Python 函数不'；t返回正确的值

我在这个问题上纠缠了好几个小时

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，一切都运行良好

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

我对具有以下结果的

simulink\u robot\u motor

变量感兴趣：

[[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，如果我调用该函数，应该返回list

simulink\u robot\u motor

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\u matlab\u sensor\u 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。我的循环是否运行不正常？

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

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)

此外，您可能不需要使用以下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)

---

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\u robot\u motor

成为函数外部不存在的局部变量。第二个是在数据文件的第二个

for行：

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

以下内容解决了这两个问题，并展示了如何使用新功能：

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

---

可能有助于提供示例输入和预期输出。谢谢@martineau。丹尼斯：不客气。请考虑一下我的答案。看，我不能，因为我的名声不好。当我的分数达到15分时，我会的。