Question

假设我们有数据框：

df = pd.DataFrame({'Animal' : ['Falcon', 'Falcon','Parrot', 'Parrot'],
                   'Max Speed' : [380.1, 370.3, 24.77, -12.55]})

我必须构造一个类似于绝对最小值的函数，它必须将元素返回更接近零的值。在本例中，通过按“动物”分组，它应返回：

   Animal  Max Speed
0  Falcon     370.30
1  Parrot     -12.55

我尝试过这样的功能：

def nearzero():
   absolute = [abs(number) for number in data]
   i = absolute.index(min(absolute))
   return data[i]

它应该返回在绝对值最小的索引中找到的元素。但这不起作用：

df.groupby(['Animal']).agg({'Max Speed': [nearzero]})

函数或groupby是否定义错误？

Answer 1

我认为您需要DataFrameGroupBy.idxmin来获得每组分钟的索引，还需要将列Max Speed转换为abs，最后一次调用loc来选择行：

df = df.loc[df['Max Speed'].abs().groupby(df['Animal']).idxmin()]
print (df)
   Animal  Max Speed
1  Falcon     370.30
3  Parrot     -12.55

另一种带有新列的解决方案：

df['Max Speed Abs'] = df['Max Speed'].abs()
df = df.loc[df.groupby('Animal')['Max Speed Abs'].idxmin()]

编辑：对于groupby乘以多个Series，请使用：

df = pd.DataFrame({'Animal' : ['Falcon', 'Falcon','Parrot', 'Parrot'],
                   'Max Speed' : [380.1, 370.3, 24.77, -12.55],
                   'Dates':['2010-10-09'] * 4})  

df = df.loc[df['Max Speed'].abs().groupby([df['Animal'], df['Dates']]).idxmin()]
print (df)
   Animal  Max Speed       Dates
1  Falcon     370.30  2010-10-09
3  Parrot     -12.55  2010-10-09

Answer 2

您可以在python中定义一个函数，

rooms

或使用生成器，

def abs_min(x):
    for elem in x:
        if abs(elem) == min(abs(x)):
            return elem

df.groupby('Animal')['Max Speed'].apply(abs_min)

Animal
Falcon    370.30
Parrot    -12.55

Answer 3

将您的功能定义为：

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请注意，此函数以 df列（系列）作为参数，但必须从基础列表中进行选择。

然后致电：

def nearzero(data):
    dat = data.tolist()
    absolute = [abs(number) for number in dat]
    return dat[absolute.index(min(absolute))]

第二种选择，而无需显式转换为基础列表：

将函数定义为：

df.groupby(['Animal'])['Max Speed'].apply(nearzero)

然后致电：

def nearzero2(data):
    return data[data.abs().idxmin()]

或者如您的问题那样获得结果：

df.groupby(['Animal'])['Max Speed'].apply(nearzero2)

具有自定义聚合功能的熊猫数据框上的groupby

3 个答案: