pandas：在特定时间窗口中选择行

Question

I have a dataset of samples covering multiple days, all with a timestamp. I want to select rows within a specific time window. Eg all rows that were generated between 1pm and 3 pm every day.

This is a sample of my data in a pandas dataframe:

22           22  2018-04-12T20:14:23Z  2018-04-12T21:14:23Z      0  6370.1   
23           23  2018-04-12T21:14:23Z  2018-04-12T21:14:23Z      0  6368.8   
24           24  2018-04-12T22:14:22Z  2018-04-13T01:14:23Z      0  6367.4   
25           25  2018-04-12T23:14:22Z  2018-04-13T01:14:23Z      0  6365.8   
26           26  2018-04-13T00:14:22Z  2018-04-13T01:14:23Z      0  6364.4   
27           27  2018-04-13T01:14:22Z  2018-04-13T01:14:23Z      0  6362.7   
28           28  2018-04-13T02:14:22Z  2018-04-13T05:14:22Z      0  6361.0   
29           29  2018-04-13T03:14:22Z  2018-04-13T05:14:22Z      0  6359.3   
..          ...                   ...                   ...    ...     ...   
562         562  2018-05-05T08:13:21Z  2018-05-05T09:13:21Z      0  6300.9   
563         563  2018-05-05T09:13:21Z  2018-05-05T09:13:21Z      0  6300.7   
564         564  2018-05-05T10:13:14Z  2018-05-05T13:13:14Z      0  6300.2   
565         565  2018-05-05T11:13:14Z  2018-05-05T13:13:14Z      0  6299.9   
566         566  2018-05-05T12:13:14Z  2018-05-05T13:13:14Z      0  6299.6   

How do I achieve that? I need to ignore the date and just evaluate the time component. I could traverse the dataframe in a loop and evaluate the date time in that way, but there must be a more simple way to do that..

I converted the messageDate which was read aa string to a dateTime by

df["messageDate"]=pd.to_datetime(df["messageDate"])

But after that I got stuck on how to filter on time only.

Any input appreciated.

Answer 1

datetime columns have DatetimeProperties object, from which you can extract datetime.time and filter on it:

import datetime

df = pd.DataFrame(
    [
        '2018-04-12T12:00:00Z', '2018-04-12T14:00:00Z','2018-04-12T20:00:00Z',
        '2018-04-13T12:00:00Z', '2018-04-13T14:00:00Z', '2018-04-13T20:00:00Z'
    ], 
    columns=['messageDate']
)
df
            messageDate
# 0 2018-04-12 12:00:00
# 1 2018-04-12 14:00:00
# 2 2018-04-12 20:00:00
# 3 2018-04-13 12:00:00
# 4 2018-04-13 14:00:00
# 5 2018-04-13 20:00:00

df["messageDate"] = pd.to_datetime(df["messageDate"])
time_mask = (df['messageDate'].dt.hour >= 13) & \
            (df['messageDate'].dt.hour <= 15)

df[time_mask]
#           messageDate
# 1 2018-04-12 14:00:00
# 4 2018-04-13 14:00:00

Answer 2

I hope the code is self explanatory. You can always ask questions.

import pandas as pd

#   Prepping data for example
dates = pd.date_range('1/1/2018', periods=7, freq='H')
data = {'A' : range(7)}
df = pd.DataFrame(index = dates, data = data)
print df
#                      A
# 2018-01-01 00:00:00  0
# 2018-01-01 01:00:00  1
# 2018-01-01 02:00:00  2
# 2018-01-01 03:00:00  3
# 2018-01-01 04:00:00  4
# 2018-01-01 05:00:00  5
# 2018-01-01 06:00:00  6

#   Creating a mask to filter the value we with to have or not.
#   Here, we use df.index because the index is our datetime.
#   If the datetime is a column, you can always say df['column_name']
mask = (df.index > '2018-1-1 01:00:00') & (df.index < '2018-1-1 05:00:00')
print mask
# [False False  True  True  True False False]

df_with_good_dates = df.loc[mask]
print df_with_good_dates
#                      A
# 2018-01-01 02:00:00  2
# 2018-01-01 03:00:00  3
# 2018-01-01 04:00:00  4

Answer 3

df=df[(df["messageDate"].apply(lambda x : x.hour)>13) & (df["messageDate"].apply(lambda x : x.hour)<15)]

您可以类似地使用x.minute，x.second。

Answer 4

try this after ensuring messageDate is indeed datetime format as you have done

df.set_index('messageDate',inplace=True)
choseInd = [ind for ind in df.index if (ind.hour>=13)&(ind.hour<=15)]
df_select = df.loc[choseInd]

you can do the same, even without making the datetime column as an index, as the answer with apply: lambda shows

it just makes your dataframe 'better looking' if the datetime is your index rather than numerical one.