[英]Organize DataFrame into columns by year and index by day-month - PYTHON - PANDAS
[英]Python: Filter DataFrame in Pandas by hour, day and month grouped by year
作为熊猫的新手,我不得不花很多时间才能找到解决该问题的方法。 考虑到我仍然需要解决边界问题,我想知道一种解决此问题的更好方法。
我有一套从2009年到2012年的10项“动力”的小量指标,并希望获得所有年份的小时和日/月窗口(即按年份,按小时,日和月分组的过滤器)。
我得出的结论如下:
import pandas as pd
import numpy as np
import datetime
dates = pd.date_range(start="08/01/2009",end="08/01/2012",freq="10min")
df = pd.DataFrame(np.random.rand(len(dates), 1)*1500, index=dates, columns=['Power'])
def filter(df, day, month, hour, daysWindow, hoursWindow):
"""
Filter a Dataframe by a date window and hour window grouped by years
@type df: DataFrame
@param df: DataFrame with dates and values
@type day: int
@param day: Day to focus on
@type month: int
@param month: Month to focus on
@type hour: int
@param hour: Hour to focus on
@type daysWindow: int
@param daysWindow: Number of days to perform the days window selection
@type hourWindow: int
@param hourWindow: Number of hours to perform the hours window selection
@rtype: DataFrame
@return: Returns a DataFrame with the
"""
df_filtered = None
grouped = df.groupby(lambda x : x.year)
for year, groupYear in grouped:
groupedMonthDay = groupYear.groupby(lambda x : (x.month, x.day))
for monthDay, groupMonthDay in groupedMonthDay:
if monthDay >= (month,day - daysWindow) and monthDay <= (month,day + daysWindow):
new_df = groupMonthDay.ix[groupMonthDay.index.indexer_between_time(datetime.time(hour - hoursWindow), datetime.time(hour + hoursWindow))]
if df_filtered is None:
df_filtered = new_df
else:
df_filtered = df_filtered.append(new_df)
return df_filtered
df_filtered = filter(df,day=8, month=10, hour=8, daysWindow=1, hoursWindow=1)
print len(df)
print len(df_filtered)
返回作为输出:
>>>
157825
117
当然,在选择像1和hoursWindow 2这样的小时时,此代码在边界问题方面需要改进。即:
>>> filter(df,day=8, month=10, hour=1, daysWindow=1, hoursWindow=2)
Traceback (most recent call last):
File "<interactive input>", line 1, in <module>
File "D:\tmp\test_filtro.py", line 40, in filter
new_df = groupMonthDay.ix[groupMonthDay.index.indexer_between_time(datetime.time(hour - hoursWindow), datetime.time(hour + hoursWindow))]
ValueError: hour must be in 0..23
选择1或30之类的日期时也会发生类似的问题。
如何改进此代码?
filter
功能的更新代码可确保没有边界问题:
import pandas as pd
import numpy as np
import datetime
dates = pd.date_range(start="08/01/2009",end="08/01/2012",freq="10min")
df = pd.DataFrame(np.random.rand(len(dates), 1)*1500, index=dates, columns=['Power'])
def filter(df, day, month, hour, minute=0, daysWindow=1, hoursWindow=1):
"""
Filter a Dataframe by a date window and hour window grouped by years
@type df: DataFrame
@param df: DataFrame with dates and values
@type day: int
@param day: Day to focus on
@type month: int
@param month: Month to focus on
@type hour: int
@param hour: Hour to focus on
@type daysWindow: int
@param daysWindow: Number of days to perform the days window selection
@type hoursWindow: int
@param hourWindow: Number of hours to perform the hours window selection
@rtype: DataFrame
@return: Returns a DataFrame with the
"""
df_filtered = None
grouped = df.groupby(lambda x : x.year)
for year, groupYear in grouped:
date = datetime.date(year, month, day)
dateStart = date - datetime.timedelta(days=daysWindow)
dateEnd = date + datetime.timedelta(days=daysWindow+1)
df_filtered_days = df[dateStart:dateEnd]
timeStart = datetime.time(0 if hour-hoursWindow < 0 else hour-hoursWindow, minute)
timeEnd = datetime.time(23 if hour+hoursWindow > 23 else hour+hoursWindow, minute)
new_df = df_filtered_days.ix[df_filtered_days.index.indexer_between_time(timeStart, timeEnd)]
if df_filtered is None:
df_filtered = new_df
else:
df_filtered = df_filtered.append(new_df)
return df_filtered
df_filtered = filter(df,day=8, month=10, hour=1, daysWindow=1, hoursWindow=2)
print len(df)
print len(df_filtered)
输出为:
>>>
157825
174
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