一般来说，@jared的答案是正确的。对于西内斯（或者，通常是任何周期函数），FFT是一种更可靠的方法，特别是如果它与二次峰值插值相结合：

import matplotlib.pyplot as plt
import numpy as np
amplitude = np.array((
    37.5, 36.4, 37.1, 37.3, 38.2, 38.4, 38.1, 36.7, 34.3, 32.2, 33.1,
    31.8, 33.4, 35.7, 37.8, 38.3, 38.1, 37. , 34.9, 32.5, 31.6, 31.7,
    33.5, 35.5, 37.8, 38.4, 38.3, 36.9, 34.8, 32.2, 33.9, 31.6, 33.3,
    35.5, 37.6, 38.3, 38.2, 36.7, 34.8, 32.4, 31.5, 32.1, 33.3, 35.5,
    37.2, 38.3, 38.3, 36.8, 34.5, 32.3, 31.6, 31.8, 33.3, 35.7, 37.8,
    38.4, 38.2))
timestep = 0.02
time = np.arange(0, timestep*(len(amplitude) - 0.5), timestep)
# first pass
spectrum = np.fft.rfft(amplitude, norm='forward')
spectral_amplitude = np.abs(spectrum)
freqs = np.fft.rfftfreq(n=time.size, d=timestep)
peak_index = 1 + spectral_amplitude[1:].argmax()
# Instead of peak_freq = freqs[peak_index]
# use parabolic peak interpolation from http://www.ericjacobsen.org/fe2/quadterp.txt
a, b, c = spectrum[peak_index-1: peak_index+2]
norm_frac = np.real((a - c)/(2*b - a - c))
peak_freq_interpolated = freqs[peak_index]*(1 + norm_frac) - freqs[peak_index - 1]*norm_frac
# second pass
# shorten the timeseries length so that the peak lands in the next-left bin
excerpt = int(np.round(time.size * freqs[peak_index-1] / peak_freq_interpolated))
spectrum = np.fft.rfft(amplitude[-excerpt:], norm='forward')
spectral_amplitude = np.abs(spectrum)
freqs = np.fft.rfftfreq(n=excerpt, d=timestep)
peak_index = 1 + spectral_amplitude[1:].argmax()
peak_freq = freqs[peak_index]
phase_correction = (1 - (len(time) - excerpt)*peak_freq*timestep)*2*np.pi
peak_phase = np.angle(spectrum[peak_index]) + phase_correction
peak_amplitude = 2*spectral_amplitude[peak_index]
print(f'Amplitude: {peak_amplitude:.3f}')
print(f'Frequency (Hz):', peak_freq)
print(f'Phase (deg): {np.rad2deg(peak_phase):.2f}')
reconstructed = peak_amplitude * np.cos(
    2*np.pi*peak_freq*time + peak_phase
) + spectral_amplitude[0]
ax_top: plt.Axes
ax_bottom: plt.Axes
fig, (ax_top, ax_bottom) = plt.subplots(nrows=2)
ax_top.plot(time, amplitude, label='experiment')
ax_top.plot(time, reconstructed, label='reconstructed')
ax_top.set_xlabel('Interpolated time (s)')
ax_top.set_ylabel('Amplitude')
ax_top.legend(loc='lower right')
ax_bottom.semilogy(freqs, spectral_amplitude)
ax_bottom.set_xlabel('Frequency (Hz)')
ax_bottom.set_ylabel('Amplitude')
plt.show()

Amplitude: 3.425
Frequency (Hz): 5.0
Phase (deg): 169.70

你可以通过给函数一个好的初始猜测并为目标函数提供一个雅可比函数来改进你的结果。

from scipy.optimize import curve_fit
import numpy as np
from matplotlib import pyplot as plt
plt.close("all")
xdata = np.array([0., 0.02, 0.04, 0.06, 0.08, 0.1, 0.12, 0.14, 0.16, 0.18, 0.2,
                  0.22, 0.24, 0.26, 0.28, 0.3, 0.32, 0.34, 0.36, 0.38, 0.4, 0.42,
                  0.44, 0.46, 0.48, 0.5, 0.52, 0.54, 0.56, 0.58, 0.6, 0.62, 0.64,
                  0.66, 0.68, 0.7, 0.72, 0.74, 0.76, 0.78, 0.8, 0.82, 0.84, 0.86,
                  0.88, 0.9, 0.92, 0.94, 0.96, 0.98, 1., 1.02, 1.04, 1.06, 1.08,
                  1.1, 1.12])
ydata = np.array([37.5, 36.4, 37.1, 37.3, 38.2, 38.4, 38.1, 36.7, 34.3, 32.2, 33.1,
                  31.8, 33.4, 35.7, 37.8, 38.3, 38.1, 37., 34.9, 32.5, 31.6, 31.7,
                  33.5, 35.5, 37.8, 38.4, 38.3, 36.9, 34.8, 32.2, 33.9, 31.6, 33.3,
                  35.5, 37.6, 38.3, 38.2, 36.7, 34.8, 32.4, 31.5, 32.1, 33.3, 35.5,
                  37.2, 38.3, 38.3, 36.8, 34.5, 32.3, 31.6, 31.8, 33.3, 35.7, 37.8,
                  38.4, 38.2])
def sin_fun(x, a, b, c, d):
    return a*np.sin(b*x+c)+d
def jac(x, a, b, c, d):
    return np.stack([np.sin(b*x+c),
                     a*x*np.cos(b*x+c),
                     a*np.cos(b*x+c),
                     np.ones_like(x)]).T
guess = [3, 28, -0.5, 35]
p_opt, p_cov = curve_fit(sin_fun, xdata, ydata, jac=jac, p0=guess)
print(p_opt)
x_fine = np.linspace(xdata.min(), xdata.max(), 1000)
y_fine = sin_fun(x_fine, *p_opt)
plt.plot(x_fine, y_fine)
plt.plot(xdata, ydata, 'x')
plt.show()

编辑：我在评论中被问到关于雅可比矩阵的问题，并给出了一个猜测。雅可比矩阵告诉曲线拟合函数，每个参数的变化将如何影响结果，它是通过对目标函数对每个参数的偏导数（按顺序）来计算的。对于猜测，我只是根据绘制的数据进行了目测。