根据行条件 python 查找最大值

Question

i have DataFrame look something like this but more data and stuff, for example,

|index | year | drinks | sold |
|------|------|--------|------|
|0     | 2010 | pepsi  | 3456 |
|1     | 2010 | spirit | 32755|
|2     | 2010 | cola   | 7854 |
|3     | 2011 | pepsi  | 6787 |
|4     | 2011 | spirit | 7899 |
|5     | 2011 | cola   | 4657 |

I want to get: **the drinks that has been sold in each year more than the year average? like

in 2010 spirit sold 32755 and it is more the 2010 average 14,688 sells

and same goes to other years. i know i have to get the average for every year first, then compare it to the sold column, but idk how** to reach it.

Answer 1

You can try:

import pandas as pd


data={'year':[2010,2010,2010,2011,2011,2011],
    'drinks':['pepsi','spirit','cola','pepsi','spirit','cola'],
    'sold':[3456,32755,7854,6787,7899,4657]}
df=pd.DataFrame(data)

for year_searched in df['year'].unique():
    df_year_searched=df[df['year']==year_searched]
    sold_avg=df_year_searched['sold'].mean()
    df_yeat_search_more_than_avg=df_year_searched[df_year_searched['sold']>sold_avg].sort_values(by='sold',ascending=False)

    for index,row in df_yeat_search_more_than_avg.iterrows():
        drink=row['drinks']
        sold=row['sold']
        print(f"in {year_searched} {drink} sold {sold} and it is more the {year_searched} average {round(sold_avg,2)} sells")

result:

in 2010 spirit sold 32755 and it is more the 2010 average 14688.33 sells
in 2011 spirit sold 7899 and it is more the 2011 average 6447.67 sells
in 2011 pepsi sold 6787 and it is more the 2011 average 6447.67 sells