如何在熊猫数据框中获取时间戳最近的行？

Question

我有一个从列表中获取的时间戳，我需要在 pandas 数据框中找到最接近我拥有的时间戳的行，即使是一组行对我来说也很好。这是示例数据框0 6160 Upper 12-7-2019 12:37:51.1235721 6162 Upper 12-7-2019 12:39:22.3557252 6175 Upper 12-7-2019 13:21:15.2241573 6180 Upper 13-7-2019 06:04:29.1571114 6263 Upper 13-7-2019 07:37:51.123572我有一个时间戳 datetime.datetime(12,7,2019,16,41,20)所以在这种情况下，我需要它在索引 2 处捕获一行。感谢您的帮助。谢谢

Answer 1

import pandas as pd

from datetime import datetime

# Input

ref_time = datetime(2019,7,12,16,41,20)

data = [[6160, 'Upper', '12-7-2019 12:37:51.123572'],

        [6162, 'Upper', '12-7-2019 12:39:22.355725'],

        [6175, 'Upper', '12-7-2019 13:21:15.224157'],

        [6180, 'Upper', '13-7-2019 06:04:29.157111'],

        [6263, 'Upper', '13-7-2019 07:37:51.123572']]

# Convert 2D list to DataFrame object

df = pd.DataFrame(data)

# Convert timestamp strings in column at index 2 to datetime objects

df.iloc[:, 2] = pd.to_datetime(df.iloc[:, 2], format='%d-%m-%Y %H:%M:%S.%f')

# Return row with minimum absolute time difference to reference time

print(df.loc[[(abs(df.iloc[:, 2]-ref_time)).idxmin()]])

Answer 2

你可以做：

import datetime

dt = datetime.datetime(2019, 12, 7,16,41,20)

# d column is the date

minidx = (dt - df['d']).idxmin()

print(df.loc[[minidx]])

   a     b        c                          d

2  2  6175   Upper  2019-12-07 13:21:15.224157