我有一个df（样本最后粘贴在这里）。
我希望找到哪个tradePrice每过去10分钟最多tradeVolume，或任何其他滚动期。
这是在xls中完成的数据透视表，它基于附加的示例数据。
minute
tradePrice Data 0 1 10 Total Result
12548 Sum - tradeVolume 3 3
Count - tradePrice 2 2
12548.5 Sum - tradeVolume 1 1
Count - tradePrice 1 1
12549 Sum - tradeVolume 1 1
Count - tradePrice 1 1
12549.5 Sum - tradeVolume 95 95
Count - tradePrice 5 5
12550 Sum - tradeVolume 6 6
Count - tradePrice 4 4
12559 Sum - tradeVolume 93 93
Count - tradePrice 1 1
12559.5 Sum - tradeVolume 1 1
Count - tradePrice 1 1
12560 Sum - tradeVolume 5 5
Count - tradePrice 4 4
12560.5 Sum - tradeVolume 3 5 8
Count - tradePrice 3 2 5
12561 Sum - tradeVolume 4 5 9
Count - tradePrice 2 3 5
12561.5 Sum - tradeVolume 3 2 5
Count - tradePrice 3 1 4
12562 Sum - tradeVolume 9 7 16
Count - tradePrice 8 1 9
12562.5 Sum - tradeVolume 6 2 8
Count - tradePrice 2 2 4
12563 Sum - tradeVolume 2 2
Count - tradePrice 1 1
Total Sum - tradeVolume 120 27 106 253
Total Count - tradePrice 20 14 13 47
结果需要像这样搜索交易量最大的价格：
Price Volume
02:00:00 AM 12559 93
02:01:00 AM 12562 7
02:10:00 AM 12549.5 95
为了获得1分钟。结果我小组和＆amp;应用了以下功能
def f(x): # function to find the POC price and volume
a = x['tradePrice'].value_counts().index[0]
b = x.loc[x['tradePrice'] == a, 'tradeVolume'].sum()
return pd.Series([a, b], ['POC_Price', 'POC_Volume'])
groupbytime = (str(Time)+"min")#ther is a column name by this
groups = df.groupby(groupbytime,as_index=True)
df_POC = groups.apply(f) #applys the function of the POC on the grouped data
我的问题是： 如何获得相同的解决方案但是每<滚动时间段（不能少于1分钟），所以过去10分钟的预期结果（以最大交易量交易的价格）是：
Price Volume
02:10:00 AM 12549.5 95
提前感谢！
样本数据：
dateTime tradePrice tradeVolume 1min time_of_day_10 time_of_day_30 date hour minute
0 2017-09-19 02:00:04 12559 93 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
49 2017-09-19 02:00:11 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
50 2017-09-19 02:00:12 12563 2 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
51 2017-09-19 02:00:12 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
122 2017-09-19 02:00:34 12561.5 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
123 2017-09-19 02:00:34 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
127 2017-09-19 02:00:34 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
129 2017-09-19 02:00:35 12561 2 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
130 2017-09-19 02:00:35 12560.5 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
131 2017-09-19 02:00:35 12561.5 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
135 2017-09-19 02:00:39 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
136 2017-09-19 02:00:39 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
137 2017-09-19 02:00:43 12561.5 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
138 2017-09-19 02:00:43 12561 2 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
139 2017-09-19 02:00:43 12560.5 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
140 2017-09-19 02:00:43 12560.5 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
152 2017-09-19 02:00:45 12562 2 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
153 2017-09-19 02:00:46 12562.5 4 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
166 2017-09-19 02:00:58 12562 1 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
167 2017-09-19 02:00:58 12562.5 2 2017-09-19 02:00:00 02:00:00 02:00:00 2017-09-19 2 0
168 2017-09-19 02:01:00 12562 7 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
169 2017-09-19 02:01:00 12562.5 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
170 2017-09-19 02:01:00 12562.5 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
171 2017-09-19 02:01:00 12561.5 2 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
175 2017-09-19 02:01:03 12561 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
176 2017-09-19 02:01:03 12561 3 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
187 2017-09-19 02:01:07 12560.5 2 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
188 2017-09-19 02:01:08 12561 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
189 2017-09-19 02:01:10 12560 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
190 2017-09-19 02:01:10 12560 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
191 2017-09-19 02:01:10 12559.5 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
192 2017-09-19 02:01:11 12560 1 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
193 2017-09-19 02:01:12 12560 2 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
194 2017-09-19 02:01:12 12560.5 3 2017-09-19 02:01:00 02:00:00 02:00:00 2017-09-19 2 1
593 2017-09-19 02:10:00 12550 1 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
594 2017-09-19 02:10:00 12549.5 12 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
604 2017-09-19 02:10:12 12548.5 1 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
605 2017-09-19 02:10:15 12549.5 22 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
606 2017-09-19 02:10:16 12549.5 21 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
636 2017-09-19 02:10:45 12548 1 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
637 2017-09-19 02:10:47 12548 2 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
638 2017-09-19 02:10:47 12549.5 23 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
639 2017-09-19 02:10:48 12549.5 17 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
640 2017-09-19 02:10:49 12549 1 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
665 2017-09-19 02:10:58 12550 1 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
666 2017-09-19 02:10:58 12550 1 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
667 2017-09-19 02:10:58 12550 3 2017-09-19 02:10:00 02:10:00 02:00:00 2017-09-19 2 10
答案 0 :(得分：3)
如果我理解你的问题，你需要选择时间粒度和时间窗口。然后你可以结合使用groupby + unstack + rolling。
首先是groupby：
time_grain = '1min'
df = df.groupby([pd.Grouper(key='dateTime', freq=time_grain),'tradePrice']).tradeVolume.sum()
dateTime tradePrice
2017-09-19 02:00:00 12559.0 93
12560.5 3
12561.0 4
12561.5 3
12562.0 9
12562.5 6
12563.0 2
2017-09-19 02:01:00 12559.5 1
12560.0 5
12560.5 5
12561.0 5
12561.5 2
12562.0 7
12562.5 2
2017-09-19 02:10:00 12548.0 3
12548.5 1
12549.0 1
12549.5 95
12550.0 6
Name: tradeVolume, dtype: int64
然后取消堆叠+滚动：
window_size = '10min'
df = df.unstack('tradePrice').rolling(window_size).sum()
tradePrice 12548.0 12548.5 12549.0 12549.5 12550.0 12559.0 \
dateTime
2017-09-19 02:00:00 NaN NaN NaN NaN NaN 93.0
2017-09-19 02:01:00 NaN NaN NaN NaN NaN 93.0
2017-09-19 02:10:00 3.0 1.0 1.0 95.0 6.0 NaN
tradePrice 12559.5 12560.0 12560.5 12561.0 12561.5 12562.0 \
dateTime
2017-09-19 02:00:00 NaN NaN 3.0 4.0 3.0 9.0
2017-09-19 02:01:00 1.0 5.0 8.0 9.0 5.0 16.0
2017-09-19 02:10:00 1.0 5.0 5.0 5.0 2.0 7.0
tradePrice 12562.5 12563.0
dateTime
2017-09-19 02:00:00 6.0 2.0
2017-09-19 02:01:00 8.0 2.0
2017-09-19 02:10:00 2.0 NaN
最后将tradePrice堆叠回索引并找到每个时间段内具有最高值的索引：
df = df.stack('tradePrice')
idx_list = df.groupby('dateTime').idxmax()
result = df.loc[idx_list]
dateTime tradePrice
2017-09-19 02:00:00 12559.0 93.0
2017-09-19 02:01:00 12559.0 93.0
2017-09-19 02:10:00 12549.5 95.0
dtype: float64
请注意，如果您使用时间偏移量，则滚动默认值为最小观测值1。这就是为什么你得到3个结果行。
我认为这种方法的最大缺点是，对于具有大量价格点的大型数据帧，这将占用大量内存（因为每个价格点都会生成一个新列）。