我在python中遇到了一些值错误的问题

Question

这是我绘制应力-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')

现在我不断收到一个值错误消息：“x 和 y 数组沿插值轴的长度必须相等。”我不明白这个......我打印了应变和应力的形状，它们是一样的 顺便说一句，这里是 csv 文件的屏幕截图： enter image description here

Answer 1

您可能正在传递一个形状为(..., N) 的数组作为第一个参数（意味着strain 的形状为(..., N)）。 SciPy 不允许这样做并抛出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。以下代码应该可以工作，考虑到您想为strain 中的每一行插入一个函数（并且该应变是一个二维数组。如果不是，您可以使用strain.reshape(-1, N) 轻松转换它）：

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：从评论中，您有 strain 形状数组 (222, 1)。这意味着你已经有了一个向量，但它的形状与 SciPy 接受的不兼容。在这种情况下，您必须重新调整应变和应力数组的形状，使其具有(N,) 形式的形状。以下代码应该可以工作：

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')