pandas Python - 从数据框熊猫中检索过去 30 天的数据
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Python - Retrieving last 30 days data from dataframe pandas
提问by Naive Babes
I've a dataframe containing six month error logs, collected every day. I want to retrieve the last 30 days records from the last date. Last date isn't today.
For example: I've data for the months May, June, July and until August 15, I want to retrieve that data from August 15to July 15making it 30 days records.
Is there a way to do this in Python Pandas?
我有一个包含六个月错误日志的数据框，每天收集。我想从最后一个日期检索过去 30 天的记录。最后一次约会不是今天。
例如：我有五月、六月、七月和直到的数据August 15，我想从中检索该数据August 15以July 15使其成为 30 天的记录。
有没有办法在 Python Pandas 中做到这一点？
This is the sample dataframe:
这是示例数据框：
Error_Description Date Weekend Type
N17739 Limit switch X- 5/1/2015 5/3/2015 Critical
N17739 Limit switch Y- 5/1/2015 5/3/2015 Critical
N938 Key non-functional 5/1/2015 5/3/2015 Non-Critical
P124 Magazine is running 5/1/2015 5/3/2015 Non-Critical
N17738 Limit switch Z+ 5/1/2015 5/3/2015 Critical
N938 Key non-functional 5/1/2015 5/3/2015 Non-Critical
... ... ... ...
P873 ENCLOSURE DOOR 8/24/2015 8/30/2015 Non-Critical
N3065 Reset M114 8/24/2015 8/30/2015 Non-Critical
N3065 Reset M114, 8/24/2015 8/30/2015 Non-Critical
N2853 Synchronization 8/24/2015 8/30/2015 Critical
P152 ENCLOSURE 8/24/2015 8/30/2015 Non-Critical
N6236 has stopped 8/24/2015 8/30/2015 Critical
采纳答案by jezrael
Date lastdayfromis used for selecting last 30 days of DataFrameby function loc.
日期lastdayfrom用于DataFrame通过函数loc选择过去 30 天。
lastdayfrom = pd.to_datetime('8/24/2015')
print lastdayfrom
#2015-08-24 00:00:00
print df
# Error_Description Date Weekend Type
#0 N17739 Limit switch X- 2015-05-01 2015-05-03 Critical
#1 N17739 Limit switch Y- 2015-05-01 2015-05-03 Critical
#2 N938 Key non-functional 2015-05-01 2015-05-03 Non-Critical
#3 P124 Magazine is running 2015-05-01 2015-05-03 Non-Critical
#4 N17738 Limit switch Z+ 2015-02-01 2015-05-03 Critical
#5 N938 Key non-functional 2015-07-25 2015-05-03 Non-Critical
#6 P873 ENCLOSURE DOOR 2015-07-24 2015-08-30 Non-Critical
#7 N3065 Reset M114 2015-07-21 2015-08-21 Non-Critical
#8 N3065 Reset M114, 2015-08-22 2015-08-22 Non-Critical
#9 N2853 Synchronization 2015-08-23 2015-08-30 Critical
#10 P152 ENCLOSURE 2015-08-24 2015-08-30 Non-Critical
#11 N6236 has stopped 2015-08-24 2015-08-30 Critical
print df.dtypes
#Error_Description object
#Date datetime64[ns]
#Weekend datetime64[ns]
#Type object
#dtype: object
#set index from column Date
df = df.set_index('Date')
#if datetimeindex isn't order, order it
df= df.sort_index()
#last 30 days of date lastday
df = df.loc[lastdayfrom - pd.Timedelta(days=30):lastdayfrom].reset_index()
print df
# Date Error_Description Weekend Type
#0 2015-07-25 N3065 Reset M114 2015-08-21 Non-Critical
#1 2015-08-22 N3065 Reset M114, 2015-08-22 Non-Critical
#2 2015-08-23 N2853 Synchronization 2015-08-30 Critical
#3 2015-08-24 P152 ENCLOSURE 2015-08-30 Non-Critical
#4 2015-08-24 N6236 has stopped 2015-08-30 Critical
回答by faltarell
You can use DataFrame.last_valid_index()to find the label of the last line, and then subtract DateOffset(30, 'D')to go back 30 days:
您可以使用DataFrame.last_valid_index()找到最后一行的标签，然后减去DateOffset(30, 'D')返回 30 天：
df[df.last_valid_index()-pandas.DateOffset(30, 'D'):]
回答by Chris
The other two answers (currently) assume the date is the index, but in python3 at least, you can solve this with just simple masking (.query(..)doesn't work).
另外两个答案（当前）假设日期是索引，但至少在 python3 中，您可以通过简单的掩码来解决这个问题（.query(..)不起作用）。
df[df["Date"] >= (pd.to_datetime('8/24/2015') - pd.Timedelta(days=30))]
df[df["Date"] >= (pd.to_datetime('8/24/2015') - pd.Timedelta(days=30))]
