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FIX: make sure we have more than 1 tick with small log ranges #18754

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Merged
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Oct 21, 2020
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FIX: make sure we have more than 1 tick with small log ranges #18754

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10 changes: 10 additions & 0 deletions lib/matplotlib/tests/test_ticker.py
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Original file line number Diff line number Diff line change
Expand Up @@ -1362,3 +1362,13 @@ def test_bad_locator_subs(sub):
ll = mticker.LogLocator()
with pytest.raises(ValueError):
ll.subs(sub)


@pytest.mark.parametrize('numticks', [1, 2, 3, 9])
@pytest.mark.style('default')
def test_small_range_loglocator(numticks):
ll = mticker.LogLocator()
ll.set_params(numticks=numticks)
for top in [5, 7, 9, 11, 15, 50, 100, 1000]:
ticks = ll.tick_values(.5, top)
assert (np.diff(np.log10(ll.tick_values(6, 150))) == 1).all()
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@tacaswell There seems to be a clear typo in the test here: you don't do anything with ticks, and the assert is the same on each loop iteration. Do you remember what you intended to do?

7 changes: 7 additions & 0 deletions lib/matplotlib/ticker.py
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Expand Up @@ -2479,6 +2479,13 @@ def tick_values(self, vmin, vmax):
if mpl.rcParams['_internal.classic_mode'] else
(numdec + 1) // numticks + 1)

# if we have decided that the stride is as big or bigger than
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Is this really the right solution to the original problem? To me, it seems that numticks is 1 because there isn't room for more ticks. Do we really want ticks that overwrite each other for very narrow axes?

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@timhoffm timhoffm Oct 17, 2020 •
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If you are very narrow, there's no good solution. Either you have only one tick, which makes your scale uninterpretable. Or you have overlap. 🤷

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Sorry, let me rephrase: We clip numticks to 2, so why don't we have 2 ticks?

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numticks is an upper bound and only "best effort" at that.

The issue is that based on numtick and a rounded "in" version of the range we pick a stride between major ticks (which is meant to catch things like 30 orders of magnitude with 9 ticks). If we propose 3 ticks, have a range of [5, 150], then we "round out" to [1, 100] for the range giving us numdec = 2 and numticks = 3 and a stride of (3 / 3) + 1 == 2 which means we put ticks at (10 ^ {-2, 0, 2, 4}) which in turn means that only the tick at 100 is actually shown. Additionally because we only show the minor ticks if the major stride is equal to 1 you end up with exactly 1 tick and 1 label shown even though the users / auto logic asked for up to 3.

I think that this fix (which fixes the values after doing a bunch of heuristics) is a better and more reliable fix than making the heuristics more complicated.

# the range, clip the stride back to the available range - 1
# with a floor of 1. This prevents getting axis with only 1 tick
# visible.
if stride >= numdec:
stride = max(1, numdec - 1)

# Does subs include anything other than 1? Essentially a hack to know
# whether we're a major or a minor locator.
have_subs = len(subs) > 1 or (len(subs) == 1 and subs[0] != 1.0)
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