在python中如何将多个列表相乘？

我一直在考虑生成一个乘法函数，用于一个名为“合并”的方法。该方法可以在下面的文章（）中找到。合并方程式如下所示：

我知道可以使用以下代码和函数将两个列表相乘：

[a*b for a,b in zip(lista,listb)]
list(map(operator.mul, lista, listb))
np.multiply(lista,listb)
ab = [lista[i]*listb[i] for i in range(len(lista))]
lista = [1,2,3,4]
listb = [2,3,4,5]
ab = []                        #Create empty list
for i in range(0, len(lista)):
     ab.append(lista[i]*listb[i])      #Adds each element to the list

但是查看2个以上的列表时，我会不断收到关于size-1数组的错误消息，或者代码会查看每个分布的第一个变量，但是，对于循环的其余部分，它会一直打印相同的值，不会转到列表中的下一个值，合并的分布是一个变量。请参阅以下代码以及部分输出和错误消息：

第1代码：

from scipy.integrate import quad
from scipy import stats
import numpy as np

def prod_pdf(x,dists):
    p_pdf=1
    print('Incoming Array:', p_pdf)
    for c,dist in enumerate(dists):
        p_pdf=p_pdf*dist[c]
        print('final:', p_pdf)
    return p_pdf

def conflate_pdf(x,dists,lb,ub):
    print('Input product pdf: ', prod_pdf(x,dists))
    denom = quad(prod_pdf, lb, ub, args=(dists,))[0]
    # denom = simps(prod_pdf)
    # denom = nquad(func=(prod_pdf), ranges=([lb, ub]), args=(dists,))[0]
    print('Denom: ', denom)
    conflated_pdf=prod_pdf(x,dists)/denom
    print('Conflated PDF: ', conflated_pdf)
    return conflated_pdf

lb=-10
ub=10
domain=np.arange(lb,ub,.01)

dist_1 = st.norm.pdf(domain, 2,1)
dist_2 = st.norm.pdf(domain, 2.5,1.5)
dist_3 = st.norm.pdf(domain, 2.2,1.6)
dist_4 = st.norm.pdf(domain, 2.4,1.3)
dist_5 = st.norm.pdf(domain, 2.7,1.5)

from matplotlib import pyplot as plt
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("Conflated PDF")
plt.legend()
plt.plot(domain, dist_1, 'r', label='Dist. 1')
plt.plot(domain, dist_2, 'g', label='Dist. 2')
plt.plot(domain, dist_3, 'b', label='Dist. 3')
plt.plot(domain, dist_4, 'y', label='Dist. 4')
plt.plot(domain, dist_5, 'c', label='Dist. 5')

dists=[dist_1, dist_2, dist_3, dist_4, dist_5]
print('distribution list: \n', dists)
graph=conflate_pdf(domain, dists,lb,ub)

plt.plot(domain,graph, 'm', label='Conflated Dist.')
plt.show()

输出的一部分：

Incoming Array: 1
final: 2.1463837356630605e-32
final: 5.0231307782193034e-48
final: 3.266239495519432e-61
final: 2.187514996217005e-81
final: 1.979657878680375e-97
Incoming Array: 1
final: 2.1463837356630605e-32
final: 5.0231307782193034e-48
final: 3.266239495519432e-61
final: 2.187514996217005e-81
final: 1.979657878680375e-97
Denom:  3.95931575736075e-96
Incoming Array: 1
final: 2.1463837356630605e-32
final: 5.0231307782193034e-48
final: 3.266239495519432e-61
final: 2.187514996217005e-81
final: 1.979657878680375e-97
Conflated PDF:  0.049999999999999996

第二代码：

import winsound
from functools import reduce
from itertools import chain
import scipy.stats as st
from glob import glob
from collections import defaultdict, Counter
from sklearn.neighbors import KDTree
import pywt
import peakutils
import scipy
import os
from scipy import signal
from scipy.fftpack import fft, fftfreq, rfft, rfftfreq, dst, idst, dct, idct
from scipy.signal import find_peaks, find_peaks_cwt, argrelextrema, welch, lfilter, butter, savgol_filter, medfilt, freqz, filtfilt
from pylab import *
import glob
import sys
import re
from numpy import NaN, Inf, arange, isscalar, asarray, array
from scipy.stats import skew, kurtosis, median_absolute_deviation
import warnings
import numpy as np
import pandas as pd
import scipy.stats as st
import matplotlib.pyplot as plt
from scipy.stats import pearsonr, kendalltau, spearmanr, ppcc_max
import matplotlib.mlab as mlab
from statsmodels.graphics.tsaplots import plot_acf
from tsfresh.feature_extraction.feature_calculators import mean_abs_change as mac
from tsfresh.feature_extraction.feature_calculators import mean_change as mc
from tsfresh.feature_extraction.feature_calculators import mean_second_derivative_central as msdc
from pyAudioAnalysis.ShortTermFeatures import energy as stEnergy
import pymannkendall as mk_test
from sklearn.preprocessing import MinMaxScaler, Normalizer, normalize, StandardScaler
from scipy.integrate import quad,simps, quad_vec, nquad

def prod_pdf(x,dists):
    i=0
    # p_pdf=np.ones(np.array(dists)[0].shape)
    dist_size = np.array(dists).shape
    print('Incoming Array:', dists)
    print('Incoming Array Size:', dist_size[1])
    print('Full Incoming Array Size:', dist_size)
    print('Number of Incoming Array Size:', dist_size[0])
    # print('Incoming Product Array:', p_pdf)
    # print('Incoming Product Array Size:', np.array(p_pdf).shape)
    if dist_size[0]==2:
        p_pdf=dists[0]*dists[1]
        print('final:', p_pdf)
        results=dists[0]*dists[1]
        i+=1
    elif dist_size[0]>2:
        results=dists[0]*dists[1]
        for i in range(2, dist_size[0]):
            p_pdf=results*dists[i]
            print('final:', p_pdf)
    return p_pdf

def conflate_pdf(x,dists,lb,ub):
    print('Input product pdf: ', prod_pdf(x,dists))
    denom = quad(prod_pdf, lb, ub, args=(dists,))[0]
    # denom = simps(prod_pdf)
    # denom = nquad(func=(prod_pdf), ranges=([lb, ub]), args=(dists,))[0]
    print('Denom: ', denom)
    conflated_pdf=prod_pdf(x,dists)/denom
    print('Conflated PDF: ', conflated_pdf)
    return conflated_pdf

lb=-10
ub=10
domain=np.arange(lb,ub,.01)

dist_1 = st.norm.pdf(domain, 2,1)
dist_2 = st.norm.pdf(domain, 2.5,1.5)
dist_3 = st.norm.pdf(domain, 2.2,1.6)
dist_4 = st.norm.pdf(domain, 2.4,1.3)
dist_5 = st.norm.pdf(domain, 2.7,1.5)

# dist_1 = list(st.norm.pdf(domain, 2,1))
# dist_2 = list(st.norm.pdf(domain, 2.5,1.5))
# dist_3 = list(st.norm.pdf(domain, 2.2,1.6))
# dist_4 = list(st.norm.pdf(domain, 2.4,1.3))
# dist_5 = list(st.norm.pdf(domain, 2.7,1.5))

from matplotlib import pyplot as plt
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("Conflated PDF")
plt.legend()
plt.plot(domain, dist_1, 'r', label='Dist. 1')
plt.plot(domain, dist_2, 'g', label='Dist. 2')
plt.plot(domain, dist_3, 'b', label='Dist. 3')
plt.plot(domain, dist_4, 'y', label='Dist. 4')
plt.plot(domain, dist_5, 'c', label='Dist. 5')

dists=[dist_1, dist_2, dist_3, dist_4, dist_5]
# print('distribution list: \n', dists)
graph=conflate_pdf(domain, dists,lb,ub)

plt.plot(domain,graph, 'm', label='Conflated Dist.')
plt.show()

错误消息：

in line 79, in conflate_pdf:
    denom = quad(prod_pdf, lb, ub, args=(dists,))[0]

  File "D:\Anaconda\lib\site-packages\scipy\integrate\quadpack.py", line 351, in quad
    retval = _quad(func, a, b, args, full_output, epsabs, epsrel, limit,

  File "D:\Anaconda\lib\site-packages\scipy\integrate\quadpack.py", line 463, in _quad
    return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)

TypeError: only size-1 arrays can be converted to Python scalars

在我看来，第一个代码是我想要的方法，第二个代码中的错误是因为集成部分需要一个标量数。如何修复这两个代码以获得以下输出

代码：

以下是所需输出的一小部分：

Incoming Array: 1
final: 0.15352177537004433
final: 0.034348669264845304
final: 0.006519131844904635
final: 0.0015040030811035296
final: 0.0003607258742065213
Incoming Array: 1
final: 0.042345986284209325
final: 0.006294747321619583
final: 0.0007651214249593444
final: 9.805307029794648e-05
final: 1.668121592516301e-05
Denom:  0.0029066671327537714
Incoming Array: 1
final: [2.14638374e-32 2.41991991e-32 2.72804284e-32 ... 6.41980576e-15
 5.92770938e-15 5.47278628e-15]
final: [4.75178372e-48 5.66328097e-48 6.74864868e-48 ... 7.03075979e-21
 6.27970218e-21 5.60806584e-21]
final: [2.80912097e-61 3.51131870e-61 4.38823989e-61 ... 1.32670185e-26
 1.14952951e-26 9.95834610e-27]
final: [1.51005552e-81 2.03116529e-81 2.73144352e-81 ... 1.76466623e-34
 1.46198598e-34 1.21092834e-34]
final: [1.09076800e-97 1.55234627e-97 2.20861552e-97 ... 3.72095218e-40
 2.98464396e-40 2.39335035e-40]
Conflated PDF:  [3.75264162e-95 5.34063998e-95 7.59844666e-95 ... 1.28014389e-37
 1.02682689e-37 8.23400219e-38]

所需绘图：

编辑1:

我设法根据@MaxPierini更新了更新代码，但是，我无法获得所需的合并分布图。请参阅以下代码和输出：

代码：

输出：

Incoming Array: [array([2.14638374e-32, 2.41991991e-32, 2.72804284e-32, ...,
       6.41980576e-15, 5.92770938e-15, 5.47278628e-15]), array([2.21385563e-16, 2.34027620e-16, 2.47380598e-16, ...,
       1.09516706e-06, 1.05938091e-06, 1.02471859e-06]), array([5.91171893e-14, 6.20014921e-14, 6.50239789e-14, ...,
       1.88699641e-06, 1.83054781e-06, 1.77571847e-06]), array([5.37554463e-21, 5.78462242e-21, 6.22446263e-21, ...,
       1.33011515e-08, 1.27181248e-08, 1.21599343e-08]), array([7.22336360e-17, 7.64263883e-17, 8.08589121e-17, ...,
       2.10858694e-06, 2.04149972e-06, 1.97645911e-06])]
Incoming Array Size: (2000,)
Incoming Product Array: [1. 1. 1. ... 1. 1. 1.]
Incoming Product Array Size: (2000,)
final: [2.14638374e-32 2.14638374e-32 2.14638374e-32 ... 2.14638374e-32
 2.14638374e-32 2.14638374e-32]
final: [5.02313078e-48 5.02313078e-48 5.02313078e-48 ... 5.02313078e-48
 5.02313078e-48 5.02313078e-48]
final: [3.2662395e-61 3.2662395e-61 3.2662395e-61 ... 3.2662395e-61 3.2662395e-61
 3.2662395e-61]
final: [2.187515e-81 2.187515e-81 2.187515e-81 ... 2.187515e-81 2.187515e-81
 2.187515e-81]
final: [1.97965788e-97 1.97965788e-97 1.97965788e-97 ... 1.97965788e-97
 1.97965788e-97 1.97965788e-97]
Input product pdf:  [1.97965788e-97 1.97965788e-97 1.97965788e-97 ... 1.97965788e-97
 1.97965788e-97 1.97965788e-97]
Conflated PDF:  [0.0005 0.0005 0.0005 ... 0.0005 0.0005 0.0005]

绘图：

编辑2:

我实现了答案中的代码（由@MaxPierini提供），它似乎起了作用，而且，如果我将

quad

更改为

fixed\u quad

并使pdf列表正常化，我设法解决了

quad

的问题。我会得到同样的结果。以下是代码：

import scipy.stats as st
import numpy as np
import scipy.stats as st
import matplotlib.pyplot as plt
from sklearn.preprocessing import MinMaxScaler, Normalizer, normalize, StandardScaler
from scipy.integrate import quad, simps, quad_vec, nquad, cumulative_trapezoid
from scipy.integrate import romberg, trapezoid, simpson, romb
from scipy.integrate import fixed_quad, quadrature, quad_explain
from scipy import stats
import time

def user_prod_pdf(x,dists):
p_list=[]
p_pdf=1
print('Incoming Array:', p_pdf)
for dist in dists:
print('Incoming Distribution Array:', dist.pdf(x))
p_pdf=p_pdf*dist.pdf(x)
print('Product PDF:', p_pdf)
p_list.append(p_pdf)
print('final Product PDF:', p_pdf)
print('Product PDF list: ', p_list)
return p_pdf

def user_conflate_pdf(x,dists,lb,ub):
print('Input product pdf: ', user_prod_pdf(x,dists))
denom = quad(user_prod_pdf, lb, ub, args=(dists,))[0]
print('Denom: ', denom)
conflated_pdf=user_prod_pdf(x,dists)/denom
print('Conflated PDF: ', conflated_pdf)
return conflated_pdf

def user_conflate_pdf_2(pdfs):
"""
Compute conflation of given pdfs.

[ARGS]
- pdfs: PDFs numpy array of shape (n, x)
where n is the number of PDFs
and x is the variable space.

[RETURN]
A 1d-array of normalized conflated PDF.
"""
# conflate
conflation = np.array(pdfs).prod(axis=0)
# normalize
conflation /= conflation.sum()
return conflation

def my_product_pdf(x,dists):
p_list=[]
p_pdf=1
print('Incoming Array:', p_pdf)
list_full_size=np.array(dists).shape
print('Full list size: ', list_full_size)
print('list size: ', list_full_size[0])
for x in range(list_full_size[1]):
p_pdf=1
for y in range(list_full_size[0]):
p_pdf=float(p_pdf)*dists[y][x]
print('Product value: ', p_pdf)
print('Product PDF:', p_pdf)
p_list.append(p_pdf)
print('final Product PDF:', p_pdf)
print('Product PDF list: ', p_list)
# return p_pdf
return p_list
# return np.array(p_list)

def my_conflate_pdf(x,dists,lb,ub):
print('\n')
# print('product pdf: ', prod_pdf(x,dists))
print('product pdf: ', my_product_pdf(x,dists))
denom = fixed_quad(my_product_pdf, lb, ub, args=(dists,), n=1)[0]
print('Denom: ', denom)
# conflated_pdf=prod_pdf(x,dists)/denom
conflated_pdf=my_product_pdf(x,dists)/denom
# conflated_pdf=[i / j for i,j in zip(my_product_pdf(x,dists), denom)]
print('Conflated PDF: ', conflated_pdf)
return conflated_pdf

lb=-10
ub=10
domain=np.arange(lb,ub,.01)

# dist_1 = st.norm(2,1)
# dist_2 = st.norm(2.5,1.5)
# dist_3 = st.norm(2.2,1.6)
# dist_4 = st.norm(2.4,1.3)
# dist_5 = st.norm(2.7,1.5)

# dist_1_pdf = st.norm.pdf(domain, 2,1)
# dist_2_pdf = st.norm.pdf(domain, 2.5,1.5)
# dist_3_pdf = st.norm.pdf(domain, 2.2,1.6)
# dist_4_pdf = st.norm.pdf(domain, 2.4,1.3)
# dist_5_pdf = st.norm.pdf(domain, 2.7,1.5)

# dist_1_pdf /= dist_1_pdf.sum()
# dist_2_pdf /= dist_2_pdf.sum()
# dist_3_pdf /= dist_3_pdf.sum()
# dist_4_pdf /= dist_4_pdf.sum()
# dist_5_pdf /= dist_5_pdf.sum()

dist_1 = st.norm(2,1)
dist_2 = st.norm(4,2)
dist_3 = st.norm(7,4)
dist_4 = st.norm(2.4,1.3)
dist_5 = st.norm(2.7,1.5)

dist_1_pdf = st.norm.pdf(domain, 2,1)
dist_2_pdf = st.norm.pdf(domain, 4,2)
dist_3_pdf = st.norm.pdf(domain, 7,4)
dist_4_pdf = st.norm.pdf(domain, 2.4,1.3)
dist_5_pdf = st.norm.pdf(domain, 2.7,1.5)

# dist_1_pdf /= dist_1_pdf.sum()
# dist_2_pdf /= dist_2_pdf.sum()
# dist_3_pdf /= dist_3_pdf.sum()
# dist_4_pdf /= dist_4_pdf.sum()
# dist_5_pdf /= dist_5_pdf.sum()

# User:
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("User Conflated PDF")
plt.plot(domain, dist_1_pdf, 'r', label='Dist. 1')
plt.plot(domain, dist_2_pdf, 'g', label='Dist. 2')
plt.plot(domain, dist_3_pdf, 'b', label='Dist. 3')
plt.plot(domain, dist_4_pdf, 'y', label='Dist. 4')
plt.plot(domain, dist_5_pdf, 'c', label='Dist. 5')

dists=[dist_1, dist_2, dist_3, dist_4, dist_5]
user_graph=user_conflate_pdf(domain,dists,lb,ub)
print('Final Conflated PDF: ', user_graph)

# user_graph /= user_graph.sum()

plt.plot(domain, user_graph, 'm', label='Conflated PDF')
plt.legend()
plt.show()

# User 2:
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("User Conflated PDF 2")
plt.plot(domain, dist_1_pdf, 'r', label='Dist. 1')
plt.plot(domain, dist_2_pdf, 'g', label='Dist. 2')
plt.plot(domain, dist_3_pdf, 'b', label='Dist. 3')
plt.plot(domain, dist_4_pdf, 'y', label='Dist. 4')
plt.plot(domain, dist_5_pdf, 'c', label='Dist. 5')

dists=[dist_1_pdf, dist_2_pdf, dist_3_pdf, dist_4_pdf, dist_5_pdf]
user_graph=user_conflate_pdf_2(dists)
print('Final User Conflated PDF 2 : ', user_graph)

# user_graph /= user_graph.sum()

plt.plot(domain, user_graph, 'm', label='Conflated PDF')
plt.legend()
plt.show()

# My Code:
# from matplotlib import pyplot as plt
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("My Conflated PDF Code")
plt.plot(domain, dist_1_pdf, 'r', label='Dist. 1')
plt.plot(domain, dist_2_pdf, 'g', label='Dist. 2')
plt.plot(domain, dist_3_pdf, 'b', label='Dist. 3')
plt.plot(domain, dist_4_pdf, 'y', label='Dist. 4')
plt.plot(domain, dist_5_pdf, 'c', label='Dist. 5')

dists=[dist_1_pdf, dist_2_pdf, dist_3_pdf, dist_4_pdf, dist_5_pdf]
my_graph=my_conflate_pdf(domain,dists,lb,ub)
print('Final Conflated PDF: ', my_graph)

my_graph /= np.array(my_graph).sum()

# my_graph = inverse_normalise(my_graph)

plt.plot(domain, my_graph, 'm', label='Conflated PDF')
plt.legend()
plt.show()

# Conflated PDF:
print('User Conflated PDF: ', user_graph)
print('My Conflated PDF: ', np.array(my_graph))

以下是输出：

Incoming Array: 1
final: 2.1463837356630605e-32
final: 5.0231307782193034e-48
final: 3.266239495519432e-61
final: 2.187514996217005e-81
final: 1.979657878680375e-97
Incoming Array: 1
final: 2.1463837356630605e-32
final: 5.0231307782193034e-48
final: 3.266239495519432e-61
final: 2.187514996217005e-81
final: 1.979657878680375e-97
Denom:  3.95931575736075e-96
Incoming Array: 1
final: 2.1463837356630605e-32
final: 5.0231307782193034e-48
final: 3.266239495519432e-61
final: 2.187514996217005e-81
final: 1.979657878680375e-97
Conflated PDF:  0.049999999999999996

我的问题是，我知道我需要规范化PDF列表。但是，假设我没有将PDF标准化，我如何修改我的合并代码以获得以下绘图

要获取上面的绘图和我的合并代码：

# user_graph /= user_graph.sum()
# dist_1_pdf /= dist_1_pdf.sum()
# dist_2_pdf /= dist_2_pdf.sum()
# dist_3_pdf /= dist_3_pdf.sum()
# dist_4_pdf /= dist_4_pdf.sum()
# dist_5_pdf /= dist_5_pdf.sum()

我的合并代码图没有标准化：

---

在第二个

prod\u pdf

中，您使用的是计算pdf，而在第一个中，您使用的是定义的分布。因此，在第二个

prod_pdf

中，您已经拥有了pdf。因此，在for循环中，您只需执行

p\u pdf=p\u pdf*pdf

从您链接的论文中，我们知道“对于离散输入分布，合并的类似定义是概率质量函数的归一化乘积”。因此，您不仅需要获取PDF的乘积，还需要对其进行规范化。因此，重写离散分布的方程，我们得到

其中，

F

是我们需要合并的分布数，

N

是离散变量

x

的长度

import numpy as np
import scipy.stats as sps
import matplotlib.pyplot as plt

# use defined distributions
def prod_pdf_1(x, dists):
    num = 1
    for dist in dists:
        num *= dist.pdf(x)
    den = 0
    for i in x:
        _mul = 1
        for dist in dists:
            _mul *= dist.pdf(i)
        den += _mul
    return num / den

# use computed PDFs
def prod_pdf_2(pdfs):
    num = 1
    for pdf in pdfs:
        num *= pdf
    den = 0
    for i in range(len(num)):
        _mul = 1
        for pdf in pdfs:
            _mul *= pdf[i]
        den += _mul
    return num / den

第一个定义使用分布，第二个定义使用PDF

现在让我们定义分布和PDF

# define x variable
x = np.linspace(-2.5, 7.5, 100)

# define distributions
dists = [
    sps.norm(2.0, 1.0),
    sps.norm(2.5, 1.5),
    sps.norm(2.2, 1.6),
    sps.norm(2.4, 1.3),
    sps.norm(2.7, 1.5),
]

# compute PDFs
pdfs = [
    sps.norm.pdf(x, 2.0, 1.0),
    sps.norm.pdf(x, 2.5, 1.5),
    sps.norm.pdf(x, 2.2, 1.6),
    sps.norm.pdf(x, 2.4, 1.3),
    sps.norm.pdf(x, 2.7, 1.5),
]

我们现在可以计算合并和绘图。注意，我们不需要规范化合并的分布，因为我们已经这样做了，但是我们需要在绘图之前规范化（求和为1）单个分布

# first method
p_pdf_1 = prod_pdf_1(x, dists)
# second method
p_pdf_2 = prod_pdf_2(pdfs)

# compare
for pdf in pdfs:
    # normalize PDF to sum to 1
    pdf /= pdf.sum()
    plt.plot(x, pdf)
plt.plot(x, p_pdf_1, label='prod 1', lw=5, color='C1')
plt.plot(x, p_pdf_2, label='prod 2', ls='--', color='k')
plt.legend();

也许这不是最优雅的解决方案，您最好使用矩阵，尤其是如果您需要计算大量分布的合并

---

更新 一个更优雅的解决方案是简单地

# use computed PDFs and matrix
def conflate(pdfs):
    # numerator (product)
    conflation = pdfs.prod(axis=0)
    # normalize (divide by the integral)
    conflation /= conflation.sum()
    return conflation

并将PDF定义为2d数组（a

F

x

N

矩阵、

F

分布PDF和

N

离散变量长度

x

）

我们也是

conflation = conflate(pdfs)

# compare
for i, pdf in enumerate(pdfs):
    # normalize PDF to sum to 1
    pdf /= pdf.sum()
    plt.plot(x, pdf, label=f'dist {i+1}')
plt.plot(x, conflation, label='conflation', ls='--', color='k')
plt.legend();

更新2 完整代码

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

def conflate(pdfs):
    """
    Compute conflation of given pdfs.
    
    [ARGS]
    - pdfs: PDFs numpy array of shape (n, x)
            where n is the number of PDFs
            and x is the variable space.
            
    [RETURN]
    A 1d-array of normalized conflated PDF.
    """
    # conflate
    conflation = pdfs.prod(axis=0)
    # normalize
    conflation /= conflation.sum()
    
    return conflation

# define x limits and size
lb = -10 
ub = 10 
size = 1000
# x linear space
x = np.linspace(lb, ub, size)

# define PDFs in x:
# these are Probability Density Functions
# evaluated in x defined linear space
pdf_1 = sps.norm.pdf(x, 2, 1)
pdf_2 = sps.norm.pdf(x, 2.5, 1.5) 
pdf_3 = sps.norm.pdf(x, 2.2, 1.6) 
pdf_4 = sps.norm.pdf(x, 2.4, 1.3) 
pdf_5 = sps.norm.pdf(x, 2.7, 1.5) 
# PDFs in (n, x) array
pdfs = np.array([pdf_1, pdf_2, pdf_3, pdf_4, pdf_5])
# compute PDFs conflation
conflated_pdf = conflate(pdfs)

# ..............................
#      >>> ========== <<<      .
# plot !!!_NORMALIZED_!!! PDFs .
#      >>> ========== <<<      .
# ..............................
for i, pdf in enumerate(pdfs):
    plt.plot(x, pdf/pdf.sum(), c=f'C{i}', label=f'PDF {i+1}', lw=1)
    # normalize =============
    #           ^^^^^^^^^^^^^
    #           PDFs really do definitely need
    #           to be normalized, i.e. they have
    #           to sum to 1, because the cumulative
    #           probability needs to be 1 (100%)

# Plot conflated PDF
# NOTE: we don't need to normalize
# because it is already normalized
plt.plot(x, conflated_pdf, 
         'k--', label='Conflated PDF', lw=2)

# Plot options here
plt.xlabel("x") 
plt.ylabel("probability density") 
plt.title("Conflated PDF") 
plt.legend(loc='upper left')

plt.show()

将numpy导入为np
将scipy.stats导入为SP
从matplotlib导入pyplot作为plt
def合并（PDF）：
"""
计算给定PDF的合并。
[ARGS]
-pdfs:pdfs形状的numpy数组（n，x）
其中n是PDF的数量
x是变量空间。
[返回]
标准化合并PDF的一维数组。
"""
#合并
合并=pdfs.prod（轴=0）
#正常化
合并/=合并.sum（）
回流合并
#定义x限制和大小
磅=-10
ub=10
尺寸=1000
#x线性空间
x=np.linspace（磅、磅、磅）
#在x中定义PDF：
#这些是概率密度函数
#在x定义的线性空间中求值
pdf_1=sps.norm.pdf（x，2，1）
pdf_2=sps.norm.pdf（x，2.5，1.5）
pdf_3=sps.norm.pdf（x，2.2，1.6）
pdf_4=sps.norm.pdf（x，2.4，1.3）
pdf_5=sps.norm.pdf（x，2.7，1.5）
#（n，x）阵列中的PDF
pdfs=np.array（[pdf_1，pdf_2，pdf_3，pdf_4，pdf_5]）
#计算PDF合并
合并\u pdf=合并（pdf）
# ..............................
#>>>======================在第二个
prod\u pdf
中，您使用的是计算pdf，而在第一个中，您使用的是定义的分发。因此，在第二个
prod_pdf
中，您已经拥有了pdf。因此，在for循环中，您只需执行
p\u pdf=p\u pdf*pdf

从您链接的论文中，我们知道“对于离散输入分布，合并的类似定义是概率质量函数的归一化乘积”。因此，您不仅需要获取PDF的乘积，还需要对其进行规范化。因此，重写离散分布的方程，我们得到


其中，
F
是我们需要合并的分布数，
N
是离散变量
x
的长度

import numpy as np
import scipy.stats as sps
import matplotlib.pyplot as plt

# use defined distributions
def prod_pdf_1(x, dists):
    num = 1
    for dist in dists:
        num *= dist.pdf(x)
    den = 0
    for i in x:
        _mul = 1
        for dist in dists:
            _mul *= dist.pdf(i)
        den += _mul
    return num / den

# use computed PDFs
def prod_pdf_2(pdfs):
    num = 1
    for pdf in pdfs:
        num *= pdf
    den = 0
    for i in range(len(num)):
        _mul = 1
        for pdf in pdfs:
            _mul *= pdf[i]
        den += _mul
    return num / den

第一个定义使用分布，第二个定义使用PDF

现在让我们定义分布和PDF

# define x variable
x = np.linspace(-2.5, 7.5, 100)

# define distributions
dists = [
    sps.norm(2.0, 1.0),
    sps.norm(2.5, 1.5),
    sps.norm(2.2, 1.6),
    sps.norm(2.4, 1.3),
    sps.norm(2.7, 1.5),
]

# compute PDFs
pdfs = [
    sps.norm.pdf(x, 2.0, 1.0),
    sps.norm.pdf(x, 2.5, 1.5),
    sps.norm.pdf(x, 2.2, 1.6),
    sps.norm.pdf(x, 2.4, 1.3),
    sps.norm.pdf(x, 2.7, 1.5),
]

我们现在可以计算合并和绘图。注意，我们不需要规范化合并的分布，因为我们已经这样做了，但是我们需要在绘图之前规范化（求和为1）单个分布

# first method
p_pdf_1 = prod_pdf_1(x, dists)
# second method
p_pdf_2 = prod_pdf_2(pdfs)

# compare
for pdf in pdfs:
    # normalize PDF to sum to 1
    pdf /= pdf.sum()
    plt.plot(x, pdf)
plt.plot(x, p_pdf_1, label='prod 1', lw=5, color='C1')
plt.plot(x, p_pdf_2, label='prod 2', ls='--', color='k')
plt.legend();


也许这不是最优雅的解决方案，你最好使用一个矩阵，特别是如果你需要计算一个大问题的合并