Streamlit（无数据图输出）

Question

我想用streamlit实现图表的输出，有一个模型和初始数据，以前图表在Speeder、PyCharn和Colab中显示，但在这里它不起作用，显示只是空的，就像一个白色 sheet.

Colab

这是它输出的 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)

正在导入：

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

Answer 1

所以你的错误是你从来没有调用 plotsir(t, S, I, R)。 plt.show() 不适用于 streamlit。请改用 st.pyplot()。工作代码：

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)

Answer 2

2020 年 12 月 1 日之后，Streamlit 将取消不带任何参数调用 st.pyplot() 的功能。需要使用Matplotlib的全局图形对象，不是线程安全的。

而是 st.pyplot(fig) 与 fig 对象。示例：

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

这意味着提供的解决方案中的“f”变量...

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

...必须像 st.pyplot() 的参数一样传递，在函数的末尾像这样：

st.pyplot(f)

Streamlit（无数据图输出）

Streamlit (graph output without data)

python

matplotlib

streamlit