Question

我有一个看起来像这样的pandas数据集

city    difference 
NY       6
SF       8
LA       8
NY       9
SF       10

我想根据difference列总结city列的值，以便我的最终数据集看起来像

city    difference    total difference
NY       6              15
NY       9
LA       8               8
SF       10             10

我试过

df['total difference'] = df.groupby('city')['difference'].sum()

但它不起作用。我甚至尝试了How to sum values of particular rows in pandas?，但新列的值为NaN。请帮忙！

Answer 1

我认为你需要transform：

df['total difference'] = df.groupby('city')['difference'].transform(sum) 
print (df)
  city  difference  total difference
0   NY           6                15
1   SF           8                18
2   LA           8                 8
3   NY           9                15
4   SF          10                18

如果还需要排序列：

df['total difference'] = df.groupby('city')['difference'].transform('sum') 
df = df.sort_values('city')
print (df)
  city  difference  total difference
2   LA           8                 8
0   NY           6                15
3   NY           9                15
1   SF           8                18
4   SF          10                18

我感兴趣的是功能和时间上的差异非常相似：

#[10000000 rows x 2 columns]
np.random.seed(100)
df = pd.DataFrame(np.random.randint(1000, size=(10000000,2)), columns=['city','difference'])
#print (df)

In [293]: %timeit (df.groupby('city')['difference'].transform('sum'))
1 loop, best of 3: 570 ms per loop

In [294]: %timeit (df.groupby('city')['difference'].transform(sum))
1 loop, best of 3: 567 ms per loop

In [295]: %timeit (df.groupby('city')['difference'].transform(np.sum))
1 loop, best of 3: 561 ms per loop

使用类似的行值对值进行求和

1 个答案: