"""
Minimal character-level Vanilla RNN model. Written by Andrej Karpathy (@karpathy)
BSD License
"""
import numpy as np
import jieba

# data I/O
data = open('/home/multiangle/download/280.txt', 'rb').read() # should be simple plain text file
data = data.decode('gbk')
data = list(jieba.cut(data,cut_all=False))
chars = list(set(data))
data_size, vocab_size = len(data), len(chars)
print ('data has %d characters, %d unique.' % (data_size, vocab_size))
char_to_ix = { ch:i for i,ch in enumerate(chars) }
ix_to_char = { i:ch for i,ch in enumerate(chars) }

# hyperparameters
hidden_size = 200   # size of hidden layer of neurons
seq_length = 25 # number of steps to unroll the RNN for
learning_rate = 1e-1

# model parameters
Wxh = np.random.randn(hidden_size, vocab_size)*0.01 # input to hidden
Whh = np.random.randn(hidden_size, hidden_size)*0.01 # hidden to hidden
Why = np.random.randn(vocab_size, hidden_size)*0.01 # hidden to output
bh = np.zeros((hidden_size, 1)) # hidden bias
by = np.zeros((vocab_size, 1)) # output bias

def lossFun(inputs, targets, hprev):
    """
    inputs,targets are both list of integers.
    hprev is Hx1 array of initial hidden state
    returns the loss, gradients on model parameters, and last hidden state
    """
    xs, hs, ys, ps = {}, {}, {}, {}
    hs[-1] = np.copy(hprev)  # hprev 中间层的值, 存作-1，为第一个做准备
    loss = 0
    # forward pass
    for t in range(len(inputs)):
        xs[t] = np.zeros((vocab_size,1)) # encode in 1-of-k representation
        xs[t][inputs[t]] = 1    # x[t] 是一个第t个输入单词的向量

        # 双曲正切, 激活函数, 作用跟sigmoid类似
        # h(t) = tanh(Wxh*X + Whh*h(t-1) + bh) 生成新的中间层
        hs[t] = np.tanh(np.dot(Wxh, xs[t]) + np.dot(Whh, hs[t-1]) + bh) # hidden state  tanh
        # y(t) = Why*h(t) + by
        ys[t] = np.dot(Why, hs[t]) + by # unnormalized log probabilities for next chars
        # softmax regularization
        # p(t) = softmax(y(t))
        ps[t] = np.exp(ys[t]) / np.sum(np.exp(ys[t])) # probabilities for next chars, 对输出作softmax
        # loss += -log(value) 预期输出是1，因此这里的value值就是此次的代价函数，使用 -log(*) 使得离正确输出越远，代价函数就越高
        loss += -np.log(ps[t][targets[t],0]) # softmax (cross-entropy loss) 代价函数是交叉熵

    # 将输入循环一遍以后，得到各个时间段的h, y 和 p
    # 得到此时累积的loss, 准备进行更新矩阵
    # backward pass: compute gradients going backwards
    dWxh, dWhh, dWhy = np.zeros_like(Wxh), np.zeros_like(Whh), np.zeros_like(Why) # 各矩阵的参数进行
    dbh, dby = np.zeros_like(bh), np.zeros_like(by)
    dhnext = np.zeros_like(hs[0])   # 下一个时间段的潜在层，初始化为零向量
    for t in reversed(range(len(inputs))): # 把时间作为维度，则梯度的计算应该沿着时间回溯
        dy = np.copy(ps[t])  # 设dy为实际输出，而期望输出（单位向量）为y, 代价函数为交叉熵函数
        dy[targets[t]] -= 1 # backprop into y. see http://cs231n.github.io/neural-networks-case-study/#grad if confused here
        dWhy += np.dot(dy, hs[t].T)  # dy * h(t).T h层值越大的项，如果错误，则惩罚越严重。反之，奖励越多（这边似乎没有考虑softmax的求导？）, y = Why * hs[t] + by
        dby += dy # 这个没什么可说的，与dWhy一样，只不过h项=1， 所以直接等于dy
        dh = np.dot(Why.T, dy) + dhnext # backprop into h  y_t = Why*H_t + b_y H_t = tanh(Whh*H_(t-1) + bh + Whx*X_t), 第一阶段求导
        # f(z) = tanh(z), f(z)' = 1 − (f(z))^2
        dhraw = (1 - hs[t] * hs[t]) * dh # backprop through tanh nonlinearity  第二阶段求导，注意tanh的求导, hs[t] = Whh*H_(t-1) + bh + Whx*X_t
        dbh += dhraw   # dbh表示传递 到h层的误差
        dWxh += np.dot(dhraw, xs[t].T)    # 对Wxh的修正，同Why
        dWhh += np.dot(dhraw, hs[t-1].T)  # 对Whh的修正
        dhnext = np.dot(Whh.T, dhraw)     # h层的误差通过Whh不停地累积
        # dh表示d_ht,H_t = tanh(Whh*H_(t-1) + bh + Whx*X_t),y对于H_(t-1)的倒数为np.dot(Whh.T, dhraw)
    for dparam in [dWxh, dWhh, dWhy, dbh, dby]:
        np.clip(dparam, -5, 5, out=dparam) # clip to mitigate exploding gradients
    return loss, dWxh, dWhh, dWhy, dbh, dby, hs[len(inputs)-1]

#基于之前的文本，生成新的文本
def sample(h, seed_ix, n):
    """
    sample a sequence of integers from the model
    h is memory state, seed_ix is seed letter for first time step
    """
    x = np.zeros((vocab_size, 1))
    x[seed_ix] = 1
    ixes = []
    for t in range(n):
        h = np.tanh(np.dot(Wxh, x) + np.dot(Whh, h) + bh)    # 更新中间层
        y = np.dot(Why, h) + by             # 得到输出
        p = np.exp(y) / np.sum(np.exp(y))   # softmax
        ix = np.random.choice(range(vocab_size), p=p.ravel())   # 根据softmax得到的结果，按概率产生下一个字符
        x = np.zeros((vocab_size, 1))       # 产生下一轮的输入
        x[ix] = 1
        ixes.append(ix)
    return ixes

n, p = 0, 0
mWxh, mWhh, mWhy = np.zeros_like(Wxh), np.zeros_like(Whh), np.zeros_like(Why)
mbh, mby = np.zeros_like(bh), np.zeros_like(by) # memory variables for Adagrad
smooth_loss = -np.log(1.0/vocab_size)*seq_length # loss at iteration 0
while True:
    # prepare inputs (we're sweeping from left to right in steps seq_length long)
    if p+seq_length+1 >= len(data) or n == 0:   # 如果 n=0 或者 p过大
        hprev = np.zeros((hidden_size,1)) # reset RNN memory 中间层内容初始化，零初始化
        p = 0 # go from start of data           # p 重置
    inputs = [char_to_ix[ch] for ch in data[p:p+seq_length]] # 一批输入seq_length个字符
    targets = [char_to_ix[ch] for ch in data[p+1:p+seq_length+1]]  # targets是对应的inputs的期望输出。

    # sample from the model now and then
    if n % 100 == 0:      # 每循环100词， sample一次，显示结果
        sample_ix = sample(hprev, inputs[0], 200)
        txt = ''.join(ix_to_char[ix] for ix in sample_ix)
        print ('----\n %s \n----' % (txt, ))

    # forward seq_length characters through the net and fetch gradient
    # hprev 有一个缺点：状态向量来自于前一个数据块
    loss, dWxh, dWhh, dWhy, dbh, dby, hprev = lossFun(inputs, targets, hprev)
    smooth_loss = smooth_loss * 0.999 + loss * 0.001   # 将原有的Loss与新loss结合起来
    if n % 100 == 0: print ('iter %d, loss: %f' % (n, smooth_loss)) # print progress

    # perform parameter update with Adagrad
    for param, dparam, mem in zip([Wxh, Whh, Why, bh, by],
                                  [dWxh, dWhh, dWhy, dbh, dby],
                                  [mWxh, mWhh, mWhy, mbh, mby]):
        mem += dparam * dparam  # 梯度的累加
        param += -learning_rate * dparam / np.sqrt(mem + 1e-8) # adagrad update 随着迭代次数增加，参数的变更量会越来越小

    p += seq_length # move data pointer
    n += 1 # iteration counter， 循环次数