熊猫：标准化不规则时间间隔

Question

I'm wondering if Pandas has some built-in functionality to take random time intervals (roughly hours) and convert them to standardized hours. Code example and non-working attempt:

import pandas as pd

df = pd.DataFrame({'start': ['2018-09-04 01:12', '2018-09-04 02:11'], 
                   'end'  : ['2018-09-04 02:10','2018-09-04 03:20'], 
                   'val'  : [500, 600]})[['start','end','val']]

df[['start','end']] = df[['start','end']].apply(pd.to_datetime)

Gives us:

           start               end  val
2018-09-04 01:12  2018-09-04 02:10  500
2018-09-04 02:11  2018-09-04 03:20  600

and:

df = df.resample('1H', on = 'start', ).reset_index()

would ideally (but doesn't) yield:

           start               end     val
2018-09-04 01:00  2018-09-04 01:59  406.78
2018-09-04 02:00  2018-09-04 02:59  513.22
2018-09-04 03:00  2018-09-04 03:59  180.00

I could code some hack to make this work, but figured Pandas would have some simple function that does this.

Answer 1

This is not a sufficiently common allotment to warrant its own method. You're doing a straightforward linear apportioning of each input interval, broken at the hour. In the first interval, you have 59 total minutes recorded, so the "value" of each minute is 500/59 (8.47+). For the second, it's 600/50 per minute (12.0).

You can do this with a relatively simple control structure, although the individual break-down is a little "wordy". As you create the new rows, use the shift operator to address both the current and previous rows of the input data frame. You need to keep track of the breakpoint (top of the hour) for each row and do that linear computation for both. Your arithmetic looks something like

TIME          VALUE
1:00 - 2:00   (1:12 - 1:00) * 0 + (2:00 - 1:12) * 500/59
2:00 - 3:00   (2:11 - 2:00) * 500/59 + (3:00 - 2:11) * 600/50
3:00 - 4:00   (3:20 - 3:00) * 600/50 + (4:00 - 3:20) * 0

Can you turn those details into the code you need?