Streamlit (graph output without data)

Question

I want to implement the output of the chart using streamlit. There is a model and initial data. Previously, the chart was displayed in Speeder, PyCharm, and Colab, but here it does not work and is displayed just empty, like a white sheet.

Colab:

Here’s what it outputs localhost streamlit

streamlit:

def SIR(y, t, N, beta, gamma):
    S, I, R = y
    dSdt = -beta * S * I / N
    dIdt = beta * S * I / N - gamma * I
    dRdt = gamma * I
    return dSdt, dIdt, dRdt

N = 1000
beta = 1.0
D = 4.0
gamma = 1.0 / D

S0, I0, R0 = 999, 1, 0

t = np.linspace(0, 49, 50)
y0 = S0, I0, R0

ret = odeint(SIR, y0, t, args=(N, beta, gamma))
S, I, R = ret.T

def plotsir(t, S, I, R):
  f, ax = plt.subplots(1,1,figsize=(10,4))
  ax.plot(t, S, 'b', alpha=0.7, linewidth=2, label='Susceptible')
  ax.plot(t, I, 'y', alpha=0.7, linewidth=2, label='Infected')
  ax.plot(t, R, 'g', alpha=0.7, linewidth=2, label='Recovered')

  ax.set_xlabel('Time (days)')

  ax.yaxis.set_tick_params(length=0)
  ax.xaxis.set_tick_params(length=0)
  ax.grid(b=True, which='major', c='w', lw=2, ls='-')
  legend = ax.legend()
  legend.get_frame().set_alpha(0.5)
  for spine in ('top', 'right', 'bottom', 'left'):
      ax.spines[spine].set_visible(False)
      plt.show()


st.pyplot(plt)

Making import:

import streamlit as st
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt

Asked By: Konstantin Ivanov

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Source

Answer 1

So your error is that you never call plotsir(t, S, I, R). And plt.show() doesn’t work with streamlit. Use instead st.pyplot(). The working code:

import streamlit as st
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt

def SIR(y, t, N, beta, gamma):
    S, I, R = y
    dSdt = -beta * S * I / N
    dIdt = beta * S * I / N - gamma * I
    dRdt = gamma * I
    return dSdt, dIdt, dRdt

N = 1000
beta = 1.0
D = 4.0
gamma = 1.0 / D

S0, I0, R0 = 999, 1, 0

t = np.linspace(0, 49, 50)
y0 = S0, I0, R0

ret = odeint(SIR, y0, t, args=(N, beta, gamma))
S, I, R = ret.T

def plotsir(t, S, I, R):
  f, ax = plt.subplots(1,1,figsize=(10,4))
  ax.plot(t, S, 'b', alpha=0.7, linewidth=2, label='Susceptible')
  ax.plot(t, I, 'y', alpha=0.7, linewidth=2, label='Infected')
  ax.plot(t, R, 'g', alpha=0.7, linewidth=2, label='Recovered')

  ax.set_xlabel('Time (days)')

  ax.yaxis.set_tick_params(length=0)
  ax.xaxis.set_tick_params(length=0)
  ax.grid(b=True, which='major', c='w', lw=2, ls='-')
  legend = ax.legend()
  legend.get_frame().set_alpha(0.5)
  for spine in ('top', 'right', 'bottom', 'left'):
      ax.spines[spine].set_visible(False)
      st.pyplot()

plotsir(t, S, I, R)

Answered By: Gledi

Answer 2

After December 1st, 2020, Streamlit will remove the ability to call st.pyplot() without any arguments. It requires the use of Matplotlib’s global figure object, which is not thread-safe.

Instead st.pyplot(fig) with fig object. Example :

>>> fig, ax = plt.subplots()
>>> ax.scatter([1, 2, 3], [1, 2, 3]) 
>>>    ... other plotting actions ...
>>> st.pyplot(fig)

That means the ‘ f ‘ variable from the solution provided…

f, ax = plt.subplots(1,1,figsize=(10,4))

…has to be passed like the argument of st.pyplot(), at the end of the function like this:

st.pyplot(f)

Answered By: Zyncho

Streamlit (graph output without data)

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