matplotlib -- interactively select points or locations? - matplotlib

In R, there is a function locator which is like Matlab's ginput where you can click on the figure with a mouse and select any x,y coordinate. In addition, there is a function called identify(x,y) where if you give it a set of points x,y that you have plotted and then click on the figure, it will return the index of the x,y point which lies nearest (within an adjustable tolerance) to the location you have selected (or multiple indices, if multiple points are selected). Is there such a functionality in Matplotlib?

You may want to use a pick event :
fig = figure()
ax1 = fig.add_subplot(111)
ax1.set_title('custom picker for line data')
line, = ax1.plot(rand(100), rand(100), 'o', picker=line_picker)
fig.canvas.mpl_connect('pick_event', onpick2)
Tolerance set by picker parameter there:
line, = ax1.plot(rand(100), 'o', picker=5) # 5 points tolerance

from __future__ import print_function
import matplotlib.pyplot as plt
from matplotlib.lines import Line2D
from matplotlib.patches import Rectangle
from matplotlib.text import Text
from matplotlib.image import AxesImage
import numpy as np
from numpy.random import rand
if 1:
fig, ax = plt.subplots()
ax.set_title('click on points', picker=True)
ax.set_ylabel('ylabel', picker=True, bbox=dict(facecolor='red'))
line, = ax.plot(rand(100), 'o', picker=5)
def onpick1(event):
if isinstance(event.artist, Line2D):
thisline = event.artist
xdata = thisline.get_xdata()
ydata = thisline.get_ydata()
ind = event.ind
print 'X='+str(np.take(xdata, ind)[0]) # Print X point
print 'Y='+str(np.take(ydata, ind)[0]) # Print Y point
fig.canvas.mpl_connect('pick_event', onpick1)

Wow many years have passed! Now matplotlib also support the ginput function which has almost the same API as Matlab. So there is no need to hack by the mpl-connect and so on any more! (https://matplotlib.org/3.1.1/api/_as_gen/matplotlib.pyplot.ginput.html) For instance,
plt.ginput(4)
will let the user to select 4 points.

The ginput() is a handy tool to select x, y coordinates of any random point from a plotted window, however that point may not belong to the plotted data. To select x, y coordinates of a point from the plotted data, an efficient tool still is to use 'pick_event' property with mpl_connect as the example given in the documentation. For example:
import matplotlib.pyplot as plt
import numpy as np
from numpy.random import rand
fig, ax = plt.subplots()
ax.plot(rand(100), rand(100), picker=3)
# 3, for example, is tolerance for picker i.e, how far a mouse click from
# the plotted point can be registered to select nearby data point/points.
def on_pick(event):
global points
line = event.artist
xdata, ydata = line.get_data()
print('selected point is:',np.array([xdata[ind], ydata[ind]]).T)
cid = fig.canvas.mpl_connect('pick_event', on_pick)
The last line above will connect the plot with the 'pick_event' and the corrdinates of the nearest plot points will keep printing after each mouse click on plot, to end this process, we need to use mpl_disconnect as:
fig.canvas.mpl_disconnect(cid)

Related

How can I place the y-axis origin at 0? [duplicate]

I want to draw a figure in matplotib where the axis are displayed within the plot itself not on the side
I have tried the following code from here:
import math
import numpy as np
import matplotlib.pyplot as plt
def sigmoid(x):
a = []
for item in x:
a.append(1/(1+math.exp(-item)))
return a
x = np.arange(-10., 10., 0.2)
sig = sigmoid(x)
plt.plot(x,sig)
plt.show()
The above code displays the figure like this:
What I would like to draw is something as follows (image from Wikipedia)
This question describes a similar problem, but it draws a reference line in the middle but no axis.
One way to do it is using spines:
import math
import numpy as np
import matplotlib.pyplot as plt
def sigmoid(x):
a = []
for item in x:
a.append(1/(1+math.exp(-item)))
return a
x = np.arange(-10., 10., 0.2)
sig = sigmoid(x)
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
# Move left y-axis and bottom x-axis to centre, passing through (0,0)
ax.spines['left'].set_position('center')
ax.spines['bottom'].set_position('center')
# Eliminate upper and right axes
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
# Show ticks in the left and lower axes only
ax.xaxis.set_ticks_position('bottom')
ax.yaxis.set_ticks_position('left')
plt.plot(x,sig)
plt.show()
shows:
Basically, I want to comment on the accepted answer (but my rep doesn't allow that).
The use of
ax.spines['bottom'].set_position('center')
draws the x-axes such that it intersect the y-axes in its center. In case of asymmetric ylim this means that x-axis passes NOT through y=0. Jblasco's answer has this drawback, the intersect is at y=0.5 (the center between ymin=0.0 and ymax=1.0)
However, the reference plot of the original question has axes that intersect each other at 0.0 (which is somehow conventional or at least common).
To achieve this behaviour,
ax.spines['bottom'].set_position('zero')
has to be used.
See the following example, where 'zero' makes the axes intersect at 0.0 despite asymmetrically ranges in both x and y.
import numpy as np
import matplotlib.pyplot as plt
#data generation
x = np.arange(-10,20,0.2)
y = 1.0/(1.0+np.exp(-x)) # nunpy does the calculation elementwise for you
fig, [ax0, ax1] = plt.subplots(ncols=2, figsize=(8,4))
# Eliminate upper and right axes
ax0.spines['top'].set_visible(False)
ax0.spines['right'].set_visible(False)
# Show ticks on the left and lower axes only
ax0.xaxis.set_tick_params(bottom='on', top='off')
ax0.yaxis.set_tick_params(left='on', right='off')
# Move remaining spines to the center
ax0.set_title('center')
ax0.spines['bottom'].set_position('center') # spine for xaxis
# - will pass through the center of the y-values (which is 0)
ax0.spines['left'].set_position('center') # spine for yaxis
# - will pass through the center of the x-values (which is 5)
ax0.plot(x,y)
# Eliminate upper and right axes
ax1.spines['top'].set_visible(False)
ax1.spines['right'].set_visible(False)
# Show ticks on the left and lower axes only (and let them protrude in both directions)
ax1.xaxis.set_tick_params(bottom='on', top='off', direction='inout')
ax1.yaxis.set_tick_params(left='on', right='off', direction='inout')
# Make spines pass through zero of the other axis
ax1.set_title('zero')
ax1.spines['bottom'].set_position('zero')
ax1.spines['left'].set_position('zero')
ax1.set_ylim(-0.4,1.0)
# No ticklabels at zero
ax1.set_xticks([-10,-5,5,10,15,20])
ax1.set_yticks([-0.4,-0.2,0.2,0.4,0.6,0.8,1.0])
ax1.plot(x,y)
plt.show()
Final remark: If ax.spines['bottom'].set_position('zero') is used but zerois not within the plotted y-range, then the axes is shown at the boundary of the plot closer to zero.
The title of this question is how to draw the spine in the middle and the accepted answer does exactly that but what you guys draw is the sigmoid function and that one passes through y=0.5. So I think what you want is the spine centered according to your data. Matplotlib offers the spine position data for that (see documentation)
import numpy as np
import matplotlib.pyplot as plt
def sigmoid(x):
return 1 / (1 + np.exp(-x))
sigmoid = np.vectorize(sigmoid) #vectorize function
values=np.linspace(-10, 10) #generate values between -10 and 10
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1)
#spine placement data centered
ax.spines['left'].set_position(('data', 0.0))
ax.spines['bottom'].set_position(('data', 0.0))
ax.spines['right'].set_color('none')
ax.spines['top'].set_color('none')
plt.plot(values, sigmoid(values))
plt.show()
Looks like this (Github):
You can simply add:
plt.axhline()
plt.axvline()
It's not fixed to the center, but it does the job very easily.
Working example:
import matplotlib.pyplot as plt
import numpy as np
def f(x):
return np.sin(x) / (x/100)
delte = 100
Xs = np.arange(-delte, +delte +1, step=0.01)
Ys = np.array([f(x) for x in Xs])
plt.axhline(color='black', lw=0.5)
plt.axvline(color='black', lw=0.5)
plt.plot(Xs, Ys)
plt.show()
If you use matplotlib >= 3.4.2, you can use Pandas syntax and do it in only one line:
plt.gca().spines[:].set_position('center')
You might find it cleaner to do it in 3 lines:
ax = plt.gca()
ax.spines[['top', 'right']].set_visible(False)
ax.spines[['left', 'bottom']].set_position('center')
See documentation here.
Check your matplotlib version with pip freeze and update it with pip install -U matplotlib.
According to latest MPL Documentation:
ax = plt.axes()
ax.spines.left.set_position('zero')
ax.spines.bottom.set_position('zero')

how to set the distance between bars and axis using matplot lib [duplicate]

So currently learning how to import data and work with it in matplotlib and I am having trouble even tho I have the exact code from the book.
This is what the plot looks like, but my question is how can I get it where there is no white space between the start and the end of the x-axis.
Here is the code:
import csv
from matplotlib import pyplot as plt
from datetime import datetime
# Get dates and high temperatures from file.
filename = 'sitka_weather_07-2014.csv'
with open(filename) as f:
reader = csv.reader(f)
header_row = next(reader)
#for index, column_header in enumerate(header_row):
#print(index, column_header)
dates, highs = [], []
for row in reader:
current_date = datetime.strptime(row[0], "%Y-%m-%d")
dates.append(current_date)
high = int(row[1])
highs.append(high)
# Plot data.
fig = plt.figure(dpi=128, figsize=(10,6))
plt.plot(dates, highs, c='red')
# Format plot.
plt.title("Daily high temperatures, July 2014", fontsize=24)
plt.xlabel('', fontsize=16)
fig.autofmt_xdate()
plt.ylabel("Temperature (F)", fontsize=16)
plt.tick_params(axis='both', which='major', labelsize=16)
plt.show()
There is an automatic margin set at the edges, which ensures the data to be nicely fitting within the axis spines. In this case such a margin is probably desired on the y axis. By default it is set to 0.05 in units of axis span.
To set the margin to 0 on the x axis, use
plt.margins(x=0)
or
ax.margins(x=0)
depending on the context. Also see the documentation.
In case you want to get rid of the margin in the whole script, you can use
plt.rcParams['axes.xmargin'] = 0
at the beginning of your script (same for y of course). If you want to get rid of the margin entirely and forever, you might want to change the according line in the matplotlib rc file:
axes.xmargin : 0
axes.ymargin : 0
Example
import seaborn as sns
import matplotlib.pyplot as plt
tips = sns.load_dataset('tips')
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10, 4))
tips.plot(ax=ax1, title='Default Margin')
tips.plot(ax=ax2, title='Margins: x=0')
ax2.margins(x=0)
Alternatively, use plt.xlim(..) or ax.set_xlim(..) to manually set the limits of the axes such that there is no white space left.
If you only want to remove the margin on one side but not the other, e.g. remove the margin from the right but not from the left, you can use set_xlim() on a matplotlib axes object.
import seaborn as sns
import matplotlib.pyplot as plt
import math
max_x_value = 100
x_values = [i for i in range (1, max_x_value + 1)]
y_values = [math.log(i) for i in x_values]
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(10, 4))
sn.lineplot(ax=ax1, x=x_values, y=y_values)
sn.lineplot(ax=ax2, x=x_values, y=y_values)
ax2.set_xlim(-5, max_x_value) # tune the -5 to your needs

Draw colorbar with twin scales

I'd like to draw a (vertical) colorbar, which has two different scales (corresponding to two different units for the same quantity) on each side. Think Fahrenheit on one side and Celsius on the other side. Obviously, I'd need to specify the ticks for each side individually.
Any idea how I can do this?
That should get you started:
import matplotlib.pyplot as plt
import numpy as np
# generate random data
x = np.random.randint(0,200,(10,10))
plt.pcolormesh(x)
# create the colorbar
# the aspect of the colorbar is set to 'equal', we have to set it to 'auto',
# otherwise twinx() will do weird stuff.
cbar = plt.colorbar()
pos = cbar.ax.get_position()
cbar.ax.set_aspect('auto')
# create a second axes instance and set the limits you need
ax2 = cbar.ax.twinx()
ax2.set_ylim([-2,1])
# resize the colorbar (otherwise it overlays the plot)
pos.x0 +=0.05
cbar.ax.set_position(pos)
ax2.set_position(pos)
plt.show()
If you create a subplot for the colorbar, you can create a twin axes for that subplot and manipulate it like a normal axes.
import matplotlib.colors as mcolors
import matplotlib.pyplot as plt
import numpy as np
x = np.linspace(-1,2.7)
X,Y = np.meshgrid(x,x)
Z = np.exp(-X**2-Y**2)*.9+0.1
fig, (ax, cax) = plt.subplots(ncols=2, gridspec_kw={"width_ratios":[15,1]})
im =ax.imshow(Z, vmin=0.1, vmax=1)
cbar = plt.colorbar(im, cax=cax)
cax2 = cax.twinx()
ticks=np.arange(0.1,1.1,0.1)
iticks=1./np.array([10,3,2,1.5,1])
cbar.set_ticks(ticks)
cbar.set_label("z")
cbar.ax.yaxis.set_label_position("left")
cax2.set_ylim(0.1,1)
cax2.set_yticks(iticks)
cax2.set_yticklabels(1./iticks)
cax2.set_ylabel("1/z")
plt.show()
Note that in newer version of matplotlib, the above answers no long work (as #Ryan Skene pointed out). I'm using v3.3.2. The secondary_yaxis function works for the colorbars in the same way as for regular plot axes and gives one colorbar with two scales: https://matplotlib.org/stable/api/_as_gen/matplotlib.axes.Axes.secondary_yaxis.html#matplotlib.axes.Axes.secondary_yaxis
import matplotlib.pyplot as plt
import numpy as np
# generate random data
x = np.random.randint(0,200,(10,10)) #let's assume these are temperatures in Fahrenheit
im = plt.imshow(x)
# create the colorbar
cbar = plt.colorbar(im,pad=0.1) #you may need to adjust this padding for the secondary colorbar label[enter image description here][1]
cbar.set_label('Temperature ($^\circ$F)')
# define functions that relate the two colorbar scales
# e.g., Celcius to Fahrenheit and vice versa
def F_to_C(x):
return (x-32)*5/9
def C_to_F(x):
return (x*9/5)+32
# create a second axes
cbar2 = cbar.ax.secondary_yaxis('left',functions=(F_to_C,C_to_F))
cbar2.set_ylabel('Temperatrue ($\circ$C)')
plt.show()
I am using an inset axis for my colorbar and, for some reason, I found the above to answers no longer worked as of v3.4.2. The twinx took up the entire original subplot.
So I just replicated the inset axis (instead of using twinx) and increased the zorder on the original inset.
axkws = dict(zorder=2)
cax = inset_axes(
ax, width="100%", height="100%", bbox_to_anchor=bbox,
bbox_transform=ax.transAxes, axes_kwargs=axkws
)
cbar = self.fig.colorbar(mpl.cm.ScalarMappable(cmap=cmap), cax=cax)
cbar.ax.yaxis.set_ticks_position('left')
caxx = inset_axes(
ax, width="100%", height="100%",
bbox_to_anchor=bbox, bbox_transform=ax.transAxes
)
caxx.yaxis.set_ticks_position('right')

Matplotlib log colorbar minor ticks disappear when range is less than a decade

My users sometimes wish to see log scaling of the values of a 2-d plot, even though the data spans less than one decade. I'm able to make plots using 'pcolormesh' or 'imshow' using the
norm=LogNorm(vmin=minimum,vmax=maximum)
parameter and accurately show log scaled 'intensity' values. I would like the 'colorbar' to show some minor ticks and tick labels, but when minimum and maximum span less than a decade, no matter what I do there is only one tick value displayed. I tried the suggestion in this SO posting:
Minor ticks in matplotlib's colorbar
As adapted in the following snippet:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.colors import LogNorm
# fill grid
x = np.linspace(1,10,10)
y = np.linspace(1,10,10)
X, Y = np.meshgrid(x,y)
Z = np.abs(X/10 + Y/10)
# plot
f, ax = plt.subplots()
p = plt.pcolormesh(X, Y, Z, norm=LogNorm(), vmin=2e-1, vmax=1)
cb = plt.colorbar(p, ax=ax)
cb.ax.minorticks_on()
plt.show()
But there are no minor ticks, labeled or otherwise:
I have also tried the following:
from matplotlib.ticker import LogFormatterMathtext
from matplotlib.ticker import LogLocator
from matplotlib.ticker import LogFormatter
import numpy as nmp
import matplotlib.pyplot as pyp
'''
<snip>
'''
ccbb=pyp.colorbar(label='ohms')
ccbb.ax.yaxis.set_minor_locator(LogLocator(subs=nmp.arange(2,10)))
# AND/OR
# ccbb.ax.yaxis.set_minor_locator(LogLocator(subs=[0.2,0.5,1.0]))
ccbb.ax.yaxis.set_minor_formatter(LogFormatterMathtext())
ccbb.update_ticks()
'''
<snip>
'''
And several other things, which I haven't saved. All of which yield the same result with the colorbar missing any but the single decade tick / label. The documentation for the ticker class is pretty impenetrable:
http://matplotlib.org/api/ticker_api.html
Especially the following statement about LogFormatter parameter labelOnlyBase:
"base is used to locate the decade tick, which will be the only one to be labeled if labelOnlyBase is False" Neither False nor True cause more than the base to be ticked, I suppose that's because this refers to the Major ticks, But why in the world can't I get the minor ticks or labels??
Any advice would be appreciated.
Matplotlib colorbars don't seem to do minor ticks in log scale. Using the method in this answer works, though it's a bit inconvenient - one day this will be automatic, but for now you have to organise the minor tick values by hand (np.arange(2, 10)/10. in this case, but you'd have to append np.arange(2, 10) if your values went up to 10)
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.colors import LogNorm
# fill grid
x = np.linspace(1,10,10)
y = np.linspace(1,10,10)
X, Y = np.meshgrid(x,y)
# Z = np.abs(X + Y)
Z = np.abs(X/10 + Y/10)
# plot
f, ax = plt.subplots()
p = plt.pcolormesh(X, Y, Z, norm=LogNorm(), vmin=2e-1, vmax=1)
cb = plt.colorbar(p, ax=ax)
# cb.ax.minorticks_on()
# We need to nomalize the tick locations so that they're in the range from 0-1...
minorticks = p.norm(np.arange(2, 10)/10.)
cb.ax.yaxis.set_ticks(minorticks, minor=True)
plt.show()
The minorticks_on() method wasn't doing anything, so I commented it out.

Enumerate plots in matplotlib figure

In a matplotlib figure I would like to enumerate all (sub)plots with a), b), c) and so on. Is there a way to do this automatically?
So far I use the individual plots' titles, but that is far from ideal as I want the number to be left aligned, while an optional real title should be centered on the figure.
import string
from itertools import cycle
from six.moves import zip
def label_axes(fig, labels=None, loc=None, **kwargs):
"""
Walks through axes and labels each.
kwargs are collected and passed to `annotate`
Parameters
----------
fig : Figure
Figure object to work on
labels : iterable or None
iterable of strings to use to label the axes.
If None, lower case letters are used.
loc : len=2 tuple of floats
Where to put the label in axes-fraction units
"""
if labels is None:
labels = string.ascii_lowercase
# re-use labels rather than stop labeling
labels = cycle(labels)
if loc is None:
loc = (.9, .9)
for ax, lab in zip(fig.axes, labels):
ax.annotate(lab, xy=loc,
xycoords='axes fraction',
**kwargs)
example usage:
from matplotlib import pyplot as plt
fig, ax_lst = plt.subplots(3, 3)
label_axes(fig, ha='right')
plt.draw()
fig, ax_lst = plt.subplots(3, 3)
label_axes(fig, ha='left')
plt.draw()
This seems useful enough to me that I put this in a gist : https://gist.github.com/tacaswell/9643166
I wrote a function to do this automatically, where the label is introduced as a legend:
import numpy
import matplotlib.pyplot as plt
def setlabel(ax, label, loc=2, borderpad=0.6, **kwargs):
legend = ax.get_legend()
if legend:
ax.add_artist(legend)
line, = ax.plot(numpy.NaN,numpy.NaN,color='none',label=label)
label_legend = ax.legend(handles=[line],loc=loc,handlelength=0,handleheight=0,handletextpad=0,borderaxespad=0,borderpad=borderpad,frameon=False,**kwargs)
label_legend.remove()
ax.add_artist(label_legend)
line.remove()
fig,ax = plt.subplots()
ax.plot([1,2],[1,2])
setlabel(ax, '(a)')
plt.show()
The location of the label can be controlled with loc argument, the distance to the axis can be controlled with borderpad argument (negative value pushes the label to be outside the figure), and other options available to legend also can be used, such as fontsize. The above script gives such figure:
A super quick way to do this is to take advantage of the fact that chr() casts integers to characters. Since a-z fall in the range 97-122, one can do the following:
import matplotlib.pyplot as plt
fig,axs = plt.subplots(2,2)
for i,ax in enumerate(axs.flat, start=97):
ax.plot([0,1],[0,1])
ax.text(0.05,0.9,chr(i)+')', transform=ax.transAxes)
which produces: