Python: How to create a step plot with offline plotly for a pandas DataFrame?

Question:

Lets say we have following DataFrame and corresponding graph generated:

import pandas as pd
import plotly
from plotly.graph_objs import Scatter

df = pd.DataFrame({"value":[10,7,0,3,8]},
index=pd.to_datetime([
"2015-01-01 00:00",
"2015-01-01 10:00",
"2015-01-01 20:00",
"2015-01-02 22:00",
"2015-01-02 23:00"]))
plotly.offline.plot({"data": [Scatter( x=df.index, y=df["value"] )]})

plotly graph

Expected results

If I use below code:

import matplotlib.pyplot as plt
plt.step(df.index, df["value"],where="post")
plt.show()

I get a step graph as below:

step graph

Question

How can I get same results as step function but using offline plotly instead?

Asked By: Cedric Zoppolo

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Source

Answers:

We can use the line parameter shape option as hv using below code:

trace1 = {
  "x": df.index,
  "y": df["value"],
  "line": {"shape": 'hv'},
  "mode": 'lines',
  "name": 'value',
  "type": 'scatter'
};

data = [trace1]
plotly.offline.plot({
    "data": data
})

Which generates below graph:

enter image description here

Answered By: Cedric Zoppolo

As your data is a pandas dataframe, alternatively to offline plotly, you could use plotly express:

import plotly.express as plx

plx.line(df,line_shape='hv')

Other line_shape options, or interpolation methods between given points:

  • ‘hv’ step ends, equivalent to pyplot’s post option;
  • ‘vh’ step starts;
  • ‘hvh’ step middles, x axis;
  • ‘vhv’ step middles, y axis;
  • ‘spline’ smooth curve between points;
  • ‘linear’ line segments between points, default value for line_shape.

Just try them…

hv = plx.line(df,line_shape='hv')
vh = plx.line(df,line_shape='vh')
vhv = plx.line(df,line_shape='vhv')
hvh = plx.line(df,line_shape='hvh')
spline = plx.line(df,line_shape='spline')
line = plx.line(df,line_shape='linear')

Selection of one of them should be commited to the nature of your data…

Answered By: ePuntel