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ax.bar doesn't work correctly when width is a timedelta64 object #11290
Description
The following example
In [1]: x = np.array(range(100), dtype='M8[D]')
In [2]: ax = plt.subplot(1,1,1)
In [3]: y = np.random.random(100)
In [4]: ax.bar(x, y)
Out[4]: <BarContainer object of 100 artists>works fine, but if I set width to be the diff of the datetime64 array, which gives a timedelta64 array, things break:
In [7]: ax.bar(x[:-1], y[:-1], width=np.diff(x))
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-7-f05692d600cb> in <module>()
----> 1 ax.bar(x[:-1], y[:-1], width=np.diff(x))
~/miniconda3/envs/dev/lib/python3.6/site-packages/matplotlib/__init__.py in inner(ax, *args, **kwargs)
1853 "the Matplotlib list!)" % (label_namer, func.__name__),
1854 RuntimeWarning, stacklevel=2)
-> 1855 return func(ax, *args, **kwargs)
1856
1857 inner.__doc__ = _add_data_doc(inner.__doc__,
~/miniconda3/envs/dev/lib/python3.6/site-packages/matplotlib/axes/_axes.py in bar(self, *args, **kwargs)
2257 if align == 'center':
2258 if orientation == 'vertical':
-> 2259 left = x - width / 2
2260 bottom = y
2261 elif orientation == 'horizontal':
TypeError: ufunc subtract cannot use operands with types dtype('float64') and dtype('<m8[D]')This should however be unambiguous, so it would be nice if it worked.
Activity
dstansby commented on May 25, 2018
Can you provide the full example with the timedelta64 array that produces that error? I've encountered similar looking problems before, that depend on when we do our unit conversions, and this might actually be fixed in the master branch.
astrofrog commented on May 25, 2018
@dstansby - the full example is:
timhoffm commented on May 26, 2018
The problem is our unit conversion. We do
However, we don't have a conversion implemented for timedelta64. A result, x is our internal float-based date format while width is still a timedelta64, which cannot be subtracted from each other.
The solution would be to implement a converter for timedelta64.
dstansby commented on May 26, 2018
Ah, so this would be fixed by something like #9120 I think.
timhoffm commented on May 26, 2018
Sort of. It needs a bit more infrastructure:
dstansby commented on Sep 18, 2018
I'm having a look at this now, but having trouble with your example @astrofrog ; when I do
I just get a bunch of zeros; dividing np.diff(x) by a number seems to just zero it out. Do you know of a way to fix this? Internally Matplotlib has to divide the width by 2, which is why I can't get your example working.
QuLogic commented on Sep 18, 2018
'M8[D]' has a single day resolution, and your array goes up by one, so dividing by two rounds down to 0. You'd need to at least change it to hourly resolution.
jklymak commented on Sep 18, 2018
@dstansby, not sure of the approach you are taking here, but I think as @timhoffm says, we should add timedelta64 to dates.py, and register it. Then all the math can be done in floating point numbers...
dstansby commented on Sep 18, 2018
Trying to take that approach, but running into the problem that np.timedelta64 is a subclass of numbers.Number, so the conversion interface assumes it doesn't need converting:
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