When trying to add a rectangle patch with a hatch pattern to a plot it seems that it is impossible to set the keyword argument edgecolor to 'none' when also specifying a hatch value.
In other words I am trying to add a hatched rectangle WITHOUT an edge but WITH a pattern filling. This doesnt seem to work. The pattern only shows up if I also allow an edge to be drawn around the rectangle patch.
Any help on how to achieve the desired behaviour?
You should use the linewidth argument, which has to be set to zero.
Example (based on your other question's answer):
import matplotlib.pyplot as plt
import matplotlib.patches as patches
import numpy as np
fig = plt.figure()
ax = fig.add_subplot(111)
# generate some data:
x,y = np.meshgrid(np.linspace(0,1),np.linspace(0,1))
z = np.ma.masked_array(x**2-y**2,mask=y>-x+1)
# plot your masked array
ax.contourf(z)
# plot a patch
p = patches.Rectangle((20,20), 20, 20, linewidth=0, fill=None, hatch='///')
ax.add_patch(p)
plt.show()
You'll get this image:
Related
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')
I would like to generate a centered figure legend for subplot(s), for which there is a single label. For my actual use case, the number of subplot(s) is greater than or equal to one; it's possible to have a 2x2 grid of subplots and I would like to use the figure-legend instead of using ax.legend(...) since the same single label entry will apply to each/every subplot.
As a brief and simplified example, consider the code just below:
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(10)
y = np.sin(x)
fig, ax = plt.subplots()
ax.plot(x, y, color='orange', label='$f(x) = sin(x)$')
fig.subplots_adjust(bottom=0.15)
fig.legend(mode='expand', loc='lower center')
plt.show()
plt.close(fig)
This code will generate the figure seen below:
I would like to use the mode='expand' kwarg to make the legend span the entire width of the subplot(s); however, doing so prevents the label from being centered. As an example, removing this kwarg from the code outputs the following figure.
Is there a way to use both mode='expand' and also have the label be centered (since there is only one label)?
EDIT:
I've tried using the bbox_to_anchor kwargs (as suggested in the docs) as an alternative to mode='expand', but this doesn't work either. One can switch out the fig.legend(...) line for the line below to test for yourself.
fig.legend(loc='lower center', bbox_to_anchor=(0, 0, 1, 0.5))
The handles and labels are flush against the left side of the legend. There is no mechanism to allow for aligning them.
A workaround could be to use 3 columns of legend handles and fill the first and third with a transparent handle.
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(10)
y = np.sin(x)
fig, ax = plt.subplots()
fig.subplots_adjust(bottom=0.15)
line, = ax.plot(x, y, color='orange', label='$f(x) = sin(x)$')
proxy = plt.Rectangle((0,0),1,1, alpha=0)
fig.legend(handles=[proxy, line, proxy], mode='expand', loc='lower center', ncol=3)
plt.show()
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')
I have created a histogram with matplotlib using the pyplot.hist() function. I would like to add a Poison error square root of bin height (sqrt(binheight)) to the bars. How can I do this?
The return tuple of .hist() includes return[2] -> a list of 1 Patch objects. I could only find out that it is possible to add errors to bars created via pyplot.bar().
Indeed you need to use bar. You can use to output of hist and plot it as a bar:
import numpy as np
import pylab as plt
data = np.array(np.random.rand(1000))
y,binEdges = np.histogram(data,bins=10)
bincenters = 0.5*(binEdges[1:]+binEdges[:-1])
menStd = np.sqrt(y)
width = 0.05
plt.bar(bincenters, y, width=width, color='r', yerr=menStd)
plt.show()
Alternative Solution
You can also use a combination of pyplot.errorbar() and drawstyle keyword argument. The code below creates a plot of the histogram using a stepped line plot. There is a marker in the center of each bin and each bin has the requisite Poisson errorbar.
import numpy
import pyplot
x = numpy.random.rand(1000)
y, bin_edges = numpy.histogram(x, bins=10)
bin_centers = 0.5*(bin_edges[1:] + bin_edges[:-1])
pyplot.errorbar(
bin_centers,
y,
yerr = y**0.5,
marker = '.',
drawstyle = 'steps-mid-'
)
pyplot.show()
My personal opinion
When plotting the results of multiple histograms on the the same figure, line plots are easier to distinguish. In addition, they look nicer when plotting with a yscale='log'.
I am quite used to working with matlab and now trying to make the shift matplotlib and numpy. Is there a way in matplotlib that an image you are plotting occupies the whole figure window.
import numpy as np
import matplotlib.pyplot as plt
# get image im as nparray
# ........
plt.figure()
plt.imshow(im)
plt.set_cmap('hot')
plt.savefig("frame.png")
I want the image to maintain its aspect ratio and scale to the size of the figure ... so when I do savefig it exactly the same size as the input figure, and it is completely covered by the image.
Thanks.
I did this using the following snippet.
#!/usr/bin/env python
import numpy as np
import matplotlib.cm as cm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt
from pylab import *
delta = 0.025
x = y = np.arange(-3.0, 3.0, delta)
X, Y = np.meshgrid(x, y)
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
Z = Z2-Z1 # difference of Gaussians
ax = Axes(plt.gcf(),[0,0,1,1],yticks=[],xticks=[],frame_on=False)
plt.gcf().delaxes(plt.gca())
plt.gcf().add_axes(ax)
im = plt.imshow(Z, cmap=cm.gray)
plt.show()
Note the grey border on the sides is related to the aspect rario of the Axes which is altered by setting aspect='equal', or aspect='auto' or your ratio.
Also as mentioned by Zhenya in the comments Similar StackOverflow Question
mentions the parameters to savefig of bbox_inches='tight' and pad_inches=-1 or pad_inches=0
You can use a function like the one below.
It calculates the needed size for the figure (in inches) according to the resolution in dpi you want.
import numpy as np
import matplotlib.pyplot as plt
def plot_im(image, dpi=80):
px,py = im.shape # depending of your matplotlib.rc you may
have to use py,px instead
#px,py = im[:,:,0].shape # if image has a (x,y,z) shape
size = (py/np.float(dpi), px/np.float(dpi)) # note the np.float()
fig = plt.figure(figsize=size, dpi=dpi)
ax = fig.add_axes([0, 0, 1, 1])
# Customize the axis
# remove top and right spines
ax.spines['right'].set_color('none')
ax.spines['left'].set_color('none')
ax.spines['top'].set_color('none')
ax.spines['bottom'].set_color('none')
# turn off ticks
ax.xaxis.set_ticks_position('none')
ax.yaxis.set_ticks_position('none')
ax.xaxis.set_ticklabels([])
ax.yaxis.set_ticklabels([])
ax.imshow(im)
plt.show()
Here's a minimal object-oriented solution:
fig = plt.figure(figsize=(8, 8))
ax = fig.add_axes([0, 0, 1, 1], frameon=False, xticks=[], yticks=[])
Testing it out with
ax.imshow([[0]])
fig.savefig('test.png')
saves out a uniform purple block.
edit: As #duhaime points out below, this requires the figure to have the same aspect as the axes.
If you'd like the axes to resize to the figure, add aspect='auto' to imshow.
If you'd like the figure to resize to be resized to the axes, add
from matplotlib import tight_bbox
bbox = fig.get_tightbbox(fig.canvas.get_renderer())
tight_bbox.adjust_bbox(fig, bbox, fig.canvas.fixed_dpi)
after the imshow call. This is the important bit of matplotlib's tight_layout functionality which is implicitly called by things like Jupyter's renderer.