How to add the subtracted date value in Python

Question

I have a data set of dates:

(u'x', datetime.datetime(2018, 1, 5, 4, 49, 30))
(u'y', datetime.datetime(2018, 1, 5, 9, 59, 10))
(u'x', datetime.datetime(2018, 1, 13, 10, 23, 17))
(u'y', datetime.datetime(2018, 1, 15, 17, 16, 34))
(u'x', datetime.datetime(2018, 1, 22, 11, 9, 55))
(u'y', datetime.datetime(2018, 1, 22, 14, 24, 59))

Here is my code to subtract the date:

for x in range(len(y)):
    if y[x][0] == 'x' and y[x+1][0] != 'y':
        #subtract the date
        print y[x+1][1] - y[x][1]

The code above prints below:

5:09:40
2 days, 6:53:17
3:15:04

My question is how do I add them all up and convert them as total in hours? Appreciate the help on this.

jpp · Accepted Answer · 2018-04-04 11:12:16Z

1

This is one way.

import datetime

lst = [(u'x', datetime.datetime(2018, 1, 5, 4, 49, 30)),
       (u'y', datetime.datetime(2018, 1, 5, 9, 59, 10)),
       (u'x', datetime.datetime(2018, 1, 13, 10, 23, 17)),
       (u'y', datetime.datetime(2018, 1, 15, 17, 16, 34)),
       (u'x', datetime.datetime(2018, 1, 22, 11, 9, 55)),
       (u'y', datetime.datetime(2018, 1, 22, 14, 24, 59))]

res = sum((y[1]-x[1]).total_seconds()/(60**2) for x, y in zip(lst[::2], lst[1::2]))

# 63.30027777777778

Explanation

Convert to seconds via timedelta.total_seconds().
Then convert to hours by dividing by 60**2.
Finally, wrap in sum and use a generator expression.

answered Apr 4, 2018 at 11:12

jpp

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2 Comments

Eseresaure Over a year ago

Is it possible not to use a generator as an example? Sometimes the data might not be sequential(like 'x' then 'y' seq) that's why I included a check for 'x' overall this makes sense but I hope to use my code above.

jpp Over a year ago

Sure, just output the list [(y[1]-x[1]).total_seconds()/(60**2) for x, y in zip(lst[::2], lst[1::2])]. sum also accepts a list, so you can sum later.

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