The following code produces a barchart with differently colored bars for each "hue" value associated with the grouping variable y:
from matplotlib import pyplot as plt
import seaborn as sns
ax = sns.barplot(x=[3, 7, 12, 10], y=list("AABB"), hue=list("XYXY"))
plt.show()
I want to change the colors so that all bars belonging to the same grouping value have the same color. That is, the first two bars where y == "A" should be orange, and the last two bars where y == "B" should be blue.
I've tried the palette argument of `sns.barplot(). However, unless I'm missing something, this allows me to specify the color of each hue level, but not separately for the values of the grouping variable.
I've also had a look at the answer to this related question, which asks something similar about box-whisker plots created by sns.boxplot(). The accepted answer suggests to access the rectangles representing the bars using the artists property of the axes. However, it seems that sns.barplot() doesn't use artists in the same way as sns.boxplot(): ax.artists appears always to be an empty list:
print(ax.artists)
[]
So, what do I have to change so that the top two bars appear in blue, and the bottom two bars appear in orange?
The bars are not in artists but in patches. So you may loop over ax.patches and set the colors to your liking.
from matplotlib import pyplot as plt
import seaborn as sns
ax = sns.barplot(x=[3, 7, 12, 10], y=list("AABB"), hue=list("XYXY"))
for i, bar in enumerate(ax.patches):
bar.set_color("C{}".format(i%2))
plt.show()
Related
I am making a scatter plot in matplotlib and need to change the background of the actual plot to black. I know how to change the face color of the plot using:
fig = plt.figure()
fig.patch.set_facecolor('xkcd:mint green')
My issue is that this changes the color of the space around the plot. How to I change the actual background color of the plot?
Use the set_facecolor(color) method of the axes object, which you've created one of the following ways:
You created a figure and axis/es together
fig, ax = plt.subplots(nrows=1, ncols=1)
You created a figure, then axis/es later
fig = plt.figure()
ax = fig.add_subplot(1, 1, 1) # nrows, ncols, index
You used the stateful API (if you're doing anything more than a few lines, and especially if you have multiple plots, the object-oriented methods above make life easier because you can refer to specific figures, plot on certain axes, and customize either)
plt.plot(...)
ax = plt.gca()
Then you can use set_facecolor:
ax.set_facecolor('xkcd:salmon')
ax.set_facecolor((1.0, 0.47, 0.42))
As a refresher for what colors can be:
matplotlib.colors
Matplotlib recognizes the following formats to specify a color:
an RGB or RGBA tuple of float values in [0, 1] (e.g., (0.1, 0.2, 0.5) or (0.1, 0.2, 0.5, 0.3));
a hex RGB or RGBA string (e.g., '#0F0F0F' or '#0F0F0F0F');
a string representation of a float value in [0, 1] inclusive for gray level (e.g., '0.5');
one of {'b', 'g', 'r', 'c', 'm', 'y', 'k', 'w'};
a X11/CSS4 color name;
a name from the xkcd color survey; prefixed with 'xkcd:' (e.g., 'xkcd:sky blue');
one of {'tab:blue', 'tab:orange', 'tab:green', 'tab:red', 'tab:purple', 'tab:brown', 'tab:pink', 'tab:gray', 'tab:olive', 'tab:cyan'} which are the Tableau Colors from the ‘T10’ categorical palette (which is the default color cycle);
a “CN” color spec, i.e. 'C' followed by a single digit, which is an index into the default property cycle (matplotlib.rcParams['axes.prop_cycle']); the indexing occurs at artist creation time and defaults to black if the cycle does not include color.
All string specifications of color, other than “CN”, are case-insensitive.
One method is to manually set the default for the axis background color within your script (see Customizing matplotlib):
import matplotlib.pyplot as plt
plt.rcParams['axes.facecolor'] = 'black'
This is in contrast to Nick T's method which changes the background color for a specific axes object. Resetting the defaults is useful if you're making multiple different plots with similar styles and don't want to keep changing different axes objects.
Note: The equivalent for
fig = plt.figure()
fig.patch.set_facecolor('black')
from your question is:
plt.rcParams['figure.facecolor'] = 'black'
Something like this? Use the axisbg keyword to subplot:
>>> from matplotlib.figure import Figure
>>> from matplotlib.backends.backend_agg import FigureCanvasAgg as FigureCanvas
>>> figure = Figure()
>>> canvas = FigureCanvas(figure)
>>> axes = figure.add_subplot(1, 1, 1, axisbg='red')
>>> axes.plot([1,2,3])
[<matplotlib.lines.Line2D object at 0x2827e50>]
>>> canvas.print_figure('red-bg.png')
(Granted, not a scatter plot, and not a black background.)
Simpler answer:
ax = plt.axes()
ax.set_facecolor('silver')
If you already have axes object, just like in Nick T's answer, you can also use
ax.patch.set_facecolor('black')
The easiest thing is probably to provide the color when you create the plot :
fig1 = plt.figure(facecolor=(1, 1, 1))
or
fig1, (ax1, ax2) = plt.subplots(nrows=1, ncols=2, facecolor=(1, 1, 1))
One suggestion in other answers is to use ax.set_axis_bgcolor("red"). This however is deprecated, and doesn't work on MatPlotLib >= v2.0.
There is also the suggestion to use ax.patch.set_facecolor("red") (works on both MatPlotLib v1.5 & v2.2). While this works fine, an even easier solution for v2.0+ is to use
ax.set_facecolor("red")
In addition to the answer of NickT, you can also delete the background frame by setting it to "none" as explain here: https://stackoverflow.com/a/67126649/8669161
import matplotlib.pyplot as plt
plt.rcParams['axes.facecolor'] = 'none'
I think this might be useful for some people:
If you want to change the color of the background that surrounds the figure, you can use this:
fig.patch.set_facecolor('white')
So instead of this:
you get this:
Obviously you can set any color you'd want.
P.S. In case you accidentally don't see any difference between the two plots, try looking at StackOverflow using darkmode.
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.
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()
I am trying to plot bar chart using seaborn. Sample data:
x=[1,1000,1001]
y=[200,300,400]
cat=['first','second','third']
df = pd.DataFrame(dict(x=x, y=y,cat=cat))
When I use:
sns.factorplot("x","y", data=df,kind="bar",palette="Blues",size=6,aspect=2,legend_out=False);
The figure produced is
When I add the legend
sns.factorplot("x","y", data=df,hue="cat",kind="bar",palette="Blues",size=6,aspect=2,legend_out=False);
The resulting figure looks like this
As you can see, the bar is shifted from the value. I don't know how to get the same layout as I had in the first figure and add the legend.
I am not necessarily tied to seaborn, I like the color palette, but any other approach is fine with me. The only requirement is that the figure looks like the first one and has the legend.
It looks like this issue arises here - from the docs searborn.factorplot
hue : string, optional
Variable name in data for splitting the plot by color. In the case of ``kind=”bar”, this also influences the placement on the x axis.
So, since seaborn uses matplotlib, you can do it like this:
import matplotlib.pyplot as plt
import numpy as np
import seaborn as sns
x=[1,1000,1001]
y=[200,300,400]
sns.set_context(rc={"figure.figsize": (8, 4)})
nd = np.arange(3)
width=0.8
plt.xticks(nd+width/2., ('1','1000','1001'))
plt.xlim(-0.15,3)
fig = plt.bar(nd, y, color=sns.color_palette("Blues",3))
plt.legend(fig, ['First','Second','Third'], loc = "upper left", title = "cat")
plt.show()
Added #mwaskom's method to get the three sns colors.
The following code:
# in ipython notebook, enable inline plotting with:
# %pylab inline --no-import-all
import matplotlib.pyplot as plt
# create some circles
circle1 = plt.Circle((-.5,0), 1, color='r', alpha=.2)
circle2 = plt.Circle(( .5,0), 1, color='b', alpha=.2)
# add them to the plot (bad form to use ;, but saving space)
# and control the display a bit
ax = plt.gca()
ax.add_artist(circle1); ax.add_artist(circle2)
ax.set_xlim(-2, 2); ax.set_ylim(-2, 2)
ax.set_aspect('equal')
# display it
plt.plot()
Produces the following plot:
I would like to specify the colors of the four regions (1) the background (currently white), (2 and 3) each individual event (the non-overlapping areas, currently blue and red), and (4) the intersection event (currently blended to purple). For example, I might color them red, green, blue, yellow -or- I might give them four different, precisely specified grayscale values (the later is more likely). [The colors will be generated based on characteristics of the underlying data.]
I specifically do not want to use alpha blending to "infer" a color in the intersection. I need to explicitly control the colors of all four regions.
I can think of a few strategies to solve this:
Ask mpl to extract the "primitive" patch objects that make up the three distinctly colored graphical regions (and do something similar to operate on the background) and then color them.
Given the circles, manually compute their intersections and color that intersection (somehow). Going point by point seems ugly.
Thanks!
I'm not 100% sure but I think matplotlib does not have the functionality to intersect polygons. But you could use shapely:
import shapely.geometry as sg
import matplotlib.pyplot as plt
import descartes
# create the circles with shapely
a = sg.Point(-.5,0).buffer(1.)
b = sg.Point(0.5,0).buffer(1.)
# compute the 3 parts
left = a.difference(b)
right = b.difference(a)
middle = a.intersection(b)
# use descartes to create the matplotlib patches
ax = plt.gca()
ax.add_patch(descartes.PolygonPatch(left, fc='b', ec='k', alpha=0.2))
ax.add_patch(descartes.PolygonPatch(right, fc='r', ec='k', alpha=0.2))
ax.add_patch(descartes.PolygonPatch(middle, fc='g', ec='k', alpha=0.2))
# control display
ax.set_xlim(-2, 2); ax.set_ylim(-2, 2)
ax.set_aspect('equal')
plt.show()