熊猫分组汇总

时间:2020-01-31 11:03:39

标签: python pandas

从我之前的问题开始:Get grouped informations from an array with Pandas

我有一个像这样的数据集,我想通过大熊猫获取此信息:对于每一天(按日分组),第二个值是“打开”,第二个值是“关闭”,最高值“高”的值和“低”的值的最小值以及体积的总和。

"Date","Time","Open","High","Low","Close","Up","Down","Volume"
01/03/2000,00:05,1481.50,1481.50,1481.00,1481.00,2,0,0.00
01/03/2000,00:10,1480.75,1480.75,1480.75,1480.75,1,0,1.00
01/03/2000,00:20,1480.50,1480.50,1480.50,1480.50,1,0,1.00
[...]
03/01/2018,11:05,2717.25,2718.00,2708.50,2709.25,9935,15371,25306.00
03/01/2018,11:10,2709.25,2711.75,2706.50,2709.50,8388,8234,16622.00
03/01/2018,11:15,2709.25,2711.50,2708.25,2709.50,4738,4703,9441.00
03/01/2018,11:20,2709.25,2709.50,2706.00,2707.25,3609,4685,8294.00

在我之前的问题中,一个用户建议我使用它:

df.groupby('Date').agg({
'Close': 'last',
'Open': 'first',
'High': 'max',
'Low': 'min',
'Volume': 'sum'

})

但是现在我想将第二个元素用于Open,将倒数第二个用于Close。我该怎么办?

1 个答案:

答案 0 :(得分:4)

您可以创建自定义函数,只有在第二个val不存在时才需要指定输出,例如NaN或第一个值x.iat[0]

def second(x):
    return x.iat[1] if len(x) > 1 else np.nan

def secondLast(x):
    return x.iat[-2] if len(x) > 1 else np.nan

df1 = df.groupby('Date').agg({
'Close': secondLast,
'Open': second,
'High': 'max',
'Low': 'min',
'Volume': 'sum'

})

print (df1)
              Close     Open    High     Low   Volume
Date                                                 
01/03/2000  1480.75  1480.75  1481.5  1480.5      2.0
03/01/2018  2709.50  2709.25  2718.0  2706.0  59663.0