Convert custom strings to datetime format

Question

I have a list of date-time data strings that looks like this:

list = ["2016-08-02T09:20:32.456Z", "2016-07-03T09:22:35.129Z"]

I want to convert this to the example format (for the first item):

"8/2/2016 9:20:32 AM"

I tried this:

from datetime import datetime
for time in list:
    date = datetime.strptime(time, '%Y %m %d %I %R')

But returns error:

ValueError: 'R' is a bad directive in format '%Y %m %d %I %R'

Thanks for any help!

The date format that you are using does not match the items of your list. And what is %R supposed to mean? — S. de Melo
– S. de Melo, Commented Oct 14, 2016 at 10:56
It's supposed to mean the time in 24hour duration: tutorialspoint.com/python/time_strptime.htm — Litwos
– Litwos, Commented Oct 14, 2016 at 11:00
Strange, it does not appear in the official documentation: docs.python.org/3.0/library/datetime.html#id1 — S. de Melo
– S. de Melo, Commented Oct 14, 2016 at 11:03

fredtantini · Accepted Answer · 2016-10-14 11:10:23Z

4

You have several problems in you code:

%Rdoesn't seem to be a correct directive (that's your error, I think you are looking for %p).
the format in strptime is the format of your input, not the output
you are using list and time as a name variable which is bad ;]

So, you could for first convert you time to a correct datetime object:

>>> t='2016-08-02T09:20:32.456Z'
>>> d=datetime.strptime(t, "%Y-%m-%dT%H:%M:%S.%fZ")
>>> d
datetime.datetime(2016, 8, 2, 9, 20, 32, 456000)

And then convert this object to a string with strftime

>>> datetime.strftime(d, "%m/%d/%Y %I:%M:%S %p")
'08/02/2016 09:20:32 AM'

answered Oct 14, 2016 at 11:10

fredtantini

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1 Comment

Thanks for the better answer and explanations!

S. de Melo · Accepted Answer · 2016-10-14 11:00:05Z

1

s = "2016-08-02T09:20:32.456Z"
d = datetime.strptime(s, "%Y-%m-%dT%H:%M:%S.%fZ")

answered Oct 14, 2016 at 11:00

S. de Melo

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