如何按天拆分 pandas 数据框或系列(可能使用迭代器)
问题描述
我有很长的时间序列,例如.
I have a long time series, eg.
import pandas as pd
index=pd.date_range(start='2012-11-05', end='2012-11-10', freq='1S').tz_localize('Europe/Berlin')
df=pd.DataFrame(range(len(index)), index=index, columns=['Number'])
现在我想提取每天的所有子数据帧,以获得以下输出:
Now I want to extract all sub-DataFrames for each day, to get the following output:
df_2012-11-05: data frame with all data referring to day 2012-11-05
df_2012-11-06: etc.
df_2012-11-07
df_2012-11-08
df_2012-11-09
df_2012-11-10
避免检查 index.date==give_date 是否非常慢的最有效方法是什么.此外,用户事先并不知道帧中的天数范围.
What is the most effective way to do this avoiding to check if the index.date==give_date which is very slow. Also, the user does not know a priory the range of days in the frame.
有什么提示可以用迭代器做到这一点吗?
Any hint do do this with an iterator?
我目前的解决方案是这样,但它不是那么优雅,并且有两个问题定义如下:
My current solution is this, but it is not so elegant and has two issues defined below:
time_zone='Europe/Berlin'
# find all days
a=np.unique(df.index.date) # this can take a lot of time
a.sort()
results=[]
for i in range(len(a)-1):
day_now=pd.Timestamp(a[i]).tz_localize(time_zone)
day_next=pd.Timestamp(a[i+1]).tz_localize(time_zone)
results.append(df[day_now:day_next]) # how to select if I do not want day_next included?
# last day
results.append(df[day_next:])
这种方法存在以下问题:
This approach has the following problems:
- a=np.unique(df.index.date) 可能需要很长时间
- df[day_now:day_next] 包含 day_next,但我需要在范围内排除它
解决方案
也许是groupby?
Perhaps groupby?
DFList = []
for group in df.groupby(df.index.day):
DFList.append(group[1])
应该给你一个数据框列表,其中每个数据框是一天的数据.
Should give you a list of data frames where each data frame is one day of data.
或者在一行中:
DFList = [group[1] for group in df.groupby(df.index.day)]
一定要爱上蟒蛇!
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