Making matplotlib's date2num and num2date perfect inverses

Question

I'm trying to write a pair of functions, plottm and unixtm, which convert back and forth between normal unix time (seconds since 1970-01-01) and Matplotlib's date representation (days since the last day of -1BC or something, a float).

If plottm and unixtm were proper inverses then this code would print the same date/time twice:

import time, datetime
import matplotlib.dates as dt

# Convert a unix time u to plot time p, and vice versa
def plottm(u): return dt.date2num(datetime.datetime.fromtimestamp(u))
def unixtm(p): return time.mktime(dt.num2date(p).timetuple())

u = 1270000000
print datetime.datetime.fromtimestamp(u), "-->", \
      datetime.datetime.fromtimestamp(unixtm(plottm(u)))

Alas, it's off by an hour (which only happens for some timestamps, otherwise I'd insert an offset and be done with it).

Probably related: Problems with Localtime

UPDATE: Related question that isn't specific to Matplotlib: Convert a unixtime to a datetime object and back again (pair of time conversion functions that are inverses)

Pedro Romano · Answer 1 · 2012-11-06T23:47:05.960

3

Based on @dreeves answer, a solution adapted to work with timezone aware datetimes:

import matplotlib.dates as dt

from calendar import timegm
from datetime import datetime

from pytz import utc


# Convert a unix time u to plot time p, and vice versa
def plottm(u):
    return dt.date2num(datetime.fromtimestamp(u, utc))

def unixtm(p):
    return timegm(dt.num2date(p, utc).utctimetuple())


u = 1270000000
print datetime.fromtimestamp(u, utc), "-->", \
      datetime.fromtimestamp(unixtm(plottm(u)), utc)

output (tested for several timezones):

2010-03-31 01:46:40+00:00 --> 2010-03-31 01:46:40+00:00

edited Nov 06 '12 at 23:47

answered Nov 06 '12 at 22:00

Pedro Romano

10,973
4
46
50

When I run this my output is `2010-03-31 01:46:40+00:00 --> 2010-03-31 06:46:40+00:00 (1270000000 --> 1270018000)` – dreeves Nov 06 '12 at 22:39
The problem is that my time zone is currently UTC, and I should have tested with different localtimes... Testing now. – Pedro Romano Nov 06 '12 at 22:48
Can you try the corrected example? It tested it with several local timezones so an it should be correct now. – Pedro Romano Nov 06 '12 at 23:29
Thanks so much, Pedro! In the meantime I asked a more general version of this question which led to what I think is a better approach than messing with tzlocal. I just posted an answer that's now working for me. – dreeves Nov 06 '12 at 23:34

score 3 · Answer 2 · answered Nov 07 '12 at 00:00

There are matplotlib.dates.epoch2num()/num2epoch functions that do exactly that:

from datetime import datetime, timedelta
import matplotlib.dates as mpl_dt

matplotlib_epoch = datetime(1, 1, 1)  # utc
posix_epoch = datetime(1970, 1, 1)  # utc
DAY = 86400  # seconds


def plottm(u):
    """posix timestamp -> plot time"""
    td = (datetime.utcfromtimestamp(u) - matplotlib_epoch)
    return td.days + 1 + (1000000 * td.seconds + td.microseconds) / 1e6 / DAY


def unixtm(p):
    """plot time -> posix timestamp"""
    td = timedelta(days=p-1)
    return (matplotlib_epoch + td - posix_epoch).total_seconds()


def main():
    f = datetime.utcfromtimestamp
    u = 1270000000.1234567890
    print(f(u))
    print(mpl_dt.epoch2num(u))
    print(plottm(u))
    print(f(mpl_dt.num2epoch(mpl_dt.epoch2num(u))))
    print(f(mpl_dt.num2epoch(plottm(u))))
    print(f(unixtm(mpl_dt.epoch2num(u))))
    print(f(unixtm(plottm(u))))

    assert abs(mpl_dt.epoch2num(u) - plottm(u)) < 1e-5

    p = 86401.234567890 / DAY
    print(f(mpl_dt.num2epoch(p)))
    print(f(unixtm(p)))
    assert abs(mpl_dt.num2epoch(p) - unixtm(p)) < 1e-5

main()

Output

2010-03-31 01:46:40.123457
733862.074076
733862.074076
2010-03-31 01:46:40.123453
2010-03-31 01:46:40.123453
2010-03-31 01:46:40.123453
2010-03-31 01:46:40.123453
0001-01-01 00:00:01.234566
0001-01-01 00:00:01.234566

Thanks J.F.! Do you see any reason to prefer this solution over the one I posted (based, confusingly, on F.J.'s answer)? — dreeves, Nov 07 '12 at 04:09
@dreeves: yes. Your code might be incorrect. I've left [appropriate comment on your answer](http://stackoverflow.com/questions/13259875/making-matplotlibs-date2num-and-num2date-perfect-inverses/13261338#comment18076602_13261110). epoch2num/num2epoch should be used unless you have specific reasons not to use them. Also you could test times around `datetime.min`, `datetime.max` to see whether the precision is lost or does it work at all. — jfs, Nov 07 '12 at 04:45

score 1 · Answer 3 · edited May 23 '17 at 12:04

1

Thanks to F.J.'s answer to a similar question, I believe the following may be the best way to deal with this:

import datetime, calendar
import matplotlib.dates as dt

def plottm(u): return dt.date2num(datetime.datetime.utcfromtimestamp(u))
def unixtm(p): return calendar.timegm(dt.num2date(p).timetuple())

edited May 23 '17 at 12:04

Community

1
1

answered Nov 06 '12 at 23:33

dreeves

26,430
45
154
229

Definitely better solution: +1 from me! :) I was sure there was a UTC equivalent for `time.mktime` but for the life of me, I couldn't remember where (and it was staring me in the face a few lines above in the `time.gmtime` documentation)! One should **always** work internally in UTC and to/from local time should only happen at user interface. – Pedro Romano Nov 06 '12 at 23:50
`num2date` depends on tz that can be configured to be any timezone, `timegm` expect utc time so `num2date` should always return time in utc. I don't know what timezone `date2num` expects (if it is not always utc; the code might fail). – jfs Nov 07 '12 at 04:44

Making matplotlib's date2num and num2date perfect inverses

3 Answers3

Output

Linked