Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
839 views
in Technique[技术] by (71.8m points)

python - Matplotlib way to annotate bar plots with lines and figures

enter image description here

I would like to create an annotation to a bar chart that compares the value of the bar to two reference values. An overlay such as shown in the picture, a kind of staff gauge, is possible, but I'm open to more elegant solutions.

The bar chart is generated with the pandas API to matplotlib (e.g. data.plot(kind="bar")), so a plus would be if the solution is playing nicely with that.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

You may use smaller bars for the target and benchmark indicators. Pandas cannot annotate bars automatically, but you can simply loop over the values and use matplotlib's pyplot.annotate instead.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt

a = np.random.randint(5,15, size=5)
t = (a+np.random.normal(size=len(a))*2).round(2)
b = (a+np.random.normal(size=len(a))*2).round(2)
df = pd.DataFrame({"a":a, "t":t, "b":b})

fig, ax = plt.subplots()


df["a"].plot(kind='bar', ax=ax, legend=True)
df["b"].plot(kind='bar', position=0., width=0.1, color="lightblue",legend=True, ax=ax)
df["t"].plot(kind='bar', position=1., width=0.1, color="purple", legend=True, ax=ax)

for i, rows in df.iterrows():
    plt.annotate(rows["a"], xy=(i, rows["a"]), rotation=0, color="C0")
    plt.annotate(rows["b"], xy=(i+0.1, rows["b"]), color="lightblue", rotation=+20, ha="left")
    plt.annotate(rows["t"], xy=(i-0.1, rows["t"]), color="purple", rotation=-20, ha="right")

ax.set_xlim(-1,len(df))
plt.show()

enter image description here


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...