如何修复有关 size-1 数组的错误？答案

【问题标题】：How to fix error regarding size-1 arrays?如何修复有关 size-1 数组的错误？
【发布时间】：2021-03-12 17:16:29
【问题描述】：

我正在研究一个问题，即使用一种名为 Conflation 的方法中的方程将几个连续概率分布组合成一个连续概率分布（可以在以下链接中找到 Conflation 方程：Conflation method question)。当我运行代码时，我收到有关 size-1 数组的错误消息。这是我的错误消息代码：

错误信息：

Traceback (most recent call last):
  File "/tmp/sessions/c903d99d60f20c3b/main.py", line 72, in <module>
    graph=conflate_pdf(domain, dists,lb,ub)
  File "/tmp/sessions/c903d99d60f20c3b/main.py", line 58, in conflate_pdf
    denom = quad(prod_pdf, lb, ub, args=(dists))[0]
  File "/usr/local/lib/python3.6/dist-packages/scipy/integrate/quadpack.py", line 341, in quad
    points)
  File "/usr/local/lib/python3.6/dist-packages/scipy/integrate/quadpack.py", line 448, 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

代码：

def prod_pdf(x,pdfs):
    prod=np.ones(pdfs[0].shape[0])
    for pdf in pdfs:
        prod=prod*pdf
    return prod

def conflate_pdf(x,dists,lb,ub):
    denom = quad(prod_pdf, lb, ub, args=(dists))[0]
    return prod_pdf(x,dists)/denom

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

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

dists=[dist_1, dist_2, dist_3, dist_4, dist_5]
graph=conflate_pdf(domain, dists,lb,ub)

from matplotlib import pyplot as plt
plt.plot(domain, dist_1)
plt.plot(domain, dist_2)
plt.plot(domain, dist_3)
plt.plot(domain, dist_4)
plt.plot(domain, dist_5)
plt.plot(domain,graph)
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("Conflated PDF")
plt.show()

从代码来看，这个错误信息是什么原因造成的？

编辑 1：

我修改了代码并进行了一些更正，但我仍然不断收到相同的错误消息。请参阅下面的代码和代码中的错误消息：

代码：

def prod_pdf(x,dists):
    prod=np.ones(np.array(dists)[0].shape)
    for dist in dists:
        print(prod)
        for c,y in enumerate(dist):
            prod[c]=prod[c]*y
        print('final:', prod)
    return prod

def conflate_pdf(x,dists,lb,ub):
    denom = quad(prod_pdf, lb, ub, dists)[0]
    # denom = simps(prod_pdf)
    print('Denom: ', denom)
    print('product pdf: ', prod_pdf(x,dists))
    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.plot(domain, dist_1, 'r')
plt.plot(domain, dist_2, 'g')
plt.plot(domain, dist_3, 'b')
plt.plot(domain, dist_4, 'y')
plt.plot(domain, dist_5, 'c')

dists=[dist_1, dist_2, dist_3, dist_4, dist_5]
graph=conflate_pdf(domain, dists,lb,ub)

plt.plot(domain,graph, 'm')
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("Conflated PDF")
plt.show()

回溯消息位置：

第 341 行：

if weight is None:
    retval = _quad(func, a, b, args, full_output, epsabs, epsrel, limit,points)

第 448 行：

def _quad(func,a,b,args,full_output,epsabs,epsrel,limit,points):
    infbounds = 0
    if (b != Inf and a != -Inf):
        pass   # standard integration
    elif (b == Inf and a != -Inf):
        infbounds = 1
        bound = a
    elif (b == Inf and a == -Inf):
        infbounds = 2
        bound = 0     # ignored
    elif (b != Inf and a == -Inf):
        infbounds = -1
        bound = b
    else:
        raise RuntimeError("Infinity comparisons don't work for you.")

    if points is None:
        if infbounds == 0:
            return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
        else:
            return _quadpack._qagie(func,bound,infbounds,args,full_output,epsabs,epsrel,limit)
    else:
        if infbounds != 0:
            raise ValueError("Infinity inputs cannot be used with break points.")
        else:
            #Duplicates force function evaluation at singular points
            the_points = numpy.unique(points)
            the_points = the_points[a < the_points]
            the_points = the_points[the_points < b]
            the_points = numpy.concatenate((the_points, (0., 0.)))
            return _quadpack._qagpe(func,a,b,the_points,args,full_output,epsabs,epsrel,limit)

编辑 2：

与@hpaulj 讨论后，似乎问题出在产品功能上。从链接Conflation method question 中，代码能够运行而不会出现 size-1 数组的错误。这是包含更多详细信息的代码：

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 dist in dists:
        p_pdf=p_pdf*dist.pdf(x)
        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]
    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)
dists=[stats.norm(2,1), stats.norm(2.5,1.5), stats.norm(2.2,1.6), stats.norm(2.4,1.3), stats.norm(2.7,1.5)]

from matplotlib import pyplot as plt
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')

graph=conflate_pdf(domain,dists,lb,ub)
plt.plot(domain,graph, 'm', label='Conflated Dist.')
plt.xlabel("domain")
plt.ylabel("pdf")
plt.title("Conflated PDF")
plt.legend()
plt.show()

这是输出的一小部分：

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]

剧情：

在我看来，它从一个中获取每个密度值并将其与其他分布相乘。但是，如果我看一下我的情况，它不是获取一个变量，而是在乘积函数之后获取整个数组，并且它对 size-1 数组产生相同的错误。请参阅以下代码和部分输出：

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 dist in dists:
        p_pdf=p_pdf*dist
        print('final:', p_pdf)
    return p_pd

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()

这是输出的一小部分：

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]
Input product pdf:  [1.09076800e-97 1.55234627e-97 2.20861552e-97 ... 3.72095218e-40
 2.98464396e-40 2.39335035e-40]
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]

编辑 3：

我设法查看代码以在 Edit 2 中实现相同的方法，但我编辑了从每个发行版中获取第一个变量的代码，但对于循环的其余部分继续打印相同的值，它不会转到列表中的下一个值，并且混合分布是单个变量。请参阅以下代码和部分输出：

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()

部分输出：

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

编辑 3：

我实现了以下代码，它似乎可以工作，而且，我设法解决了 quad 的问题，如果我将 quad 更改为 fixed_quad 并规范化 pdf 列表。我会得到同样的结果。下面是代码：

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))

这是输出：

我的问题是，我知道我需要标准化 PDF 列表。但是，假设我没有规范化 PDF，如何修改我的合并代码以获得以下图？

要获得上面的情节和我的混合代码：

# 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()

我没有规范化的混合代码图：

【问题讨论】：

在 scipy 文档中对这一行的语法（接受的参数等）进行一些研究：denom = quad(prod_pdf, lb, ub, args=(dists))[0]
您正在将一个数组（具有多个值）传递给一个需要标量、单个数字的函数。检查回溯并确定哪个参数是错误的！
@S3DEV 我了解这条线是如何工作的，但在问题 (stackoverflow.com/questions/64346110/…) 中，答案表明可以实现集成。我尝试修改获得多个分布列表（可能不是规范分布）的代码并执行相同的方法，但我不断收到相同的错误。
@hpaulj 我了解这条线的工作原理，但在问题 (stackoverflow.com/questions/64346110/…) 中，答案表明可以实现集成。我尝试修改获得多个分布列表（可能不是规范分布）的代码并执行相同的方法，但我不断收到相同的错误。
这个小错误发生在哪里哦？如果没有追溯，我们将无法帮助您。我无法在脑海中运行这样的代码！

标签： python arrays list numpy scipy

【解决方案1】：

在ipython 会话中，我复制了您的代码。在我之前的阅读中，它看起来并不完整且可运行。但是通过几个导入它确实会运行并产生您的错误：

In [1]: from scipy.integrate import quad
In [2]: from scipy import stats
In [3]: def prod_pdf(x,pdfs):
   ...:     prod=np.ones(pdfs[0].shape[0])
   ...:     for pdf in pdfs:
   ...:         prod=prod*pdf
   ...:     return prod
   ...: 
   ...: def conflate_pdf(x,dists,lb,ub):
   ...:     denom = quad(prod_pdf, lb, ub, args=(dists))[0]
   ...:     return prod_pdf(x,dists)/denom
   ...: 
   ...: lb=-10
   ...: ub=10
   ...: domain=np.arange(lb,ub,.01)
   ...: 
   ...: dist_1 = stats.norm.pdf(domain, 2,1)
   ...: dist_2 = stats.norm.pdf(domain, 2.5,1.5)
   ...: dist_3 = stats.norm.pdf(domain, 2.2,1.6)
   ...: dist_4 = stats.norm.pdf(domain, 2.4,1.3)
   ...: dist_5 = stats.norm.pdf(domain, 2.7,1.5)
   ...: 
   ...: dists=[dist_1, dist_2, dist_3, dist_4, dist_5]
   ...: graph=conflate_pdf(domain, dists,lb,ub)
   ...:

回溯

Traceback (most recent call last):
  File "<ipython-input-3-a55a08f59f08>", line 22, in <module>
    graph=conflate_pdf(domain, dists,lb,ub)
  File "<ipython-input-3-a55a08f59f08>", line 8, in conflate_pdf
    denom = quad(prod_pdf, lb, ub, args=(dists))[0]
  File "/usr/local/lib/python3.8/dist-packages/scipy/integrate/quadpack.py", line 351, in quad
    retval = _quad(func, a, b, args, full_output, epsabs, epsrel, limit,
  File "/usr/local/lib/python3.8/dist-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

错误出现在编译代码中，因此不提供更多信息。所以我们不得不求助于检查输入和输出。

我首先想到quad 的参数有问题。所以可以肯定的是我扔了一些照片。

In [5]: def prod_pdf(x,pdfs):
   ...:     prod=np.ones(pdfs[0].shape[0])
   ...:     for pdf in pdfs:
   ...:         prod=prod*pdf
   ...:     return prod
   ...: 
   ...: def conflate_pdf(x,dists,lb,ub):
   ...:     print(lb, ub, dists)
   ...:     denom = quad(prod_pdf, lb, ub, args=(dists,))[0]
   ...:     return prod_pdf(x,dists)/denom
   ...: 

 ....
-10 10 [array([2.14638374e-32, 2.41991991e-32, 2.72804284e-32, ...,
       6.41980576e-15, 5.92770938e-15, 5.47278628e-15]), .

上下界看起来没问题，dists 是一个列表。在一个更简单的情况下，我测试了数组的边界并得到了不同的错误。

那么prod_pdf 会被调用吗？

In [6]: def prod_pdf(x,pdfs):
   ...:     print(x, pdfs)
   ...:     prod=np.ones(pdfs[0].shape[0])
   ...:     for pdf in pdfs:
   ...:         prod=prod*pdf
   ...:     return prod
   ...: 
   ....

0.0 [array([2.14638374e-32, 2.41991991e-32, 2.72804284e-32, ...,
       6.41980576e-15, 5.92770938e-15, 5.47278628e-15]), ....

所以 prod_pdf 被调用，带有 0 和 dists 列表；期望值。

那么返回的prod 是什么？

In [8]: def prod_pdf(x,pdfs):
   ...:     prod=np.ones(pdfs[0].shape[0])
   ...:     for pdf in pdfs:
   ...:         prod=prod*pdf
   ...:     print(prod)
   ...:     return prod
   ...
[1.09076800e-97 1.55234627e-97 2.20861552e-97 ... 3.72095218e-40
 2.98464396e-40 2.39335035e-40]

它的形状是什么？

In [9]: def prod_pdf(x,pdfs):
   ...:     prod=np.ones(pdfs[0].shape[0])
   ...:     for pdf in pdfs:
   ...:         prod=prod*pdf
   ...:     print(prod.shape)
   ...:     return prod
   ...: 
....
(2000,)

哎呀！ quad 集成了一个标量函数。 prod_pdf 应该返回一个标量，一个数字，而不是 2000 个值。这就是_quadpack._qagse 抱怨的错误。问题不在于quad 的输入，而在于您的函数返回的值。

现在我知道要查找什么，错误很明显。

如果我们人为地将其截断为 1 个值，quad 会运行而不会出错。

In [10]: def prod_pdf(x,pdfs):
    ...:     prod=np.ones(pdfs[0].shape[0])
    ...:     for pdf in pdfs:
    ...:         prod=prod*pdf
    ...:     print(prod.shape)
    ...:     return prod[0]

我将完成所有这些调试步骤（进行大量编辑），以便您看到我必须经历的步骤。

【讨论】：

感谢您的回答。好吧，我将代码修改为 EDIT 1。但是，我不断收到同样的错误。此外，如果您在以下链接 (stackoverflow.com/questions/64346110/…) 中查看问题的答案，答案能够集成并生成合并 PDF 的图，我只是想达到该步骤，但使用列表而不是获取 pdf在功能上。看看链接，你就会明白我在这条评论中的意思
我没有看过你更大的问题或那个链接。我的重点是标题错误 - size-1 问题。如果函数返回多个值，quad 将不起作用。在我测试的prod_pdf 中，它返回了一个与您的dists, (2000,) 之一大小相同的数组。那是错误的。 quad 的目的是执行一维积分，只返回一个浮点数。
我明白你的意思，我通过 Scipy 中的函数进一步了解它及其变量。我提到的链接没有产生这个错误。我想如果您看一下它，它可能会对您有所帮助（了解我在代码中犯的错误，以便它可以类似于用户的答案以及我的代码中可能缺少的内容）和我（进一步了解发生此错误的原因以及我的代码和用户代码有什么不同）你能看看吗？
链接的问题有同样的错误。答案使用不同的prod_pdf，它返回一个数字。只要你返回prod=np.ones(pdfs[0].shape[0]) 或相同大小的东西，你就会得到这个size-1 错误。 denom=quad(...) 行应该产生 one 值，而不是 2000。
好的，所以我认为这里的问题是 prod_pdf 函数是它返回数组而不是导致此问题的单个变量。我查看了他的代码并单独运行它，我设法发现它正在打印出单个变量并设法得到一个变量作为分母。我试图调整我的代码来做到这一点，但相反，它继续获得整个数组的产品。请参阅已编辑的问题 Edit 2。