我怎么能在 python 中 plot 我自己的 function ？

Question

I create a function in python and I like to plot it on set (0:100). So I have defined an vector x , but when I want to calculate y for each x , the python returns me this error:

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Can anyone help me?

g = 9.81
Qave = 0.05
def efficiency(h):
    Qn = 3e-5 *np.sqrt(2*g*h)
    NN = np.ceil(Qave/Qn)
    teta = 0.9 * np.sqrt(2*g*h)
    if (teta>=3.025):
        Kl = 2e-5
    else:
        Kl = 2*np.sqrt(1e-9/np.pi*teta)
    inverse_efficiency = np.exp(-Kl*1.2e4*teta)
    return(inverse_efficiency)
# plot data
x = np.arange(0.01, 100, 0.01)
y=efficiency(x)
plot.plot(time, amplitude)

Answer 1

Because your function efficieny can only be applied on a single value, not an array x .

Try to set y as:

y=[efficiency (i) for i in x]
#print (y)
plot.plot(x, y)

Answer 2

You can vectorize your function. At the moment the function only works for scalar inputs. That is why your function fails for a numpy array, because the if (teta >= 3.025): for arrays.

You can fix your code by using numpys vectorization.

import numpy as np
import matplotlib.pyplot as plt

g = 9.81
Qave = 0.05


def efficiency(h):
    Qn = 3e-5 * np.sqrt(2 * g * h)
    NN = np.ceil(Qave / Qn)
    teta = 0.9 * np.sqrt(2 * g * h)

    if teta >= 3.025:
        Kl = 2e-5
    else:
        Kl = 2 * np.sqrt(1e-9 / np.pi * teta)
    inverse_efficiency = np.exp(-Kl * 1.2e4 * teta)
    return (inverse_efficiency)

efficiency_vector = np.vectorize(efficiency)

# plot data
h = np.arange(0.01, 100, 0.01)
efficiency = efficiency_vector(h)
plt.plot(h, efficiency)