我对列表索引很生气，无法解释我做错了什么.

I'm getting mad with list indexes, and can't explain what I'm doing wrong.

我有这段代码，我想在其中创建一个列表列表，每个列表都包含我从 csv 读取的相同电路参数(电压、电流等)的值code> 文件如下所示:

I have this piece of code in which I want to create a list of lists, each one containing values of the same circuit parameter (voltage, current etc..) that I'm reading from a csv file that looks like this:

Sample, V1, I1, V2, I2
0, 3, 0.01, 3, 0.02
1, 3, 0.01, 3, 0.03

等等.我想要的是创建一个列表，例如包含 V1 和 I1(但我想以交互方式选择)的形式 [[V1], [I1]]，所以:

And so on. What I want is to create a list that for example contains V1 and I1 (but I want to chose interactively) in the form [[V1], [I1]], so:

[[3,3], [0.01, 0.01]]

我使用的代码是这样的:

The code that I'm using is this:

plot_data = [[]]*len(positions)    
for row in reader:
    for place in range(len(positions)):
        value = float(row[positions[place]])
        plot_data[place].append(value)

plot_data 是包含所有值的列表，而 positions 是一个列表，其中包含我要从 .csv 复制的列的索引 文件.问题是，如果我在 shell 中尝试命令，似乎可以工作，但是如果我运行脚本而不是将每个值附加到正确的子列表，它会将所有值附加到所有列表，所以我获得 2(或更多) 相同的列表.

plot_data is the list that contains all the values, while positions is a list with the indexes of the columns that I want to copy from the .csv file. The problem is that if I try the commands in the shell, seems to work, but if I run the script instead of appending each value to the proper sub-list, it appends all values to all lists, so I obtain 2 (or more) identical lists.

推荐答案

Python 列表是可变对象，这里:

Python lists are mutable objects and here:

plot_data = [[]] * len(positions) 

您正在重复相同的列表 len(positions) 次.

you are repeating the same list len(positions) times.

>>> plot_data = [[]] * 3
>>> plot_data
[[], [], []]
>>> plot_data[0].append(1)
>>> plot_data
[[1], [1], [1]]
>>> 

列表中的每个列表都是对同一对象的引用.你修改一个，你会看到所有的修改.

Each list in your list is a reference to the same object. You modify one, you see the modification in all of them.

如果你想要不同的列表，你可以这样做:

If you want different lists, you can do this way:

plot_data = [[] for _ in positions]

例如:

>>> pd = [[] for _ in range(3)]
>>> pd
[[], [], []]
>>> pd[0].append(1)
>>> pd
[[1], [], []]