python - 关于格式化 DataFrame 以与 matplotlib 一起使用的问题

Question

我有一个偏微分方程的解，我想绘制它。我在文档中看到了两种方法，其中一种适合我，另一种则不适合我。不会产生错误。一种方法只是生成正确的绘图(正弦波)，另一种方法生成一条斜率为 1 的线。即使我现在有可以运行的代码，第二种方法在将来可能会有用。提前致谢。

工作解决方案:

plt.plot(arange(0, 16*pi, Dt), u[:, index])
plt.show()

这太棒了，而且 super 简单!下面的方法也在 matplotlib 文档中找到，但它产生了错误的绘图。我想知道我的错误:

无效解决方案:

df = pd.DataFrame({'t':arange(0, 16*pi, Dt), 'u':u[:,index]})
plt.plot('t', 'u', data=df)
plt.show()

<小时/>

上下文的完整代码

from math import sin, cos, pi, fabs, log10, ceil, floor
from numpy import arange, zeros
import pandas as pd
from matplotlib import pyplot as plt


#function applies periodic boundary condition where h is the period
def apply_pbc(f, i, Dx, M, h):
    f[i][0] = f[i][int(h/Dx)]
    f[i][int((M + Dx)/Dx)] = f[i][int((M + Dx)/Dx - 1)]

    return f

# function for finding an index associated with
# a particular data point of interest for plotting
# or other analysis
def find_index(start, stop, step, x):
    counter = len(arange(start, stop, step))
    for i in arange(counter):
        x_i = start + i*step
        if abs(x - x_i) < pow(10, -15):
            index = i
            print("x = ", x_i, "@index j = ", i)
            break

    return index

#main body
if __name__ == "__main__":

    #constants
    a = 0.25
    b = 0.25
    c = 1

    #period of boundary conditions
    h = 4*pi

    #space and time endpoints
    M = 4*pi
    N = 16*pi

    #mesh
    Dx = 0.005*4*pi
    Dt = (0.25*Dx)/c

    #simplification of numeric method
    r = (Dt*pow(c,2))/pow(Dx,2)

    #get size of data set
    rows = len(arange(0, N, Dt))
    cols = len(arange(-Dx, M, Dx))

    #initiate solution arrays
    u = zeros((rows, cols))
    v = zeros((rows, cols))

    #apply initial conditions
    for j in range(cols):
        x = -Dx + j*Dx
        u[0][j] = cos(x)
        v[0][j] = 0


    #solve
    for i in range(1, rows):
        for j in range(1, cols - 1):
            u[i][j] = u[i-1][j] + v[i-1][j]*Dt \
                    + (a/2)*(u[i-1][j+1] - 2*u[i-1][j] + u[i-1][j-1])

            v[i][j] = v[i-1][j] \
                    + r*(u[i-1][j+1] - 2*u[i-1][j] + u[i-1][j-1]) \
                    + (b/2)*(v[i-1][j+1] - 2*v[i-1][j] + v[i-1][j-1])
        apply_pbc(u, i, Dx, M, h)
        apply_pbc(v, i, Dx, M, h)

    print("done")

    #we want to plot the solution u(t,x), where x = pi
    index = find_index(-Dx, M + Dx, Dx, pi)

    df = pd.DataFrame({'t':arange(0,16*pi, Dt), 'u':u[:,index]})
    plt.plot('t', 'x', data=df)
    # plt.plot(arange(0, 16*pi, Dt), u[:, index])
    plt.show()

Answer 1

来自documentation of plt.plot() :

Plotting labelled data

There's a convenient way for plotting objects with labelled data (i.e. data that can be accessed by index obj['y']). Instead of giving the data in x and y, you can provide the object in the data parameter and just give the labels for x and y:

plot('xlabel', 'ylabel', data=obj)

我认为您的代码中有一个拼写错误。在您提供的完整代码中，这是绘制绘图的行:

plt.plot('t', 'x', data=df)

这确实给出了

更改为

plt.plot('t', 'u', data=df)

按预期工作:

代码的最后一点:

df = pd.DataFrame({'t':arange(0,16*pi, Dt), 'u':u[:,index]})
plt.plot('t', 'x', data=df)  # <-- 'x' instead of 'u'
# plt.plot(arange(0, 16*pi, Dt), u[:, index])
plt.show()