使用Groupby的Python Pandas条件求和

Question

使用样本数据：

df = pd.DataFrame({'key1' : ['a','a','b','b','a'],
               'key2' : ['one', 'two', 'one', 'two', 'one'],
               'data1' : np.random.randn(5),
               'data2' : np. random.randn(5)})

df

    data1        data2     key1  key2
0    0.361601    0.375297    a   one
1    0.069889    0.809772    a   two
2    1.468194    0.272929    b   one
3   -1.138458    0.865060    b   two
4   -0.268210    1.250340    a   one

我试图弄清楚如何按key1分组数据并仅对key2等于“ one”的data1值求和。

这是我尝试过的

def f(d,a,b):
    d.ix[d[a] == b, 'data1'].sum()

df.groupby(['key1']).apply(f, a = 'key2', b = 'one').reset_index()

但这给了我一个数值为“无”的数据框

index   key1    0
0       a       None
1       b       None

这里有什么想法吗？我正在寻找与以下SQL等效的Pandas：

SELECT Key1, SUM(CASE WHEN Key2 = 'one' then data1 else 0 end)
FROM df
GROUP BY key1

提前致谢

Answer 1

首先按key1列分组：

In [11]: g = df.groupby('key1')

然后为每个组取subDataFrame，其中key2等于“ one”并求和data1列：

In [12]: g.apply(lambda x: x[x['key2'] == 'one']['data1'].sum())
Out[12]:
key1
a       0.093391
b       1.468194
dtype: float64

为了解释发生了什么，让我们看一下“ a”组：

In [21]: a = g.get_group('a')

In [22]: a
Out[22]:
      data1     data2 key1 key2
0  0.361601  0.375297    a  one
1  0.069889  0.809772    a  two
4 -0.268210  1.250340    a  one

In [23]: a[a['key2'] == 'one']
Out[23]:
      data1     data2 key1 key2
0  0.361601  0.375297    a  one
4 -0.268210  1.250340    a  one

In [24]: a[a['key2'] == 'one']['data1']
Out[24]:
0    0.361601
4   -0.268210
Name: data1, dtype: float64

In [25]: a[a['key2'] == 'one']['data1'].sum()
Out[25]: 0.093391000000000002

通过将数据帧限制为仅那些key2等于1的数据帧，可能会更容易/更清楚地做到这一点：

In [31]: df1 = df[df['key2'] == 'one']

In [32]: df1
Out[32]:
      data1     data2 key1 key2
0  0.361601  0.375297    a  one
2  1.468194  0.272929    b  one
4 -0.268210  1.250340    a  one

In [33]: df1.groupby('key1')['data1'].sum()
Out[33]:
key1
a       0.093391
b       1.468194
Name: data1, dtype: float64