Reputation: 20007
I'm creating loglog plots with matplotlib. As can be seen in the figure below, the default ticks are chosen badly (at best); the right y-axis doesn't even have any at all (it does in the linear equivalent) and both x-axis have only one.
Is there a way to get a reasonable number of ticks with labels, without specifying them by hand for every plot?
EDIT: the exact code is too long, but here's a short example of the problem:
x = linspace(4, 18, 20)
y = 1 / (x ** 4)
fig = figure()
ax = fig.add_axes([.1, .1, .8, .8])
ax.loglog(x, y)
ax.set_xlim([4, 18])
ax2 = ax.twiny()
ax2.set_xlim([4 / 3., 18 / 3.])
ax2.set_xscale('log')
show()
Upvotes: 3
Views: 3363
Reputation: 20007
In the end, this is the best I could come up with with the help of other answers here and elsewere is this:
On the left, x and y vary over only a part of an order of magnitude, with labels working out fairly well. On the left, x varies between 1 and 2 orders of magnitude. It works okay, but the method is reaching it's limit. The y values vary many orders of magnitude and the standard labels are used automatically.
from matplotlib import ticker
from numpy import linspace, logspace, log10, floor
from warnings import warn
def round_to_n(x, n):
''' http://stackoverflow.com/questions/3410976/how-to-round-a-number-to-significant-figures-in-python '''
return round(x, -int(floor(log10(abs(x)))) + (n - 1))
def ticks_log_format(value, index):
''' http://stackoverflow.com/questions/19239297/matplotlib-bad-ticks-labels-for-loglog-twin-axis '''
pwr = floor(log10(value))
base = value / (10 ** pwr)
if pwr == 0 or pwr == 1:
return '${0:d}$'.format(int(value))
if -3 <= pwr < 0:
return '${0:.3g}$'.format(value)
if 0 < pwr <= 3:
return '${0:d}$'.format(int(value))
else:
return '${0:d}\\times10^{{{1:d}}}$'.format(int(base), int(pwr))
def calc_ticks(domain, tick_count, equidistant):
if equidistant:
ticks = logspace(log10(domain[0]), log10(domain[1]), num = tick_count, base = 10)
else:
ticks = linspace(domain[0], domain[1], num = tick_count)
for n in range(1, 6):
if len(set(round_to_n(tick, n) for tick in ticks)) == tick_count:
break
return list(round_to_n(tick, n) for tick in ticks)
''' small domain log ticks '''
def sdlt_x(ax, domain, tick_count = 4, equidistant = True):
''' http://stackoverflow.com/questions/3410976/how-to-round-a-number-to-significant-figures-in-python '''
if min(domain) <= 0:
warn('domain %g-%g contains values lower than 0' % (domain[0], domain[1]))
domain = [max(value, 0.) for value in domain]
ax.set_xscale('log')
ax.set_xlim(domain)
ax.xaxis.set_major_formatter(ticker.FuncFormatter(ticks_log_format))
if log10(max(domain) / min(domain)) > 1.7:
return
ticks = calc_ticks(domain, tick_count = tick_count, equidistant = equidistant)
ax.set_xticks(ticks)
''' any way to prevent this code duplication? '''
def sdlt_y(ax, domain, tick_count = 5, equidistant = True):
''' http://stackoverflow.com/questions/3410976/how-to-round-a-number-to-significant-figures-in-python '''
if min(domain) <= 0:
warn('domain %g-%g contains values lower than 0' % (domain[0], domain[1]))
domain = [max(value, 1e-8) for value in domain]
ax.set_yscale('log')
ax.set_ylim(domain)
ax.yaxis.set_major_formatter(ticker.FuncFormatter(ticks_log_format))
if log10(max(domain) / min(domain)) > 1.7:
return
ticks = calc_ticks(domain, tick_count = tick_count, equidistant = equidistant)
ax.set_yticks(ticks)
''' demo '''
fig, (ax1, ax2,) = plt.subplots(1, 2)
for mi, ma, ax in ((100, 130, ax1,), (10, 400, ax2,), ):
x = np.linspace(mi, ma, 50)
y = 1 / ((x + random(50) * 0.1 * (ma - mi)) ** 4)
ax.scatter(x, y)
sdlt_x(ax, (mi, ma, ))
sdlt_y(ax, (min(y), max(y), ))
show()
EDIT: updated with an option to make labels equidistant (so the values are logarithmic, but the visible positions are equidistant).
Upvotes: 0
Reputation: 8668
I've been fighting with something like what you show (only one major tick in the axis range). None of the matplotlib tick formatter satisfied me, so I use matplotlib.ticker.FuncFormatter
to achieve what I wanted. I haven't tested with twin axes, but my feeling is that it should work anyway.
import matplotlib.pyplot as plt
from matplotlib import ticker
import numpy as np
#@Mark: thanks for the suggestion :D
mi, ma, conv = 4, 8, 1./3.
x = np.linspace(mi, ma, 20)
y = 1 / (x ** 4)
fig, ax = plt.subplots()
ax.plot(x, y) # plot the lines
ax.set_xscale('log') #convert to log
ax.set_yscale('log')
ax.set_xlim([0.2, 1.8]) #large enough, but should show only 1 tick
def ticks_format(value, index):
"""
This function decompose value in base*10^{exp} and return a latex string.
If 0<=value<99: return the value as it is.
if 0.1<value<0: returns as it is rounded to the first decimal
otherwise returns $base*10^{exp}$
I've designed the function to be use with values for which the decomposition
returns integers
"""
exp = np.floor(np.log10(value))
base = value/10**exp
if exp == 0 or exp == 1:
return '${0:d}$'.format(int(value))
if exp == -1:
return '${0:.1f}$'.format(value)
else:
return '${0:d}\\times10^{{{1:d}}}$'.format(int(base), int(exp))
# here specify which minor ticks per decate you want
# likely all of them give you a too crowed axis
subs = [1., 3., 6.]
# set the minor locators
ax.xaxis.set_minor_locator(ticker.LogLocator(subs=subs))
ax.yaxis.set_minor_locator(ticker.LogLocator(subs=subs))
# remove the tick labels for the major ticks:
# if not done they will be printed with the custom ones (you don't want it)
# plus you want to remove them to avoid font missmatch: the above function
# returns latex string, and I don't know how matplotlib does exponents in labels
ax.xaxis.set_major_formatter(ticker.NullFormatter())
ax.yaxis.set_major_formatter(ticker.NullFormatter())
# set the desired minor tick labels using the above function
ax.xaxis.set_minor_formatter(ticker.FuncFormatter(ticks_format))
ax.yaxis.set_minor_formatter(ticker.FuncFormatter(ticks_format))
The figure that I get is the following :
Of course you can set different minor locators for x and y axis and you can wrap everything from ticks_format
to the end into a function that accepts an axes instance ax
and subs
or subsx
and subsy
as input parameters.
I hope that this helps you
Upvotes: 1