I was running seaborn ver. 0.10.1 on my jupyter notebook. This morning I upgraded to the latest version 0.11.0. Now, my kde jointplot doesn't give the color mapping that it used to. The code is the same. Only the versions are different.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib notebook
np.random.seed(1234)
v1 = pd.Series(np.random.normal(0,10,1000), name='v1')
v2 = pd.Series(np.random.normal(60,15,1000), name='v2')
v3 = pd.Series(2*v1 + v2, name='v3')
# set the seaborn style for all the following plots
sns.set_style('white')
sns.jointplot(v1, v3, kind='kde', space=0);
The function kdeplot (which is used internally by jointplot()to draw the bivariate density plot) has been extensively changed in v.0.11. See What's new and the documentation.
You now have to pass fill=True to get a filled KDE, and you need to specify thresh=0 if you want to fill the available space with color.
sns.jointplot(x=v1, y=v3, kind='kde', space=0, fill=True, thresh=0, cmap='Blues');
Related
I managed to make a displot as I intended with seaborn and the only thing I want to change is the bars' outline width. Specifically, I want to make it thinner. Here's the code and a sample of how the dataframe is composed.
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
data_final = pd.merge(data, data_filt)
q = sns.displot(data=data_final[data_final['cond_state'] == True], y='Brand', hue='Style', multiple='stack')
plt.title('Sample of brands and their offering of ramen styles')
I'm specifying that the plot should only use rows where the cond_state is True. Here is a sample of the data_final dataframe.
Here is how the plot currently looks like.
I've tried various ways published online, but most of them use the deprecated distplot instead of displot. There also doesn't seem to be a parameter for changing the bars' outline width in the seaborn documentation for displot and FacetGrid
The documentation for the seaborn displot function doesn't have this parameter listed, but you can pass matplotlib axes arguments, such as linewidth = 0.25, to the seaborn.displot function to solve your problem.
I am currently trying to generate visualizations in zeppelin (0.8.1) notebooks using the pyspark interpreter with python 3.7.3.
Generating the following simple plot with seaborn (0.9.0) takes around 5 minutes (with very high CPU usage throughout the duration):
%pyspark
import seaborn as sns
import numpy as np
import pandas as pd
data = pd.DataFrame(np.random.rand(100,3))
sns.pairplot(data)
This behavior is rather inconsistent as the following (much more data intensive) plot is rendered instantly
%pyspark
import seaborn as sns
import numpy as np
import pandas as pd
df = pd.DataFrame(data = np.random.rand(10000,2))
sns.lineplot(x = 0, y = 1, data = df)
I noticed that using matplotlib (3.1.0) is generally much faster for and almost as snappy as I am used to from jupyter notebook environments.
I have already read about issue ZEPPELIN-1894 but I can render the mentioned scatterplot instantly as well.
Ok, after posting here the solution is to use the %spark.ipyspark interpreter, this might require installing additional packages:
pip install protobuf grpcio
I'm generating a heatmap from a pandas dataframe using a code that looks like this on my apple computer.
import matplotlib.pyplot as plt
import seaborn as sns
fig, ax = plt.subplots(figsize=(14,14))
sns.set(font_scale=1.4)
sns_plot = sns.heatmap(df, annot=True, linewidths=.5, fmt='g', ax=ax).set_yticklabels(ax.get_yticklabels(), rotation=0)
ax.set_ylabel('Product')
ax.set_xlabel('Manufacturer')
ax.xaxis.set_ticks_position('top')
ax.xaxis.set_label_position('top')
fig.savefig('output.png')
And I get a heatmap looking like this:
I then put my code in a docker container with an ubuntu image and I install the same version of seaborn. The only difference is that I need to add a matplotlib configuration so that TCL doesn't scream:
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
import seaborn as sns
And I get a heatmap that looks like this (I use the same code and the same pandas dataframe):
I'm unable to find why the color gradient is inverted and would love to hear if you have any idea.
Thank you !
The default colormap has changed to 'rocket' for sequential data with 0.8 release of seaborn, see the release notes. The colormap looks this way now:
You can always use the cmap argument and specify which colormap you prefer to use. For example, to get the pre-0.8 colormap for non-divergent data use: cmap=sns.cubehelix_palette(light=.95, as_cmap=True).
I'm attempting to follow the violin plot example in bokeh, but am unable to add x-axis labels to my violins. According to the Seaborn documentation it looks like I should be able to add x-axis labels via the "names" argument, however, the following code does not add x-axis labels:
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
from bokeh import mpl
from bokeh.plotting import show
# generate some random data
data = 1 + np.random.randn(20, 6)
# Use Seaborn and Matplotlib normally
sns.violinplot(data, color="Set3", names=["kirk","spock","bones","scotty","uhura","sulu"])
plt.title("Seaborn violin plot in Bokeh")
# Convert to interactive Bokeh plot with one command
show(mpl.to_bokeh(name="violin"))
I believe that the issue is that I'm converting a figure from seaborn to matplotlib to bokeh, but I'm not sure at what level the x-axis labels go in.
I've confirmed that the labels are showing up in matplotlib before conversion to bokeh. I've also tried adding the labels to bokeh after conversion, but this results in a weird plot. I've created an issue for this problem with the bokeh developers here.
Since Bokeh 12.5 (April 2017), support for Matplotlib has been deprecated, so mpl.to_bokeh() is no longer available.
Hi I have a text file containing three columns of numbers; one column each for the x,y,z coordinates of a bunch of points. All numbers are between 0 ad 1.
I want to plot all these points in the unit cube [0,1]x[0,1]x[0,1].
Please let me know how I can do this in Octave or MatPlot lib, whichever prduces a better quality image.
If I understand your question correctly, this is how it looks in Matplotlib:
This is the code to produce this plot:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
import numpy as np
np.random.seed(101)
x,y,z = np.random.rand(3,20)
fig = plt.figure()
# version 1.0.x syntax:
#ax = fig.add_subplot(111, projection='3d')
# version 0.99.x syntax: (accepted by 1.0.x as well)
ax = Axes3D(fig)
ax.scatter(x,y,z)
fig.savefig('scatter3d.png')
As the code suggests, there are slight differences in how matplotlib version 0.99.1.1 and version 1.0.1 behave, as noted in this SO question/answer. I am using 0.99.1.1, and I had trouble using all the options available to 2D scatter plots, which should be the same for 3D plots as well. The full list of scatter features are listed here.
The above code resulted from looking at the matplotlib tutorial on 3D plotting.