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如果您想了解TypeError：“ ML模型”无法迭代“ numpy.float64”对象的相关知识，那么本文是一篇不可错过的文章，我们将为您提供关于'numpy.float64' 对象不可迭代——独立工作但不在 for 循环中、'numpy.float64' 对象不能解释为整数，Python 中的 ODE 求解、'numpy.float64' 对象在曲线拟合时不可调用、'numpy.float64' 类型的对象没有 len()：我该如何解决这个问题？的有价值的信息。

• TypeError：“ ML模型”无法迭代“ numpy.float64”对象
• 'numpy.float64' 对象不可迭代——独立工作但不在 for 循环中
• 'numpy.float64' 对象不能解释为整数，Python 中的 ODE 求解
• 'numpy.float64' 对象在曲线拟合时不可调用
• 'numpy.float64' 类型的对象没有 len()：我该如何解决这个问题？

TypeError：“ ML模型”无法迭代“ numpy.float64”对象

如何解决TypeError：“ ML模型”无法迭代“ numpy.float64”对象

我如何避免在突出显示的行上出现错误，它使我不断产生关于float64类型不可迭代的错误。本质上，我试图在第一个函数中计算SGD的成本，然后在第二个函数中计算SGD。

def calculate_cost_gradient(W,X_batch,Y_batch):
    # if only one example is passed (eg. in case of SGD)
    if type(Y_batch) == np.float64:
        Y_batch = np.array([Y_batch])
        X_batch = np.array([X_batch])  # gives multidimensional array
    distance = 1 - (Y_batch * np.dot(X_batch,W))
    dw = np.zeros(len(W))
######## Error is here ######## 
    for ind,d in enumerate(distance):
        if max(0,d) == 0:
            di = W
        else:
            di = W - (regularization_strength * Y_batch[ind] * X_batch[ind])
        dw += di
    dw = dw/len(Y_batch)  # average
    return dw
def sgd(features,outputs):
    max_epochs = 5000
    weights = np.zeros(features.shape[1])
    nth = 0
    prev_cost = float("inf")
    cost_threshold = 0.01  # in percent
    # stochastic gradient descent
    for epoch in range(1,max_epochs):
        # shuffle to prevent repeating update cycles
        X,Y = shuffle(features,outputs)
        for ind,x in enumerate(X):
            ascent = calculate_cost_gradient(weights,x,Y[ind])
            weights = weights - (learning_rate * ascent)
        # convergence check on 2^nth epoch
        if epoch == 2 ** nth or epoch == max_epochs - 1:
            cost = compute_cost(weights,features,outputs)
            print("Epoch is:{} and Cost is: {}".format(epoch,cost))
            # stoppage criterion
            if abs(prev_cost - cost) < cost_threshold * prev_cost:
                return weights
            prev_cost = cost
            nth += 1
    return weights

解决方法

for ind,d in enumerate(range(distance)):

'numpy.float64' 对象不可迭代——独立工作但不在 for 循环中

如何解决''numpy.float64'' 对象不可迭代——独立工作但不在 for 循环中

当我把代码写成：

{ [key in string]: ItemType }

我收到一个错误：

for i in range(1,k):
        
        idx = np.random.choice(data.shape[0],1,p=squared_distances/sum(squared_distances))
        
        centroids[i]=data.[idx,:].toarray()
        
        squared_distances=np.min(pairwise_distances(data,centroids[0:i+1],metric=''euclidean'')**2  
  

但是当我运行单独的行时：

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-71-dff0c6ebb89d> in <module>
      1 for i in range(1,k):
      2 
----> 3        idx = np.random.choice(data.shape[0],p=squared_distances/sum(squared_distances))
      4 
      5        centroids[i]=data[idx,:].toarray()
TypeError: ''numpy.float64'' object is not iterable

它给出的输出为：

idx = np.random.choice(data.shape[0],p=squared_distances/sum(squared_distances))

如果代码在触发上述错误的 for 循环中，我想知道有什么问题。

供参考：

idx
array([5147])

解决方法

'numpy.float64' 对象不能解释为整数，Python 中的 ODE 求解

如何解决''numpy.float64'' 对象不能解释为整数，Python 中的 ODE 求解

我正在做一个热 Simulink 工作，我收到一条错误消息，说“''numpy.float64'' 对象不能被解释为一个整数”。我找不到那个错误以及如何解决这个问题。

首先我读取一个文本文档并为初始条件创建一个数组。(ycut)

接下来，我将创建一个函数“EPCMchg”

然后我正在使用“odeint”进行模拟。实际上我正在转换 MATLAB 代码，它具有 ODE45 微分方程求解器。它在 python 中的等价物是 odeint。 我阅读了许多与错误“''numpy.float64'' 对象不能被解释为整数”和 odeint 相关的文档，但无法捕获错误。

如果有人可以帮助我，非常感谢。提前致谢 这是代码和错误。

import numpy as np
import pandas as pd
import matplotlib
from math import sqrt
from scipy.integrate import odeint

ICHG = 1 #charge or ICHG=-1 discharge
print(ICHG)

df2 = pd.read_csv(ycut_file,sep=r''\\s{2,}'',engine=''python'',header=None,names=[''Value''])
ycut = df2.to_numpy().flatten()

**Output:**
ycut:  [-4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01
  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02
 -4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01
  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02
 -4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01
  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02
 -4.000000e+01  5.500000e+02 -4.000000e+01  5.500000e+02 -4.000000e+01
  5.500000e+02 -3.999990e+01  5.500001e+02 -3.999980e+01  5.500003e+02
 -3.999950e+01  5.500011e+02 -3.999840e+01  5.500034e+02 -3.999540e+01
  5.500098e+02 -3.998740e+01  5.500267e+02 -3.996830e+01  5.500675e+02
 -3.992660e+01  5.501580e+02 -3.984240e+01  5.503420e+02 -3.968600e+01
  5.506877e+02 -3.941640e+01  5.512901e+02 -3.898240e+01  5.522699e+02
 -3.832740e+01  5.537646e+02 -3.739550e+01  5.559138e+02 -3.613990e+01
  5.588393e+02 -3.453170e+01  5.626245e+02 -3.256610e+01  5.672970e+02
 -3.026550e+01  5.728186e+02 -2.767850e+01  5.790859e+02 -2.487450e+01
  5.859400e+02 -2.193520e+01  5.931861e+02 -1.894460e+01  6.006170e+02
 -1.597890e+01  6.080382e+02 -1.309800e+01  6.152899e+02 -1.034030e+01
  6.222625e+02 -7.719300e+00  6.289058e+02 -5.224500e+00  6.352294e+02
 -2.824200e+00  6.412981e+02 -4.698000e-01  6.472213e+02  1.899100e+00
  6.531412e+02  4.350000e+00  6.592214e+02  6.956000e+00  6.656405e+02
  9.798900e+00  6.725974e+02  1.298260e+01  6.803386e+02  1.666380e+01
  6.892234e+02  2.110970e+01  6.998522e+02  2.679910e+01  7.132907e+02
  3.458400e+01  7.314356e+02  4.593790e+01  7.575796e+02  6.331840e+01
  7.972448e+02  9.068090e+01  8.593479e+02  1.355955e+02  9.271171e+02]


def EPCMchg(x,t,NSTAGES,Tf1,Tf2,Tf3,Hf1,Hf2,Hf3,TankD,TankH,deltaz,porosity,dp,Cpf,kf,rhof,Cpb,kb,rhob,ufluid,mdotsaltflow,hotHTF):
    #Models an encapsulated Thermal Energy Storage system with multiple salt PCM Assumed salt properties 
     #for NaNO3 with freezing temp=frzHTF
    #(650F),cold HTF fluid from steam generator=coldHTF (550f),and hot temp from solar=hotHTF (1050F).
    #Intermediate Calculations of parameters used in differential equations
    # Surface area of Tank for calculating Mass veLocity of fluid
    pi = 3.14159
    atank = pi*TankD*TankD/4 #ft^2
    #Volume of each element of height deltaz
    deltaV = pi*TankD*TankD*deltaz/4 # ft^3
    #Properties of heat transfer fluid used in differential eqns
    rhocpf=rhof*Cpf*porosity # btu/ft3/F
    
    #Particle surface area per unit volume ft2/ft3
    dparticle=dp/304.8 #[ft] dp mm particle
    av=(6/dparticle) #ft2/ft3
    #heat transfer coefficnet inside particle (use a simple thermal conductivit/particle radius
    hsl0=kb/dparticle #heat transfer coeff particle solid,Btu/h/ft2/F
    #heat transfer coefficnet of liquid inside particle (use a simple thermal conductivit/particle radius 
     #150% of solid to account for convection
    hl0=hsl0*1.5 #heat transfer coeff particle liquid,Btu/h/ft2/F
    
    #Input fluid Temperature at top of the tank during charging from the solar field
    prevTf= hotHTF #deg F
    
    #Bottom Salt 1 is the bottom,NSTAGES=top
    
    rany =NSTAGES+1
    #xdot = np.zeros(rany)
    for STAGE in range(rany,-1):
        HbSTAGE = x[2*STAGE-1] #Solid Enthalpy (btu/lb)
        TfSTAGE = x[2*STAGE] # FLuid Temperature,F
        #[ Tf,Hf,dh ] = saltstage( STAGE )
        
        #Bottom Stages
        Tf,dh=Tf1,0
        
        #Middle stages
        if STAGE > NSTAGES/3:
            Tf,dh=Tf3,Cpb*(Tf-Tf1)
            
        #Top stages
        if STAGE > 2 * (NSTAGES/3):
            Tf,dh=Tf2,Cpb*(Tf-Tf1)
            
        # Determine conditions inside capsules (molten,solid or fractional 
        #solid
        if HbSTAGE<(Hf+dh):
            xsolid=1-(HbSTAGE-dh)/Hf #fraction solidification inside capsule
            if xsolid<=0:
                xsolid=0
            else:
                xsolid=1
                
            TbSTAGE=Tf
            hsl=hsl0/sqrt(xsolid) #heat transfer coeff of solid varies as sqrt of soli farction
            hl=hl0/(1-sqrt(xsolid)) #heat transfer coeff of liquid remaining
            u=1/((1/ufluid)+(1/hsl)+(1/hl))

            uav=u*av
            tauconst=(mdotsaltflow*Cpf/atank)/rhocpf #constnts in diff equations
            aconst=uav/rhocpf #constnts in diff equations
            cconst=uav/rhob/(1-porosity) #constnts in diff equations
            
            
        #solid cooling
        if HbSTAGE<dh:
            xsolid=1
            TbSTAGE = Tf+(HbSTAGE-dh)/Cpf
            hsl=hsl0/sqrt(xsolid)
            u=1/((1/ufluid)+(1/hsl))
            
            uav=u*av
            tauconst=(mdotsaltflow*Cpf/atank)/rhocpf
            aconst=uav/rhocpf
            cconst=uav/rhob/(1-porosity)
            
        
        # sensible heat in liquid free convection hl=90
        if HbSTAGE>(Hf+dh):
            xsolid=0
            TbSTAGE=Tf+(HbSTAGE-Hf-dh)/Cpf
            hl=hl0/(1-sqrt(xsolid))
            u=1/((1/ufluid)+(1/hl))
            
            uav=u*av
            tauconst=(mdotsaltflow*Cpf/atank)/rhocpf
            aconst=uav/rhocpf
            cconst=uav/rhob/(1-porosity)
            
        
            
        xdot[2*STAGE-1] = cconst*(TfSTAGE - TbSTAGE) # capsule enthalpy
        xdot[2*STAGE] = tauconst*(prevTf - TfSTAGE)+ aconst * (TbSTAGE-TfSTAGE) #HTF Temp
        
        #PREPARE FOR NEXT STAGE :
        prevTf = TfSTAGE
        
        xs[STAGE]=xsolid
        
    xs4=xs
    xdot = xdot
    return xs4,xdot;

## Start simulation:
t_final=4.0
tspan = np.linspace(0,t_final)
tspan
**Output:**
<class ''numpy.ndarray''>
array([0.,0.08163265,0.16326531,0.24489796,0.32653061,0.40816327,0.48979592,0.57142857,0.65306122,0.73469388,0.81632653,0.89795918,0.97959184,1.06122449,1.14285714,1.2244898,1.30612245,1.3877551,1.46938776,1.55102041,1.63265306,1.71428571,1.79591837,1.87755102,1.95918367,2.04081633,2.12244898,2.20408163,2.28571429,2.36734694,2.44897959,2.53061224,2.6122449,2.69387755,2.7755102,2.85714286,2.93877551,3.02040816,3.10204082,3.18367347,3.26530612,3.34693878,3.42857143,3.51020408,3.59183673,3.67346939,3.75510204,3.83673469,3.91836735,4.        ])

if ICHG == 1:
      [tchg,ychg]=odeint(EPCMchg,ycut,tspan,args=(NSTAGES,hotHTF))
       plt.plot(tchg,ychg)
       plt.show()

**Output:**
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-13-4b38269782f5> in <module>
      1 #call ode45 with function EPCMchg to charge
      2 if ICHG == 1:
----> 3     [tchg,hotHTF))
      4     plt.plot(tchg,ychg)
      5     plt.show()

C:\\ProgramData\\Anaconda3\\lib\\site-packages\\scipy\\integrate\\odepack.py in odeint(func,y0,args,Dfun,col_deriv,full_output,ml,mu,rtol,atol,tcrit,h0,hmax,hmin,ixpr,mxstep,mxhnil,mxordn,mxords,printmessg,tfirst)
    239     t = copy(t)
    240     y0 = copy(y0)
--> 241     output = _odepack.odeint(func,242                              full_output,243                              ixpr,<ipython-input-11-6746119c69d3> in EPCMchg(x,hotHTF)
     26     rany =NSTAGES+1
     27     #xdot = np.zeros(rany)
---> 28     for STAGE in range(rany,-1):
     29         HbSTAGE = x[2*STAGE-1] #Solid Enthalpy (btu/lb)
     30         TfSTAGE = x[2*STAGE] # FLuid Temperature,F

TypeError: ''numpy.float64'' object cannot be interpreted as an integer

'numpy.float64' 对象在曲线拟合时不可调用

如何解决''numpy.float64'' 对象在曲线拟合时不可调用

我正在尝试在 CSV 数据中拟合正弦曲线以找到曲线的最大值。数据类似于

      Time    Voltage Current
0   -0.02500    -1.4    0.38
1   -0.02498    -1.6    0.32
2   -0.02496    -1.8    0.40
3   -0.02494    -1.4    0.44
4   -0.02492    -1.6    0.30
5   -0.02490    -2.0    0.40
6   -0.02488    -1.8    0.44
7   -0.02486    -1.4    0.32
8   -0.02484    -1.4    0.42
9   -0.02482    -1.6    0.46
10  -0.02480    -1.8    0.36

这是代码：

from numpy import sin
from scipy.optimize import curve_fit
from matplotlib import pyplot
# input and output variables
x,y = dfi["Time"],dfi["Voltage"] 
# defining the function
def func(x,a,b,c,d):
   return a * sin(b(x-c)) + d
# curve fit
popt,_ = curve_fit(func,x,y)

它给了我这个错误：

<ipython-input-20-f95822761dd9> in objective(x,d)
      9 # defining the function
     10 def objective(x,d):
---> 11         return a * sin(b(x-c)) + d
     12 
     13 # curve fit
TypeError: ''numpy.float64'' object is not callable

这是我第一次尝试曲线拟合，我不明白这个错误。

'numpy.float64' 类型的对象没有 len()：我该如何解决这个问题？

如何解决''numpy.float64'' 类型的对象没有 len()：我该如何解决这个问题？

我正在尝试计算我的 10,000 个数字列表中大于 1.6 的值总数。我尝试了几种方法：

for value in chi_sq_values:
    if value > 1.6:
        print(len(value))

给了我标题中的错误，但以下有效

greater = [i for i in chi_sq_values if i > 1.6]
total = len(greater)
print(total)

我想尝试两种方法来验证我的答案，我该如何修复第一种方法以使其有效？

解决方法

counter = 0
for value in chi_sq_values:
    if value > 1.6:
        counter += 1
print(counter)

chi_sq_values = [1,2,3]
greater = []
for value in chi_sq_values:
    if value > 1.6:
        print(value)
        greater.append(value)

print(len(greater))
    

np.count_nonzero(np.asarray(chi_sq_values) > 1.6)

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