Python:* args解压缩，如何从test_train_split重新打包? [英] Python: *args unpacks, how to repack from test_train_split?

问题描述

我在列表x中放置了多个(类型)输入，并且正在使用

x0_train, x0_test, x1_train, x1_test, ... , y_train, y_test = 
   train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

    # But I have to manually repack
    x_train = [x0_train, x1_train]
    x_test = [x0_test, x1_test]

进行操作:

x = [some_matrix, scalar_value, something_else, ...]
x0_train, x0_test, x1_train, x1_test, ... , y_train, y_test = 
   train_test_split(x[0],x[1],... , y, test_size=0.2, random_state=np.random, shuffle=True)

我设法将输入参数x[0], x[1], ...更改为*x:

x0_train, x0_test, x1_train, x1_test, ... , y_train, y_test = 
   train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

    # But I have to manually repack
    x_train = [x0_train, x1_train]
    x_test = [x0_test, x1_test]

但是有一种无需手动重新包装即可接收它的方法吗?等价于什么:

*x_train, *x_test, y_train, y_test = 
   train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

还是有其他方法可以做到这一点?例如:构造一个字典并使用**解包，但是我仍然遇到同样的问题.反正有什么约定(如果有的话)?

解决方案

这解决了锯齿形分割问题:

recv = [None for i in range(2*len(x))]
*recv, y_train, y_test = train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

# Edit: credits hpaulj
x_train = recv[::2]
x_test = recv[1::2]

---

此外，如果有一种方法可以复制引用，那么它也会起作用

x_train = [ None for _ in range(len(x))]
x_test = [ None for _ in range(len(x))]

recv = [item for sublist in zip(x_train, x_test) for item in sublist]
# But unfortunately the above line gives only the values and not references
# Hence doesn't work

*recv, y_train, y_test = train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

I have multiple (type) inputs put inside a list x and I'm doing the test train split using:

x = [some_matrix, scalar_value, something_else, ...]
x0_train, x0_test, x1_train, x1_test, ... , y_train, y_test = 
   train_test_split(x[0],x[1],... , y, test_size=0.2, random_state=np.random, shuffle=True)

I managed to change the input parameters x[0], x[1], ... to *x:

x0_train, x0_test, x1_train, x1_test, ... , y_train, y_test = 
   train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

    # But I have to manually repack
    x_train = [x0_train, x1_train]
    x_test = [x0_test, x1_test]

But is there a way to receive it without having to manually repack? What is the equivalent of:

*x_train, *x_test, y_train, y_test = 
   train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

Or is there any other way to do this? For eg: constructing a dictionary and using ** to unpack, but I still have the same problem. What is the convention anyway (if one exists)?

解决方案

This solves the zigzag split problem:

recv = [None for i in range(2*len(x))]
*recv, y_train, y_test = train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

# Edit: credits hpaulj
x_train = recv[::2]
x_test = recv[1::2]

---

Also, if there's a way to copy references this will work too

x_train = [ None for _ in range(len(x))]
x_test = [ None for _ in range(len(x))]

recv = [item for sublist in zip(x_train, x_test) for item in sublist]
# But unfortunately the above line gives only the values and not references
# Hence doesn't work

*recv, y_train, y_test = train_test_split(*x, y, test_size=0.2, random_state=np.random, shuffle=True)

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