I have two ndarrays: Mat, labels
Currently I display Mat:
plt.imshow(Mat, cmap='gray', vmin=0, vmax=1, interpolation='None')
labels has the same shape as Mat, and lables[(i,j)] contains a label of Mat[(i,j)].
How can I show the label on each pixel?
The easiest approach uses Seaborn's heatmap. When annot=True it prints the data values into the cells. But annot= can also be a matrix of labels. In that case it is important to set the print format to string (fmt='s'). annot_kws= can set additional keywords, such as fontsize or color. x and yticklabels can be incorporated in the call to heatmap(), or be set afterwards using matplotlib.
An important benefit of the default coloring is that Sorn uses black on the light colored cells and white on the dark cells.
Here is an example that uses some utf8 characters as labels.
from matplotlib import pyplot as plt
import numpy as np
import seaborn as sns
M, N = 5, 10
mat = np.random.rand(M, N)
labels = np.random.choice(['X', '☀', '★', '♛'], size=(M, N))
ax = sns.heatmap(mat, cmap="inferno", annot=labels, annot_kws={'fontsize': 16}, fmt='s')
plt.show()
PS: There is a matplotlib example in the documentation to create something similar without Seaborn. It can be easily adapted to print strings from a different matrix, and also a test can be added to change the color depending on the cell darkness.
Related
Suppose we have this:
import seaborn as sns
import pandas as pd
import numpy as np
samples = 2**13
data = pd.DataFrame({'Values': list(np.random.normal(size=samples)) + list(np.random.uniform(size=samples)),
'Kind': ['Normal'] * samples + ['Uniform'] * samples})
sns.displot(data, hue='Kind', x='Values', fill=True)
I want my Normal's histogram (or KDE) emphasized. I'd like it in red and non transparent in the background. Uniform should have alpha = .5.
How do I specify these style parameters in a "per hue" manner?
It's possible to do it with two separate histplots on the same Axes, as #Redox suggested. We can basically recreate the same plot, but with fine-grade control over colours and alpha. However I had to explicitly pass the number of bins in to get the same plot as yours. I also needed to define the colour for Uniform otherwise a ghost element would be added to the legend! I used C1, meaning the first default colour.
_, ax = plt.subplots()
sns.histplot(data=data[data.Kind=='Normal'], x="Values", ax=ax, label='Normal', color='tab:red',bins=130,alpha=1)
sns.histplot(data=data[data.Kind=='Uniform'], x="Values", ax=ax, label='Uniform', color='C1',bins=17, alpha=.5)
ax.set_xlabel('')
ax.legend()
Note that if you just want to set the colour without alpha you can already do this on a displot via the palette argument - pass in a dictionary of your unique hue values to colour names. However, the alpha that you pass in must be a scalar. I tried to use this clever answer to set colours as RGBA colours which include alpha, which seems to work with other figure level plots in Seaborn. However, displot overrides this and sets the alpha separately!
I don't understand why pyplot print always red color, but variable changing
def showDot(dot):
classColormap = ListedColormap(['#FF0000', '#00FF00', '#000000'])
pl.scatter(dot[0][0],dot[0][1],c=dot[1],cmap=classColormap)
pl.show()
Also when i write c = 2(constant color, but not red) pyplot print red
It is not entirely clear to me what you want the end result to look like, but it seems to me that c has to be some number specifying the intensity. To me, it looks like c=dot[1] would return an array in your code. The following code does produce different colors. Here I have simply defined the intensity acording to the distance from origo. You probably have something else.
I have never used colormaps before but my guess is that it needs to operate on a range of values in order to assign different colors. If you add one and one dot, my guess is that it always assigns the same color from the colormap (the mid value perhaps?)
from matplotlib import pyplot as pl
from matplotlib import colors as cl
import numpy as np
mean = (1, 2)
cov = [[1, 0.2], [.2, 1]]
dots = np.random.multivariate_normal(mean, cov, (2, 20))
# The distance from origo, which will define the intensity
dots_intensity = np.hypot(dots[0,:], dots[1,:])
classColormap = cl.ListedColormap(['#FF0000', '#00FF00', '#000000'])
pl.scatter(dots[0,:], dots[1,:],c=dots_intensity,cmap=classColormap)
hope it helps.
I found this article useful https://medium.com/better-programming/how-to-use-colormaps-with-matplotlib-to-create-colorful-plots-in-python-969b5a892f0c
I want to create a bar-plot in python. I want this plot to be beautiful though and I don't like the looks of python's axes.bar() function. Therefore, I have decided to use plt.vlines(). The challenge here is that my x-data is a list that contains strings and not numerical data. When I plot my graph, the spacing between the two columns (in my example column 2 = 0) is pretty big:
Furthermore, I want a grid. However, I would like to have minor grid lines as well. I know how to get all of this if my data was numerical. But since my x-data contains strings, I don't know how to set x_max. Any suggestions?
Internally, the positions of the labels are numbered 0,1,... So setting the x-limits a bit before 0 and after the last, shows them more centered.
Usually, bars are drawn with their 'feet' on the ground, which can be set via plt.ylim(0, ...). Minor ticks can be positioned for example at multiples of 0.2. Setting the length of the ticks to zero lets the position count for the grid, but suppresses the tick mark.
from matplotlib import pyplot as plt
from matplotlib.ticker import MultipleLocator
import numpy as np
labels = ['Test 1', 'Test 2']
values = [1, 0.7]
fig, ax = plt.subplots()
plt.vlines(labels, 0, values, colors='dodgerblue', alpha=.4, lw=7)
plt.xlim(-0.5, len(labels) - 0.5) # add some padding left and right of the bars
plt.ylim(0, 1.1) # bars usually have their 0 at the bottom
ax.xaxis.set_minor_locator(MultipleLocator(.2))
plt.tick_params(axis='x', which='both', length=0) # ticks not shown, but position serves for gridlines
plt.grid(axis='both', which='both', ls=':') # optionally set the linestyle of the grid
plt.show()
How would you go about changing the style of only some boxes in a matplotlib boxplot? Below, you can see an example of styling, but I would like the style to only apply to one of the boxes.
The same question has been asked for seaborn boxplots already. For matplotlib boxplots this is even easier, since the boxplot directly returns a dictionary of the involved artists, see boxplot documentation.
This means that if bplot = ax.boxplot(..) is your boxplot, you may access the boxes via bplot['boxes'], select one of them and set its linestyle to your desire. E.g.
bplot['boxes'][2].set_linestyle("-.")
Modifying the boxplot_color example
import matplotlib.pyplot as plt
import numpy as np
# Random test data
np.random.seed(19680801)
all_data = [np.random.normal(0, std, size=100) for std in range(1, 4)]
labels = ['x1', 'x2', 'x3']
fig, ax = plt.subplots()
# notch shape box plot
bplot = ax.boxplot(all_data, vert=True, patch_artist=True, labels=labels)
# Loop through boxes and colorize them individually
colors = ['pink', 'lightblue', 'lightgreen']
for patch, color in zip(bplot['boxes'], colors):
patch.set_facecolor(color)
# Make the third box dotted
bplot['boxes'][2].set_linestyle("-.")
plt.show()
I'd like to NOT specify a color for each plotted line, and have each line get a distinct color. But if I run:
from matplotlib import pyplot as plt
for i in range(20):
plt.plot([0, 1], [i, i])
plt.show()
then I get this output:
If you look at the image above, you can see that matplotlib attempts to pick colors for each line that are different, but eventually it re-uses colors - the top ten lines use the same colors as the bottom ten. I just want to stop it from repeating already used colors AND/OR feed it a list of colors to use.
I usually use the second one of these:
from matplotlib.pyplot import cm
import numpy as np
#variable n below should be number of curves to plot
#version 1:
color = cm.rainbow(np.linspace(0, 1, n))
for i, c in zip(range(n), color):
plt.plot(x, y, c=c)
#or version 2:
color = iter(cm.rainbow(np.linspace(0, 1, n)))
for i in range(n):
c = next(color)
plt.plot(x, y, c=c)
Example of 2:
matplotlib 1.5+
You can use axes.set_prop_cycle (example).
matplotlib 1.0-1.4
You can use axes.set_color_cycle (example).
matplotlib 0.x
You can use Axes.set_default_color_cycle.
You can use a predefined "qualitative colormap" like this:
from matplotlib.cm import get_cmap
name = "Accent"
cmap = get_cmap(name) # type: matplotlib.colors.ListedColormap
colors = cmap.colors # type: list
axes.set_prop_cycle(color=colors)
Tested on matplotlib 3.0.3. See https://github.com/matplotlib/matplotlib/issues/10840 for discussion on why you can't call axes.set_prop_cycle(color=cmap).
A list of predefined qualititative colormaps is available at https://matplotlib.org/gallery/color/colormap_reference.html :
prop_cycle
color_cycle was deprecated in 1.5 in favor of this generalization: http://matplotlib.org/users/whats_new.html#added-axes-prop-cycle-key-to-rcparams
# cycler is a separate package extracted from matplotlib.
from cycler import cycler
import matplotlib.pyplot as plt
plt.rc('axes', prop_cycle=(cycler('color', ['r', 'g', 'b'])))
plt.plot([1, 2])
plt.plot([2, 3])
plt.plot([3, 4])
plt.plot([4, 5])
plt.plot([5, 6])
plt.show()
Also shown in the (now badly named) example: http://matplotlib.org/1.5.1/examples/color/color_cycle_demo.html mentioned at: https://stackoverflow.com/a/4971431/895245
Tested in matplotlib 1.5.1.
I don't know if you can automatically change the color, but you could exploit your loop to generate different colors:
for i in range(20):
ax1.plot(x, y, color = (0, i / 20.0, 0, 1)
In this case, colors will vary from black to 100% green, but you can tune it if you want.
See the matplotlib plot() docs and look for the color keyword argument.
If you want to feed a list of colors, just make sure that you have a list big enough and then use the index of the loop to select the color
colors = ['r', 'b', ...., 'w']
for i in range(20):
ax1.plot(x, y, color = colors[i])
You can also change the default color cycle in your matplotlibrc file.
If you don't know where that file is, do the following in python:
import matplotlib
matplotlib.matplotlib_fname()
This will show you the path to your currently used matplotlibrc file.
In that file you will find amongst many other settings also the one for axes.color.cycle. Just put in your desired sequence of colors and you will find it in every plot you make.
Note that you can also use all valid html color names in matplotlib.
As Ciro's answer notes, you can use prop_cycle to set a list of colors for matplotlib to cycle through. But how many colors? What if you want to use the same color cycle for lots of plots, with different numbers of lines?
One tactic would be to use a formula like the one from https://gamedev.stackexchange.com/a/46469/22397, to generate an infinite sequence of colors where each color tries to be significantly different from all those that preceded it.
Unfortunately, prop_cycle won't accept infinite sequences - it will hang forever if you pass it one. But we can take, say, the first 1000 colors generated from such a sequence, and set it as the color cycle. That way, for plots with any sane number of lines, you should get distinguishable colors.
Example:
from matplotlib import pyplot as plt
from matplotlib.colors import hsv_to_rgb
from cycler import cycler
# 1000 distinct colors:
colors = [hsv_to_rgb([(i * 0.618033988749895) % 1.0, 1, 1])
for i in range(1000)]
plt.rc('axes', prop_cycle=(cycler('color', colors)))
for i in range(20):
plt.plot([1, 0], [i, i])
plt.show()
Output:
Now, all the colors are different - although I admit that I struggle to distinguish a few of them!