轴等于一个语法错误 [英] axis equals one syntax error

问题描述

我想使用txt文件中的值制作步进函数图，以运行将生成图的方程式.

I want to make a a step function plot using values from a txt file to run equations that will generate a plot.

我的 txt 文件看起来像

My txt file looks something like

      x0     x1        y0        y1
   0.000  0.110 -0.000050 -0.046207
   0.110  0.143 -0.046206 -0.049596
   0.143  1.000 -0.049596  0.000000

我想使用的方程是

1. (((y1-yo)/(x1-x0)*(1-x0))+y0)
2. (-u + K/2)/e

我在axis=1 上收到一个错误，说

I get an error on axis=1 saying,

df['u'] = df.apply(lambda row: (((row.y1-row.y0)/(row.x1-row.x0)*(1-row.x0))+row.y0, axis=1)
                                                                                         ^
SyntaxError: invalid syntax

下面是我的脚本

import numpy as np
import matplotlib.pyplot as plt 
import csv

import pandas as pd

x=[]
y=[]
K=-2.69028905
e=1

df = pd.read_csv('vol_values.txt', delim_whitespace=True)
print(df)
df['u'] = df.apply(lambda row: (((row.y1-row.y0)/(row.x1-row.x0)*(1-row.x0))+row.y0, axis=1)
df['V'] = df.apply(lambda row: -row.u + K/2, axis=1)
df
print(df)
x=df['x in KxC']
y=df['Voltage']

plt.step(x,y)

如果我的代码有任何可疑之处，请告诉我.我只想使用我上面使用 apply 和 lambda 行函数生成的代码来生成方程！

If there's anything fishy on my codes, please let me know. I would only like to use the code i generated above using apply and lambda row function to generate equations!

推荐答案

您在 "+row.y0 " 之后错过了匹配的大括号.它应该是+row.y0)":

You have missed a matching brace there after "+row.y0 " . It should be "+row.y0)":

df['u'] = df.apply(lambda row: (((row.y1-row.y0)/(row.x1-row.x0)*(1-row.x0))+row.y0 ) , axis=1)

这必须解决该语法错误

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