Python3 - 列表索引超出范围答案

【问题标题】：Python3 - List Index out of rangePython3 - 列表索引超出范围
【发布时间】：2018-07-05 03:19:35
【问题描述】：

我正在研究字符级循环神经网络。为了训练网络，我从互联网上复制了一个文本语料库。这是其中有错误的代码块：

X = np.zeros((int(len(data)/SEQ_LENGTH), SEQ_LENGTH, VOCAB_SIZE))
y = np.zeros((int(len(data)/SEQ_LENGTH), SEQ_LENGTH, VOCAB_SIZE))
for i in range(0, int(len(data)/SEQ_LENGTH)):
    X_sequence = data[i*SEQ_LENGTH:(i+1)*SEQ_LENGTH]
    X_sequence_ix = [char_to_ix[value] for value in X_sequence]
    input_sequence = np.zeros((SEQ_LENGTH, VOCAB_SIZE))
    for j in range(SEQ_LENGTH):
        input_sequence[j][X_sequence_ix[j]] = 1.
    X[i] = input_sequence

    y_sequence = data[i*SEQ_LENGTH+1:(i+1)*SEQ_LENGTH+1]
    y_sequence_ix = [char_to_ix[value] for value in y_sequence]
    target_sequence = np.zeros((SEQ_LENGTH, VOCAB_SIZE))
    for j in range(SEQ_LENGTH):

        target_sequence[j][y_sequence_ix[j]] = 1       
    y[i] = target_sequence

基本上，我所做的只是将字符转换为它们的 ASCII 等价物。 y_sequence 是字符序列，y_sequence_ix 是其对应的 ASCII 序列。 VOCAB_SIZE 变量包含文本语料库中唯一字符的数量。错误发生在这一行：

target_sequence[j][y_sequence_ix[j]] = 1

完整的源代码和文本语料库：https://github.com/tanmay-edgelord/charRNN

请询问您回答问题所需的任何信息。

编辑

调用函数traceback.print_stack()时的TRACEBACK

File "/usr/lib/python3.5/runpy.py", line 184, in _run_module_as_main
    "__main__", mod_spec)
  File "/usr/lib/python3.5/runpy.py", line 85, in _run_code
    exec(code, run_globals)
  File "/usr/local/lib/python3.5/dist-packages/ipykernel_launcher.py", line 16, in <module>
    app.launch_new_instance()
  File "/usr/local/lib/python3.5/dist-packages/traitlets/config/application.py", line 658, in launch_instance
    app.start()
  File "/usr/local/lib/python3.5/dist-packages/ipykernel/kernelapp.py", line 478, in start
    self.io_loop.start()
  File "/usr/local/lib/python3.5/dist-packages/zmq/eventloop/ioloop.py", line 177, in start
    super(ZMQIOLoop, self).start()
  File "/usr/local/lib/python3.5/dist-packages/tornado/ioloop.py", line 888, in start
    handler_func(fd_obj, events)
  File "/usr/local/lib/python3.5/dist-packages/tornado/stack_context.py", line 277, in null_wrapper
    return fn(*args, **kwargs)
  File "/usr/local/lib/python3.5/dist-packages/zmq/eventloop/zmqstream.py", line 440, in _handle_events
    self._handle_recv()
  File "/usr/local/lib/python3.5/dist-packages/zmq/eventloop/zmqstream.py", line 472, in _handle_recv
    self._run_callback(callback, msg)
  File "/usr/local/lib/python3.5/dist-packages/zmq/eventloop/zmqstream.py", line 414, in _run_callback
    callback(*args, **kwargs)
  File "/usr/local/lib/python3.5/dist-packages/tornado/stack_context.py", line 277, in null_wrapper
    return fn(*args, **kwargs)
  File "/usr/local/lib/python3.5/dist-packages/ipykernel/kernelbase.py", line 283, in dispatcher
    return self.dispatch_shell(stream, msg)
  File "/usr/local/lib/python3.5/dist-packages/ipykernel/kernelbase.py", line 233, in dispatch_shell
    handler(stream, idents, msg)
  File "/usr/local/lib/python3.5/dist-packages/ipykernel/kernelbase.py", line 399, in execute_request
    user_expressions, allow_stdin)
  File "/usr/local/lib/python3.5/dist-packages/ipykernel/ipkernel.py", line 208, in do_execute
    res = shell.run_cell(code, store_history=store_history, silent=silent)
  File "/usr/local/lib/python3.5/dist-packages/ipykernel/zmqshell.py", line 537, in run_cell
    return super(ZMQInteractiveShell, self).run_cell(*args, **kwargs)
  File "/usr/local/lib/python3.5/dist-packages/IPython/core/interactiveshell.py", line 2728, in run_cell
    interactivity=interactivity, compiler=compiler, result=result)
  File "/usr/local/lib/python3.5/dist-packages/IPython/core/interactiveshell.py", line 2856, in run_ast_nodes
    if self.run_code(code, result):
  File "/usr/local/lib/python3.5/dist-packages/IPython/core/interactiveshell.py", line 2910, in run_code
    exec(code_obj, self.user_global_ns, self.user_ns)
  File "<ipython-input-15-b47f6c9a5577>", line 2, in <module>
    traceback.print_stack()

【问题讨论】：

仔细检查target_sequence 的形状和大小。如果我没记错的话，它应该有两个维度。因此，调用target_sequence[j] 应该返回另一个列表（您再次使用[y_sequence[j]] 对其进行索引，但如果[y_sequence[j]] 的结果大于VOCAB_SIZE，您将超出范围。回溯一下，您的y_sequence 似乎是基于 SEQ_LENGTH 而不是 VOCAB_SIZE，所以这很有可能是问题的根源。
target_sequence 的尺寸为 100x80。此外，y_sequence 的长度不是问题，因为我有兴趣访问与 y_sequence 变量中第 j 个字符的 ASCII 码对应的索引并设置 target_sequence[j][y_sequence_ix[j] = 1
请发布完整的 Traceback。
当您caught the error 并检查数据和变量时，您发现了什么？
变量的长度符合预期，必须匹配的变量是匹配的。这就是我感到困惑的原因，另外我从这个链接复制了这段代码：chunml.github.io/ChunML.github.io/project/…。这显然对他有用

标签： python python-3.x list deep-learning recurrent-neural-network

【解决方案1】：

问题在于y_sequence_ix[j]，因为您的j 范围函数正在使用 SEQ_LENGTH+1。这就是您想要的，因为您要预测列表中的 next 值，但它也会产生一个问题，即 y_sequence_ix 在 i 的最后一次迭代中太短了 1 个样本，而 data 与 @ 均分987654327@。

import string
import random
import numpy as np

ix_to_char = list(string.ascii_lowercase)
data = [random.choice(string.ascii_lowercase) for x in range(1000)]
char_to_ix = {v:i for i,v in enumerate(ix_to_char)}
VOCAB_SIZE  = len(char_to_ix)
SEQ_LENGTH = 5

usable_len = len(data)-1    #Note: don't try to use the last entry in data
X = np.zeros((int(usable_len/SEQ_LENGTH), SEQ_LENGTH, VOCAB_SIZE))
y = np.zeros((int(usable_len/SEQ_LENGTH), SEQ_LENGTH, VOCAB_SIZE))
for i in range(0, int(usable_len/SEQ_LENGTH)):       
    X_sequence = data[i*SEQ_LENGTH:(i+1)*SEQ_LENGTH]
    X_sequence_ix = [char_to_ix[value] for value in X_sequence]

    input_sequence = np.zeros((SEQ_LENGTH, VOCAB_SIZE))
    for j in range(SEQ_LENGTH):
        input_sequence[j][X_sequence_ix[j]] = 1.
    X[i] = input_sequence

    y_sequence = data[i*SEQ_LENGTH+1:(i+1)*SEQ_LENGTH+1]
    y_sequence_ix = [char_to_ix[value] for value in y_sequence]
    target_sequence = np.zeros((SEQ_LENGTH, VOCAB_SIZE))
    for j in range(SEQ_LENGTH):
        target_sequence[j][y_sequence_ix[j]] = 1
    y[i] = target_sequence

除了顶部的数据创建，这里唯一修改的是在数据创建中使用len(data)-1而不是len(data)。通过截断单个值，您可以确保不会收到 index out of range 类型错误。这是最容易做的事情，但是您也可以将 SEQ_LENGTH 设置为不是 len(data) 的偶数倍的值，即..6 适用于上述示例。

【讨论】：

关于最后一个案例，我不明白你想说什么
如何用 0 填充数据？
我重新考虑了一下。填充会起作用，但教网络一个不正确的值可能是个坏主意。以上会更好，并且是一个非常简单的mod。试一试。顺便说一句..这可能最初是从某人那里工作的，因为他们的数据长度不同或设置的 SEQ_LENGTH 不同。