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Cannot use a timedelta Rectangle width with a datetime axis #4916

brandon-rhodes opened this issue Aug 13, 2015 · 5 comments


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commented Aug 13, 2015

I would like to draw a rectangle to represent the weekend on a plot whose x-axis is time. In an IPython Notebook:

%pylab inline
dt = datetime.datetime
dates = [
    dt(2015, 1, 1),
    dt(2015, 1, 2),
    dt(2015, 1, 3),
    dt(2015, 1, 4),
    dt(2015, 1, 5),
data = [56.0, 53.2, 57.8, 59.1, 55.5]
    xy=(dt(2015, 1, 3), 50.0),
plot(dates, data)

The big problem is that the above code dies with an exception.

AttributeError                            Traceback (most recent call last)
<ipython-input-13-b4545adc3ff2> in <module>()
     12     xy=(dt(2015, 1, 3), 50.0),
     13     width=datetime.timedelta(days=2),
---> 14     height=1.0,
     15     ))
     16 plot(dates, data)


/home/brhodes/.v/misc/lib/python2.7/site-packages/matplotlib/dates.pyc in _to_ordinalf(dt)
    202             dt -= delta
--> 204     base = float(dt.toordinal())
    205     if hasattr(dt, 'hour'):
    206         base += (dt.hour / HOURS_PER_DAY + dt.minute / MINUTES_PER_DAY +

AttributeError: 'datetime.timedelta' object has no attribute 'toordinal'

The other problem is that my backup plan, of replacing the timedelta() with the floating-point value 2.0, destroys the date-ness of the x axis. It reverts to being a float that does not display as a date.

Why is a timedelta not a valid increment or offset with which I can express width along a datetime x-axis?

Thanks for any help you can provide!

@tacaswell tacaswell added this to the proposed next point release milestone Aug 13, 2015


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commented Aug 13, 2015

As a work around I would try using ax.fill_betweenx or one of the lower-level polygons which lets you specify all of the vertex coordinates.


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commented Dec 23, 2016

@brandon-rhodes I faced the same problem as you, to annotate a graph using a rectangle.

My workaround is to define a function, it can automatically use timestamp as x axis.

# draw a rectangle
def annote_rect(ax, x, y, w, h):
    t = [x, x+w, x+w, x, x]
    d = [y, y, y+h, y+h, y]
    ax.plot(t, d, 'm--', lw=2)

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commented May 11, 2017

Another workaround is to subclass timezone.timedelta to implement a toordinal method
(which has no true meaning).
Basically I used

class OrdinalTimedelta(datetime.timedelta):
    _origin =, 1, 1)
    def toordinal(self):
        return (self._origin + self).toordinal()

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commented May 12, 2017

Clever! Would you like to make an example show-casing this?


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commented May 15, 2017

@dstansby dstansby modified the milestones: 2.2 (next next feature release), 2.1 (next point release) Aug 22, 2017

@QuLogic QuLogic modified the milestones: needs sorting, v2.2.0 Feb 12, 2018

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