Dave的答案是不正确的，因为scipy的vonmises没有环绕[-pi, pi]。
你可以使用下面的代码，它基于同样的原理，基于numpy中描述的方程。

def vonmises_kde(data, kappa, n_bins=100):
    from scipy.special import i0
    bins = np.linspace(-np.pi, np.pi, n_bins)
    x = np.linspace(-np.pi, np.pi, n_bins)
    # integrate vonmises kernels
    kde = np.exp(kappa*np.cos(x[:, None]-data[None, :])).sum(1)/(2*np.pi*i0(kappa))
    kde /= np.trapz(kde, x=bins)
    return bins, kde

下面是一个例子

import matplotlib.pyplot as plt
import numpy as np
from numpy.random import vonmises

# generate complex circular distribution
data = np.r_[vonmises(-1, 5, 1000), vonmises(2, 10, 500), vonmises(3, 20, 100)]

# plot data histogram
fig, axes = plt.subplots(2, 1)
axes[0].hist(data, 100)

# plot kernel density estimates
x, kde = vonmises_kde(data, 20)
axes[1].plot(x, kde)

以下是@kingjr给出的更准确答案的快速近似：

def vonmises_pdf(x, mu, kappa):
    return np.exp(kappa * np.cos(x - mu)) / (2. * np.pi * scipy.special.i0(kappa))

def vonmises_fft_kde(data, kappa, n_bins):
    bins = np.linspace(-np.pi, np.pi, n_bins + 1, endpoint=True)
    hist_n, bin_edges = np.histogram(data, bins=bins)
    bin_centers = np.mean([bin_edges[1:], bin_edges[:-1]], axis=0)
    kernel = vonmises_pdf(
        x=bin_centers,
        mu=0,
        kappa=kappa
    )
    kde = np.fft.fftshift(np.fft.irfft(np.fft.rfft(kernel) * np.fft.rfft(hist_n)))
    kde /= np.trapz(kde, x=bin_centers)
    return bin_centers, kde

测试（使用tqdm进行进度条和计时，使用matplotlib验证结果）：

import numpy as np
from tqdm import tqdm
import scipy.stats
import matplotlib.pyplot as plt

n_runs = 1000
n_bins = 100
kappa = 10

for _ in tqdm(xrange(n_runs)):
    bins1, kde1 = vonmises_kde(
        data=np.r_[
            np.random.vonmises(-1, 5, 1000),
            np.random.vonmises(2, 10, 500),
            np.random.vonmises(3, 20, 100)
        ],
        kappa=kappa,
        n_bins=n_bins
    )

for _ in tqdm(xrange(n_runs)):
    bins2, kde2 = vonmises_fft_kde(
        data=np.r_[
            np.random.vonmises(-1, 5, 1000),
            np.random.vonmises(2, 10, 500),
            np.random.vonmises(3, 20, 100)
        ],
        kappa=kappa,
        n_bins=n_bins
    )

plt.figure()
plt.plot(bins1, kde1, label="kingjr's solution")
plt.plot(bins2, kde2, label="dolf's FFT solution")
plt.legend()
plt.show()

结果：

100%|██████████| 1000/1000 [00:07<00:00, 135.29it/s]
100%|██████████| 1000/1000 [00:00<00:00, 1945.14it/s]

(1945/ 135 =快14倍）

为了获得更快的速度，使用2的整数幂作为bin的数量。它的伸缩性也更好（即，它在许多bin和大量数据的情况下保持快速）。在我的PC上，它比原始答案（n_bins=1024）快118倍。

为什么会成功

两个信号（* 无 * 零填充）的FFT的乘积等于两个信号的circular (or cyclic) convolution。kernel density estimation基本上是与在每个数据点的位置处具有脉冲的信号卷积的核。
"为什么不准确"
由于我使用直方图来均匀地分布数据，因此我丢失了每个样本的确切位置，而只使用它所属的bin的中心。每个bin中的样本数用作该点的脉冲幅度。* 例如：* 暂时忽略归一化，如果您有一个从0到1的bin，以及该bin中的两个样本，分别位于0.1和0.2，则exact KDE将为the kernel centred around 0.1 + the kernel centred around 0.2。近似值将为2x '以0.5为中心的内核，这是bin的中心。

所以我有一个我认为合理的解决方案。基本上我使用冯米塞斯分布作为核密度估计的基函数。代码如下，以防对其他人有用。

def vonmises_KDE(data, kappa, plot=None):       

    """    
    Create a kernal densisity estimate of circular data using the von mises 
    distribution as the basis function.

    """

    # imports
    from scipy.stats import vonmises
    from scipy.interpolate import interp1d

    # convert to radians
    data = np.radians(data)

    # set limits for von mises
    vonmises.a = -np.pi
    vonmises.b = np.pi
    x_data = np.linspace(-np.pi, np.pi, 100)

    kernels = []

    for d in data:

        # Make the basis function as a von mises PDF
        kernel = vonmises(kappa, loc=d)
        kernel = kernel.pdf(x_data)
        kernels.append(kernel)

        if plot:
            # For plotting
            kernel /= kernel.max()
            kernel *= .2
            plt.plot(x_data, kernel, "grey", alpha=.5)

    vonmises_kde = np.sum(kernels, axis=0)
    vonmises_kde = vonmises_kde / np.trapz(vonmises_kde, x=x_data)
    f = interp1d( x_data, vonmises_kde )

    if plot:
        plt.plot(x_data, vonmises_kde, c='red')  

    return x_data, vonmises_kde, f

baz = np.array([-92.29061004, -85.42607874, -85.42607874, -70.01689348,
               -63.43494882, -63.43494882, -70.01689348, -70.01689348,
               -59.93141718, -63.43494882, -59.93141718, -63.43494882,
               -63.43494882, -63.43494882, -57.52880771, -53.61564818,
               -57.52880771, -63.43494882, -63.43494882, -92.29061004,
               -16.92751306, -99.09027692, -99.09027692, -16.92751306,
               -99.09027692, -16.92751306,  -9.86580694,  -8.74616226,
                -9.86580694,  -8.74616226,  -8.74616226,  -2.20259816,
                -2.20259816,  -2.20259816,  -9.86580694,  -2.20259816,
                -2.48955292,  -2.48955292,  -2.48955292,  -2.48955292,
                 4.96974073,   4.96974073,   4.96974073,   4.96974073,
                -2.48955292,  -2.48955292,  -2.48955292,  -2.48955292,
                -2.48955292,  -9.86580694,  -9.86580694,  -9.86580694,
               -16.92751306, -19.29004622, -19.29004622, -26.56505118,
               -19.29004622, -19.29004622, -19.29004622, -19.29004622])   
kappa = 12
x_data, vonmises_kde, f = vonmises_KDE(baz, kappa, plot=1)

作为@kingjr回答的一个小而重要的补充：我预计许多开发人员会希望以极坐标投影显示此KDE，如

import matplotlib.pyplot as plt
import numpy as np
from numpy.random import vonmises

def vonmises_kde(data, kappa, n_bins=100):
    from scipy.special import i0
    bins = np.linspace(-np.pi, np.pi, n_bins)
    x = np.linspace(-np.pi, np.pi, n_bins)
    # integrate vonmises kernels
    kde = np.exp(kappa*np.cos(x[:, None]-data[None, :])).sum(1)/(2*np.pi*i0(kappa))
    kde /= np.trapz(kde, x=bins)
    return bins, kde

# generate complex circular distribution
data = np.r_[vonmises(-1, 5, 1000), vonmises(2, 10, 500), vonmises(3, 20, 100)]

# plot data histogram
fig, axes = plt.subplots(2, 1, subplot_kw={'projection': 'polar'})
axes[0].hist(data, 100)

# plot kernel density estimates
x, kde = vonmises_kde(data, 20)
axes[1].plot(x, kde)

plt.show()