Question

我想获取数据集中每年排名前10位的“净利润”的平均值，并将其存储在新的DF中。这是我到目前为止所拥有的。

#loading the csv file
df = pd.read_csv('tmdb-movies.csv')

#Narrows down data to only include the last 25 years
df = df[df['release_year'] > 1992]

#Create column for calculated profit (revenue - budget)
df['Net Profit'] = (df['revenue_adj'] - df['budget_adj']) / 1000000

df = df['Net Profit'].groupby(df['release_year']).nlargest(10)

df.head(20)

这是我到目前为止的输出：

release_year       
1993          10223    1293.766704
              10225     628.371507
              10226     490.390201
              10222     451.732687
              10312     345.086055
              10260     344.534054
              10240     312.158213
              10234     279.252261
              10228     272.728678
              10250     226.825117
1994          4180     1093.391570
              4179      916.422179
              4187      483.382849
              4181      471.414971
              4203      434.906488
              4186      388.200466
              4197      365.866669
              4182      338.760765
              4177      302.944183
              4194      246.901592
Name: Net Profit, dtype: float64

我想要的输出如下：

release_year      Average Net Profit
1993              400
1994              459
1995              480

Answer 1

按level=0分组以进行索引并使用.mean()

df.groupby(df.index.get_level_values(level=0)).mean()

release_year
1993    464.484548
1994    504.219173
1995    511.392629
1996    424.185431
1997    543.448686
Name: Net Profit, dtype: float64

Answer 2

您可以通过mean通过level

df.mean(level=0)

计算python中每年的前10个平均值（groupby，nlargest）

2 个答案: