups by date and calculate the minimum value of a 'quantity' column for all the groups respectively and subtract the value from all the values of a 'quantity' column of that group. The desired output is:

day         value
2018-02-03  32                #(because, 42-10 = 32), 10 is minimum for date 2018-02-03.
2018-02-03  12
2018-02-03  0
2018-02-03  22
2018-02-03  0
2018-02-04  6
2018-02-04  0
2018-02-04  10
2018-02-03  10
2018-02-05  25
2018-02-05  0
2018-02-05  10

Now, this is what I tried:

df = df.groupby('Date', as_index = True)
datamin = df.groupby('Date')['quantity'].min()

Bu this creates a dataframe with the first quantity by Date ana I also do not know, how to proceed after this!!

pandas-groupby pandas group-by
2 Answers

try via groupby() and transform():

df['value']=df.groupby('date')['quantity'].transform(lambda x:x-x.min())

output of df:

    date         quantity   value
0   2018-02-03      42      32
1   2018-02-03      22      12
2   2018-02-03      10      0
3   2018-02-03      32      22
4   2018-02-03      10      0
5   2018-02-04      8       6
6   2018-02-04      2       0
7   2018-02-04      12      10
8   2018-02-03      20      10
9   2018-02-05      30      25
10  2018-02-05      5       0
11  2018-02-05      15      10
user_142898920

For improve performance use GroupBy.transform without lambda function, better is subtract all values of column like:

df['value'] = df['quantity'].sub(df.groupby('date')['quantity'].transform('min'))
user_29010020
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