bmu*_*bmu 30 python numpy time-series pandas
我有大熊猫温度和辐射的时间序列dataframe.时间分辨率是常规步骤1分钟.
import datetime
import pandas as pd
import numpy as np
date_times = pd.date_range(datetime.datetime(2012, 4, 5, 8, 0),
datetime.datetime(2012, 4, 5, 12, 0),
freq='1min')
tamb = np.random.sample(date_times.size) * 10.0
radiation = np.random.sample(date_times.size) * 10.0
frame = pd.DataFrame(data={'tamb': tamb, 'radiation': radiation},
index=date_times)
frame
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 241 entries, 2012-04-05 08:00:00 to 2012-04-05 12:00:00
Freq: T
Data columns:
radiation 241 non-null values
tamb 241 non-null values
dtypes: float64(2)
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我怎样才能将其下采样dataframe到一小时的分辨率,计算温度的每小时平均值和辐射的每小时总和?
bmu*_*bmu 53
使用pandas 0.18,重新采样API已更改(请参阅文档).所以对于pandas> = 0.18,答案是:
In [31]: frame.resample('1H').agg({'radiation': np.sum, 'tamb': np.mean})
Out[31]:
tamb radiation
2012-04-05 08:00:00 5.161235 279.507182
2012-04-05 09:00:00 4.968145 290.941073
2012-04-05 10:00:00 4.478531 317.678285
2012-04-05 11:00:00 4.706206 335.258633
2012-04-05 12:00:00 2.457873 8.655838
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旧答案:
我正在回答我的问题以反映时间序列相关的变化pandas >= 0.8(所有其他答案都已过时).
使用pandas> = 0.8,答案是:
In [30]: frame.resample('1H', how={'radiation': np.sum, 'tamb': np.mean})
Out[30]:
tamb radiation
2012-04-05 08:00:00 5.161235 279.507182
2012-04-05 09:00:00 4.968145 290.941073
2012-04-05 10:00:00 4.478531 317.678285
2012-04-05 11:00:00 4.706206 335.258633
2012-04-05 12:00:00 2.457873 8.655838
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