函数不返回正确的值

2024-06-26 05:43:51 发布

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我在这个问题上纠缠了好几个小时。你知道吗

我已经编写了一个Python代码来读取和转换文本文件中的数据,一切运行正常。你知道吗

simulink_robot_motor1=[]
with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
    rows=0
    for line in data_file:
        rows=rows+1
    columns=len(line.split(","))
    simulink_robot_motor=[[0 for x in range(columns)] for y in range(rows)]
    i=0
    with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
        for line in data_file:
            current_line = line.split(",")
            current_line = list(map(float, current_line))
            simulink_robot_motor[i]=current_line
            i=i+1

我对simulink_robot_motor变量感兴趣,它有以下结果:

[[0.0, 3.6],
[1.6e-06, 3.6],
[4.57e-06, 3.6],
[7.67e-06, 3.6],
[1.09e-05, 3.6],
...

现在,我想在函数中使用这段代码。因此,如果调用函数,应该返回listsimulink_robot_motor。你知道吗

def get_matlab_sensor_data():        
    simulink_robot_motor1=[]
    with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
        rows=0
        for line in data_file:
            rows=rows+1
        columns=len(line.split(","))
        simulink_robot_motor=[[0 for x in range(columns)] for y in range(rows)]
        i=0
        with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
            for line in data_file:
                current_line = line.split(",")
                current_line = list(map(float, current_line))
                simulink_robot_motor[i]=current_line
                i=i+1
                return (simulink_robot_motor)

但是如果我运行get_matlab_sensor_data(),我会得到以下结果:

   [[0, 3.6],
   [0, 0],
   [0, 0],
   [0, 0],
   [0, 0],
    ...

我尝试了更小的数据集,也关闭了科学十进制风格。然而,它仍然不起作用。我的循环运行不正常吗?你知道吗


Tags: intxtfordatawithlinerobotopen
2条回答

首先,在原始代码中,您将整个文件读取两次,这是不必要的。这里有一个简化:

simulink_robot_motor=[]
with open(filename,'r') as data_file:
    for line in data_file:
        current_line = list(map(float, line.split(',')))
        simulink_robot_motor.append(current_line)

print(simulink_robot_motor)  # -> [[0.0, 3.6], [1.6e-06, 3.6], [4.57e-06, 3.6], [7.67e-06, 3.6], [1.09e-05, 3.6]]

当您试图将代码转换为函数时,出现了两个问题,一个是simulink_robot_motor变成了函数外不存在的局部变量。第二个是在第二个for line in data_file:循环中有一个return语句,这意味着它将在只读取一行之后返回。你知道吗

下面修复了这两个问题,并展示了如何使用新函数:

def get_matlab_sensor_data():
    results=[]
    with open(filename,'r') as data_file:
        for line in data_file:
            current_line = list(map(float, line.split(',')))
            results.append(current_line)

    return results

simulink_robot_motor = get_matlab_sensor_data()
print(simulink_robot_motor)  # -> same results as before

您的return语句处于不正确的缩进级别,因此它在for循环中执行得太快,永远不会完成循环。你知道吗

def get_matlab_sensor_data():        
    simulink_robot_motor1=[]
    with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
        rows=0
        for line in data_file:
            rows=rows+1
            columns=len(line.split(","))
        simulink_robot_motor=[[0 for x in range(columns)] for y in range(rows)]
        i=0
        with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
            for line in data_file:
                current_line = line.split(",")
                current_line = list(map(float, current_line))
                simulink_robot_motor[i]=current_line
                i=i+1
    return (simulink_robot_motor)

另外,您可能不需要用语句缩进第二个语句:

def get_matlab_sensor_data():        
    simulink_robot_motor1=[]
    with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
        rows=0
        for line in data_file:
            rows=rows+1
            columns=len(line.split(","))
        simulink_robot_motor=[[0 for x in range(columns)] for y in range(rows)]
        i=0
    with open('C:\\Users\...\sensor_data.txt',"r") as data_file:
         for line in data_file:
             current_line = line.split(",")
             current_line = list(map(float, current_line))
             simulink_robot_motor[i]=current_line
             i=i+1
    return (simulink_robot_motor)

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