Is it possible to display the venn diagram within a universal set? - python-3.x

Is it possible to display the universal set using matplotlib-venn? I'm new to both python and the matplotlib package, so I'm not actually sure what's possible and what's not.
I'm trying to create a venn diagram generator that accepts the values for each circle and then an argument (ex. A intersection B), then highlights only the intersection of the two circles. Basically, this is what I want the output to be.


Python has package matplotlib_venn but it needs some tricks in your case.
import matplotlib.pyplot as plt
from matplotlib_venn import venn3
A = set([9,3,6])
B = set([2,4,6,8])
C = set([0,5,1,7])
v = venn3([A,B,C], ('P', 'Q', 'U'))
v.get_label_by_id('100').set_text('\n'.join(map(str,A-B)))
v.get_label_by_id('110').set_text('\n'.join(map(str,A&B)))
v.get_label_by_id('010').set_text('\n'.join(map(str,B-A)))
v.get_label_by_id('001').set_text('\n'.join(map(str,C)))
v.get_patch_by_id('001').set_color('white')
plt.axis('on')
plt.show()
You can use plt.annotate for configuring the position of values

Related

How to add traces in plotly.express

I am very new to python and plotly.express, and I find it very confusing...
I am trying to use the principle of adding different traces to my figure, using example code shown here https://plotly.com/python/line-charts/, Line Plot Modes, #Create traces.
BUT I get my data from a .CSV file.
import plotly.express as px
import plotly as plotly
import plotly.graph_objs as go
import pandas as pd
data = pd.read_csv(r"C:\Users\x.csv")
fig = px.scatter(data, x="Time", y="OD", color="C-source", size="C:A 1 ratio")
fig = px.line(data, x="Time", y="OD", color="C-source")
fig.show()
The above lines produces scatter/line plots with the correct data, but the data is mixed together. I have data from 2 different sources marked by a column named "Strain" in my .csv file that I would like the chart to reflect.
Is the traces option a possible way to do it, or is there another way?

You can add traces using an Express plot by using .select_traces(). Something like:
fig.add_traces(
list(px.line(...).select_traces())
)
Note the need to convert to list, since .select_traces() returns a generator.

It looks like you probably want the lines with the scatter dots as well on a single plot?
You're setting fig to equal px.scatter() and then setting (changing) it to equal px.line(). When set to line, the scatter plot is overwritten.
You're already importing graph objects so you can use add_trace with go, something like this:
fig.add_trace(go.Scatter(x=data["Time"], y=data["OD"], mode='markers', marker=dict(color=data["C-source"], size=data["C:A 1 ratio"])))
Depending on how your data is set up, you may need to add each C-source separately doing something like:
x=data.query("C-source=='Term'")["Time"], ... , name='Term'`
Here's a few references with examples and options you can use to set up your scatter:
Scatter plot examples　　
Marker styles　　
Scatter arguments and attributes

You can use the apporach stated in Plotly: How to combine scatter and line plots using Plotly Express?
fig3 = go.Figure(data=fig1.data + fig2.data)
or a more convenient and scalable approach:
fig1.data and fig2.data are common tuples that hold all the info needed for a plot and the + just concatenates them.
# this will hold all figures until they are combined
all_figures = []
# data_collection: dictionary with Pandas dataframes
for df_label in data_collection:
df = data_collection[df_label]
fig = px.line(df, x='Date', y=['Value'])
all_figures.append(fig)
import operator
import functools
# now you can concatenate all the data tuples
# by using the programmatic add operator
fig3 = go.Figure(data=functools.reduce(operator.add, [_.data for _ in all_figures]))
fig3.show()

thanks for taking the time to help me out. I ended up with two solutions that worked, of which using "facet_col" to divide the plot into two subplots (1 for each strain) was the most simple solution.
https://plotly.com/python/axes/

Thanks. this worked for me also where Fig_Set_B is a list of scatter plots
# create a tuple of first line plots in first 6 plots from plot set Fig_Set_B`
fig_combined = go.Figure(data= tuple(Fig_Set_B[x].data[0] for x in range(6)) )
fig_combined.show()

Using RGB values control individual data points matplotlib

I'm trying to be able to control the colour of an individual data point using a corresponding rgb tuple. I've tried looping through the data set and plotting individual data points however I get the same effect as the code I have below; all that happens is it refuses to produce a graph.
This is an example of the data type I'm working with
Any tips?
import matplotlib.pyplot as plt
y=[(0.200,0.1100,0.520)]
for i in range(4):
y.append(y)
plt.plot([1,2,3,4], [3,4,5,2],c=y)
plt.show()

One problem is that you are appending the list to the new list. Instead, try appending the tuple to the list. Moreover, you need to use scatter plot for the color argument which contains rgb tuple for each point. However, in oyur case, I see only a single color for all the scatter points.
tup=(0.200,0.1100,0.520)
y = []
for i in range(4):
y.append(tup)
plt.scatter([1,2,3,4], [3,4,5,2], c=y)
A rather short version to your code is using a list comprehension
tup=(0.200,0.1100,0.520)
y = [tup for _ in range(4)]
plt.scatter([1,2,3,4], [3,4,5,2], c=y)

Creating a structured grid of subplots with Seaborn FacetGrid

My attempt to use FacetGrid in Seaborn does not produces the expected results.
Moreover, I would like to control the white space in the grid.
My data and code is the following:
toy.to_json()
'{"has_cus_id_but_not_acc_id":{"0":0,"1":0,"2":0,"3":0,"4":0,"5":0,"6":0,"7":0,"8":0,"9":0,"10":0,"11":0,"12":0,"13":0,"14":0,"15":0,"16":0,"17":0,"18":1,"19":0,"20":0,"21":0,"22":1,"23":0,"24":0,"25":1,"26":0,"27":1,"28":0,"29":1,"30":0,"31":1,"32":0,"33":1,"34":0,"35":1,"36":0,"37":1,"38":0,"39":0,"40":1,"41":1,"42":0,"43":1,"44":0,"45":1,"46":0,"47":1,"48":0,"49":1,"50":0,"51":1,"52":0,"53":1,"54":0,"55":1,"56":0,"57":1,"58":0,"59":1,"60":0,"61":1,"62":0,"63":1,"64":0,"65":1,"66":0,"67":1,"68":0,"69":1,"70":0,"71":1,"72":0,"73":1,"74":0,"75":1,"76":0,"77":0,"78":1,"79":0,"80":1,"81":0,"82":0,"83":1,"84":0,"85":1},"reg_year":{"0":2014.0,"1":2014.0,"2":2014.0,"3":2014.0,"4":2014.0,"5":2014.0,"6":2014.0,"7":2014.0,"8":2015.0,"9":2015.0,"10":2015.0,"11":2015.0,"12":2015.0,"13":2015.0,"14":2015.0,"15":2015.0,"16":2015.0,"17":2016.0,"18":2016.0,"19":2016.0,"20":2016.0,"21":2016.0,"22":2016.0,"23":2016.0,"24":2016.0,"25":2016.0,"26":2016.0,"27":2016.0,"28":2016.0,"29":2016.0,"30":2016.0,"31":2016.0,"32":2016.0,"33":2016.0,"34":2016.0,"35":2016.0,"36":2016.0,"37":2016.0,"38":2017.0,"39":2017.0,"40":2017.0,"41":2017.0,"42":2017.0,"43":2017.0,"44":2017.0,"45":2017.0,"46":2017.0,"47":2017.0,"48":2017.0,"49":2017.0,"50":2017.0,"51":2017.0,"52":2017.0,"53":2017.0,"54":2017.0,"55":2017.0,"56":2017.0,"57":2017.0,"58":2017.0,"59":2017.0,"60":2018.0,"61":2018.0,"62":2018.0,"63":2018.0,"64":2018.0,"65":2018.0,"66":2018.0,"67":2018.0,"68":2018.0,"69":2018.0,"70":2018.0,"71":2018.0,"72":2018.0,"73":2018.0,"74":2018.0,"75":2018.0,"76":2018.0,"77":2018.0,"78":2018.0,"79":2018.0,"80":2018.0,"81":2018.0,"82":2019.0,"83":2019.0,"84":2019.0,"85":2019.0},"reg_month":{"0":3.0,"1":5.0,"2":6.0,"3":7.0,"4":9.0,"5":10.0,"6":11.0,"7":12.0,"8":1.0,"9":3.0,"10":5.0,"11":6.0,"12":7.0,"13":8.0,"14":9.0,"15":11.0,"16":12.0,"17":1.0,"18":1.0,"19":2.0,"20":3.0,"21":4.0,"22":4.0,"23":5.0,"24":6.0,"25":6.0,"26":7.0,"27":7.0,"28":8.0,"29":8.0,"30":9.0,"31":9.0,"32":10.0,"33":10.0,"34":11.0,"35":11.0,"36":12.0,"37":12.0,"38":1.0,"39":2.0,"40":2.0,"41":3.0,"42":4.0,"43":4.0,"44":5.0,"45":5.0,"46":6.0,"47":6.0,"48":7.0,"49":7.0,"50":8.0,"51":8.0,"52":9.0,"53":9.0,"54":10.0,"55":10.0,"56":11.0,"57":11.0,"58":12.0,"59":12.0,"60":1.0,"61":1.0,"62":2.0,"63":2.0,"64":3.0,"65":3.0,"66":4.0,"67":4.0,"68":5.0,"69":5.0,"70":6.0,"71":6.0,"72":7.0,"73":7.0,"74":8.0,"75":8.0,"76":9.0,"77":10.0,"78":10.0,"79":11.0,"80":11.0,"81":12.0,"82":1.0,"83":1.0,"84":2.0,"85":2.0},"Total_Revenue":{"0":35852.02,"1":2623.97,"2":3526.67,"3":21466.71,"4":72784.1200000003,"5":103921.2899999999,"6":10852.87,"7":16522.07,"8":7443.76,"9":68962.1600000002,"10":10956.38,"11":193856.8799999985,"12":110766.6099999997,"13":123861.8599999987,"14":2722.34,"15":303488.6900000007,"16":6876.58,"17":17729.5,"18":4687.93,"19":26914.06,"20":2228.12,"21":15708.93,"22":859.58,"23":19164.89,"24":163164.4799999995,"25":33180.7300000001,"26":10033.01,"27":1114.48,"28":462613.2900000042,"29":9822.95,"30":70901.4400000003,"31":22370.29,"32":46711.8900000002,"33":2335.02,"34":7259.28,"35":11.83,"36":13590.51,"37":7677.77,"38":282.01,"39":358522.7900000003,"40":5844.0,"41":7027.28,"42":1908.71,"43":4032.35,"44":11072.6,"45":3973.15,"46":30706.23,"47":2644.13,"48":23831.75,"49":670.12,"50":6949.54,"51":4687.7,"52":9672.69,"53":7333.01,"54":12814.33,"55":689.39,"56":6962.86,"57":2283.16,"58":1259.5,"59":224.84,"60":12812.12,"61":247.68,"62":25452.65,"63":1245.02,"64":24211.36,"65":5255.25,"66":28402.76,"67":9148.55,"68":14822.61,"69":345.37,"70":12408.13,"71":989.93,"72":10601.33,"73":730.32,"74":169020.5000000001,"75":697.54,"76":3862038.6799997138,"77":6148750.9899984254,"78":194.06,"79":2379382.4500000761,"80":1174.11,"81":1729567.9000000793,"82":889650.029999995,"83":95.8,"84":415996.6999999974,"85":654.78}}'
g = sns.FacetGrid(toy, col='has_cus_id_but_not_acc_id', hue='reg_year')
g.map(sns.barplot, 'reg_month', 'Total_Revenue')
g.add_legend();
If I use bar in pyplot I get this:
g = sns.FacetGrid(toy, col='has_cus_id_but_not_acc_id', hue='reg_year')
g.map(plt.bar, 'reg_month', 'Total_Revenue')
g.add_legend();
Again, I would like to be able to define the white space of the grid.
In addition I would not like to have the bars stacked one over the other but rather one next to the other.

Some values of the year 2018 are really large compared to the any of the values where has_cus_id_but_not_acc_id is 1. Hence the right plot is almost empty. It might make sense to use a logarithmic scale.
Now you have 6 years, so each month would need to show 6 bars next to each other. That will make bars pretty small and does not let the chart be easily readable. Still it's possible.
The following does not use seaborn, but pandas and matplotlib:
import matplotlib.pyplot as plt
import pandas as pd
toy = '{"has_cus_id_but_not_acc_id":{"0":0,"1":0,"2":0,"3":0,"4":0,"5":0,"6":0,"7":0,"8":0,"9":0,"10":0,"11":0,"12":0,"13":0,"14":0,"15":0,"16":0,"17":0,"18":1,"19":0,"20":0,"21":0,"22":1,"23":0,"24":0,"25":1,"26":0,"27":1,"28":0,"29":1,"30":0,"31":1,"32":0,"33":1,"34":0,"35":1,"36":0,"37":1,"38":0,"39":0,"40":1,"41":1,"42":0,"43":1,"44":0,"45":1,"46":0,"47":1,"48":0,"49":1,"50":0,"51":1,"52":0,"53":1,"54":0,"55":1,"56":0,"57":1,"58":0,"59":1,"60":0,"61":1,"62":0,"63":1,"64":0,"65":1,"66":0,"67":1,"68":0,"69":1,"70":0,"71":1,"72":0,"73":1,"74":0,"75":1,"76":0,"77":0,"78":1,"79":0,"80":1,"81":0,"82":0,"83":1,"84":0,"85":1},"reg_year":{"0":2014.0,"1":2014.0,"2":2014.0,"3":2014.0,"4":2014.0,"5":2014.0,"6":2014.0,"7":2014.0,"8":2015.0,"9":2015.0,"10":2015.0,"11":2015.0,"12":2015.0,"13":2015.0,"14":2015.0,"15":2015.0,"16":2015.0,"17":2016.0,"18":2016.0,"19":2016.0,"20":2016.0,"21":2016.0,"22":2016.0,"23":2016.0,"24":2016.0,"25":2016.0,"26":2016.0,"27":2016.0,"28":2016.0,"29":2016.0,"30":2016.0,"31":2016.0,"32":2016.0,"33":2016.0,"34":2016.0,"35":2016.0,"36":2016.0,"37":2016.0,"38":2017.0,"39":2017.0,"40":2017.0,"41":2017.0,"42":2017.0,"43":2017.0,"44":2017.0,"45":2017.0,"46":2017.0,"47":2017.0,"48":2017.0,"49":2017.0,"50":2017.0,"51":2017.0,"52":2017.0,"53":2017.0,"54":2017.0,"55":2017.0,"56":2017.0,"57":2017.0,"58":2017.0,"59":2017.0,"60":2018.0,"61":2018.0,"62":2018.0,"63":2018.0,"64":2018.0,"65":2018.0,"66":2018.0,"67":2018.0,"68":2018.0,"69":2018.0,"70":2018.0,"71":2018.0,"72":2018.0,"73":2018.0,"74":2018.0,"75":2018.0,"76":2018.0,"77":2018.0,"78":2018.0,"79":2018.0,"80":2018.0,"81":2018.0,"82":2019.0,"83":2019.0,"84":2019.0,"85":2019.0},"reg_month":{"0":3.0,"1":5.0,"2":6.0,"3":7.0,"4":9.0,"5":10.0,"6":11.0,"7":12.0,"8":1.0,"9":3.0,"10":5.0,"11":6.0,"12":7.0,"13":8.0,"14":9.0,"15":11.0,"16":12.0,"17":1.0,"18":1.0,"19":2.0,"20":3.0,"21":4.0,"22":4.0,"23":5.0,"24":6.0,"25":6.0,"26":7.0,"27":7.0,"28":8.0,"29":8.0,"30":9.0,"31":9.0,"32":10.0,"33":10.0,"34":11.0,"35":11.0,"36":12.0,"37":12.0,"38":1.0,"39":2.0,"40":2.0,"41":3.0,"42":4.0,"43":4.0,"44":5.0,"45":5.0,"46":6.0,"47":6.0,"48":7.0,"49":7.0,"50":8.0,"51":8.0,"52":9.0,"53":9.0,"54":10.0,"55":10.0,"56":11.0,"57":11.0,"58":12.0,"59":12.0,"60":1.0,"61":1.0,"62":2.0,"63":2.0,"64":3.0,"65":3.0,"66":4.0,"67":4.0,"68":5.0,"69":5.0,"70":6.0,"71":6.0,"72":7.0,"73":7.0,"74":8.0,"75":8.0,"76":9.0,"77":10.0,"78":10.0,"79":11.0,"80":11.0,"81":12.0,"82":1.0,"83":1.0,"84":2.0,"85":2.0},"Total_Revenue":{"0":35852.02,"1":2623.97,"2":3526.67,"3":21466.71,"4":72784.1200000003,"5":103921.2899999999,"6":10852.87,"7":16522.07,"8":7443.76,"9":68962.1600000002,"10":10956.38,"11":193856.8799999985,"12":110766.6099999997,"13":123861.8599999987,"14":2722.34,"15":303488.6900000007,"16":6876.58,"17":17729.5,"18":4687.93,"19":26914.06,"20":2228.12,"21":15708.93,"22":859.58,"23":19164.89,"24":163164.4799999995,"25":33180.7300000001,"26":10033.01,"27":1114.48,"28":462613.2900000042,"29":9822.95,"30":70901.4400000003,"31":22370.29,"32":46711.8900000002,"33":2335.02,"34":7259.28,"35":11.83,"36":13590.51,"37":7677.77,"38":282.01,"39":358522.7900000003,"40":5844.0,"41":7027.28,"42":1908.71,"43":4032.35,"44":11072.6,"45":3973.15,"46":30706.23,"47":2644.13,"48":23831.75,"49":670.12,"50":6949.54,"51":4687.7,"52":9672.69,"53":7333.01,"54":12814.33,"55":689.39,"56":6962.86,"57":2283.16,"58":1259.5,"59":224.84,"60":12812.12,"61":247.68,"62":25452.65,"63":1245.02,"64":24211.36,"65":5255.25,"66":28402.76,"67":9148.55,"68":14822.61,"69":345.37,"70":12408.13,"71":989.93,"72":10601.33,"73":730.32,"74":169020.5000000001,"75":697.54,"76":3862038.6799997138,"77":6148750.9899984254,"78":194.06,"79":2379382.4500000761,"80":1174.11,"81":1729567.9000000793,"82":889650.029999995,"83":95.8,"84":415996.6999999974,"85":654.78}}'
df = pd.read_json(toy)
df['reg_year'].astype(int)
u = df["has_cus_id_but_not_acc_id"].unique()
y = df['reg_year'].unique()
fig, axes = plt.subplots(1,len(u), sharey=True)
axes[0].set_yscale("log")
for ax, (n, grp) in zip(axes.flat, df.groupby("has_cus_id_but_not_acc_id")):
piv = grp.pivot('reg_month', 'reg_year', 'Total_Revenue')
empty = pd.DataFrame(index=range(1,12), columns=y)
empty.combine_first(piv).plot.bar(ax=ax, width=0.8, legend=False)
axes[1].legend()
plt.show()

healpy: Formatting subplots

I want to plot many subplots in one figure using healpy. How to:
Set the position of the colorbar?
Set the tick and ticklabel of colorbar?
Set the position and size of the subplots?
I want to generate a plot such as figure 1, which is plotted in MATLAB, based on general coordinates
Right now, I only plot it as follows using healpy:
A similar code to produce figure 3 (similar to figure2) is as follow:
import numpy as np
import healpy as hp
degree = 4
nside = 2**degree
num_Pixel = hp.nside2npix(nside)
m = np.arange(num_Pixel)
margins = [[0.02,0,0,0],[0.01,0,0,0],[0.01,0,0.01,0],
[0.02,0,0,0],[0.01,0,0,0],[0.01,0,0.01,0],
[0.02,0.05,0,0],[0.01,0.05,0,0],[0.01,0.05,0.01,0]]
title = [
'Equinox', 'Jun. Solstice', 'Dec. Solstice',
'','','','','','']
for ifig in range(1,10):
if ifig < 7:
hp.cartview(
m, sub=330+ifig, margins=margins[ifig1],
cbar=False, title=title[ifig-1])
else:
hp.cartview(
m, sub=330+ifig, margins=margins[ifig-1],
cbar=True, title=title[ifig-1])
The code produced the figure 3

I'm afraid healpy doesn't come with a good way to handle the colorbar, ticks, ticklabels, the axes etc.
The best way forward would be to generate FITS images, based on your HEALPix map (e.g. using hp.cartview(..., return_projected_map=True) or using the reproject package).
You also need to generate the right FITS header for that, astropy would be the right tool for that (how-to manipulate FITS headers).
Once you have that, you can use the excellent WCSAxes framework within astropy, which gives you plenty of well-documented customization options.

Need help in creating a function to plot a Matplotlib GridSpec

I have a dataset with 80 variables. I am interested in creating a function that will automate the creation of a 20 X 4 GridSpec in Matplotlib. Each subplot would either contain a histogram or a barplot for each of the 80 variables in the data. As a first step, I successfully created two functions (I call them 'counts' and 'histogram') that contain the layout of the plot that I want. Both of them work when tested on individual variables. As a next step, I attempted to create a function that would take the column names, loop through a conditional to test whether the data type is an object or otherwise and call the right function based on the datatype as a new subplot. Here is the code that I have so far:
Creates list of coordinates we will need for subplot specification:
A = np.arange(21)
B = np.arange(4)
coords = []
for i in A:
for j in B:
coords.append([A[i], B[j]])
#Create the gridspec and layout the figure
import matplotlib.gridspec as gridspec
fig = plt.figure(figsize=(12,6))
gs = gridspec.GridSpec(2,4)
#Function that relies on what we've done above:
def grid(cols=['MSZoning', 'LotFrontage', 'LotArea', 'Street', 'Alley']):
for i in cols:
for vals in coords:
if str(train[i].dtype) == 'object':
plt.subplot('gs'+str(vals))
counts(cols)
else:
plt.subplot('gs'+str(vals))
histogram(cols)
When attempted, this code returns an error:
ValueError: Single argument to subplot must be a 3-digit integer
For purposes of helping you visualize, what I am hoping to achieve, I attach the screen shot below, which was produced by the line by line coding (with my created helper functions) I am trying to avoid:
Can anyone help me figure out where I am going wrong? I would appreciate any advice. Thank you!

The line plt.subplot('gs'+str(vals)) cannot work; which is also what the error tells you.
As can be seen from the matplotlib GridSpec tutorial, it needs to be
ax = plt.subplot(gs[0, 0])
So in your case you may use the values from the list as
ax = plt.subplot(gs[vals[0], vals[1]])
Mind that you also need to make sure that the coords list must have the n*m elements, if the gridspec is defined as gs = gridspec.GridSpec(n,m).

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