我有一些关于python中的值错误的问题。

Question

下面是我绘制应力-应变曲线的代码

import matplotlib.pyplot as plt
import numpy as np
import math
from scipy.interpolate import interp1d
from matplotlib.offsetbox import AnchoredText
import pandas as pd


#both strain is a column in the given dataframe, and I manually calculated stress
df_1 = pd.read_csv('1045.csv',skiprows=25,header=[0,1])
print(df_1.head())

A1 = 40.602*(10e-6)
stress1 = ((df_1.Load)/A1)


plt.figure(figsize=(12,9))
plt.plot(df_1.Strain1.values,df_1.Load.values,'g')
plt.ylabel('stress(Pa)',fontsize=13)
plt.xlabel('Strain(%)',fontsize=13)
plt.xticks(np.arange(-6e-5,0.15,step=0.005),rotation = 45)
plt.yticks(np.arange(0,42000,step=1000))

strain = df_1.Strain1.values
stress = np.array(((df_1.Load.values)/A1))
strain = np.array((df_1.Strain1.values))

LinearLimit=1
Strain_values_linear = np.linspace(strain[0], strain[LinearLimit], num=50, endpoint=True)
Strain_values_eng = np.linspace(strain[LinearLimit], strain[-1], num=50, endpoint=True)
f1 = interp1d(strain, stress, fill_value='extrapolate')
f2 = interp1d(strain, stress, kind=3, fill_value='extrapolate')

现在，我一直收到一个值错误："x和y数组沿插值轴的长度必须相等。“我不明白……我打印了应变和应力的形状，它们是一样的。这里是csv文件的截图：enter image description here

Answer 1

您可能会传递一个形状数组(..., N)作为第一个参数(这意味着strain具有(..., N)形式的形状)。SciPy不允许这样做，并抛出一个ValueError。有关详细信息，请参阅documentation。如果strain数组中有多个向量，则应该运行for循环。考虑到您想要为strain中的每一行插入一个函数(该应变是一个二维数组)，下面的代码应该可以工作。如果不是，您可以使用strain.reshape(-1, N)很容易地转换它)：

import matplotlib.pyplot as plt
import numpy as np
import math
from scipy.interpolate import interp1d
from matplotlib.offsetbox import AnchoredText
import pandas as pd


#both strain is a column in the given dataframe, and I manually calculated stress
df_1 = pd.read_csv('1045.csv',skiprows=25,header=[0,1])
print(df_1.head())

A1 = 40.602*(10e-6)
stress1 = ((df_1.Load)/A1)


plt.figure(figsize=(12,9))
plt.plot(df_1.Strain1.values,df_1.Load.values,'g')
plt.ylabel('stress(Pa)',fontsize=13)
plt.xlabel('Strain(%)',fontsize=13)
plt.xticks(np.arange(-6e-5,0.15,step=0.005),rotation = 45)
plt.yticks(np.arange(0,42000,step=1000))

strain = df_1.Strain1.values
stress = np.array(((df_1.Load.values)/A1))
strain = np.array((df_1.Strain1.values))

LinearLimit=1
Strain_values_linear = np.linspace(strain[0], strain[LinearLimit], num=50, endpoint=True)
Strain_values_eng = np.linspace(strain[LinearLimit], strain[-1], num=50, endpoint=True)

f1, f2 = [], []
for row in range(len(strain)):
    f1.append(interp1d(strain[row], stress, fill_value='extrapolate'))
    f2.append(interp1d(strain[row], stress, kind=3, fill_value='extrapolate'))

编辑:从注释中可以看到形状为(222, 1)的strain数组。这意味着您已经有了一个向量，但该形状与SciPy接受的内容不兼容。在这种情况下，您必须重塑应变和应力数组，使其具有表单(N,)的形状。下面的代码应该可以工作：

import matplotlib.pyplot as plt
import numpy as np
import math
from scipy.interpolate import interp1d
from matplotlib.offsetbox import AnchoredText
import pandas as pd


#both strain is a column in the given dataframe, and I manually calculated stress
df_1 = pd.read_csv('1045.csv',skiprows=25,header=[0,1])
print(df_1.head())

A1 = 40.602*(10e-6)
stress1 = ((df_1.Load)/A1)


plt.figure(figsize=(12,9))
plt.plot(df_1.Strain1.values,df_1.Load.values,'g')
plt.ylabel('stress(Pa)',fontsize=13)
plt.xlabel('Strain(%)',fontsize=13)
plt.xticks(np.arange(-6e-5,0.15,step=0.005),rotation = 45)
plt.yticks(np.arange(0,42000,step=1000))

strain = df_1.Strain1.values
stress = np.array(((df_1.Load.values)/A1))
strain = np.array((df_1.Strain1.values))
strain = strain.reshape(-1,)
stress = stress.reshape(-1,)

LinearLimit=1
Strain_values_linear = np.linspace(strain[0], strain[LinearLimit], num=50, endpoint=True)
Strain_values_eng = np.linspace(strain[LinearLimit], strain[-1], num=50, endpoint=True)
f1 = interp1d(strain, stress, fill_value='extrapolate')
f2 = interp1d(strain, stress, kind=3, fill_value='extrapolate')