Prioritized Experience Replay Buffer

This implements paper Prioritized experience replay, using a binary segment tree.

13import random
15import numpy as np

Buffer for Prioritized Experience Replay

Prioritized experience replay samples important transitions more frequently. The transitions are prioritized by the Temporal Difference error (td error), $\delta$.

We sample transition $i$ with probability, where $\alpha$ is a hyper-parameter that determines how much prioritization is used, with $\alpha = 0$ corresponding to uniform case. $p_i$ is the priority.

We use proportional prioritization $p_i = |\delta_i| + \epsilon$ where $\delta_i$ is the temporal difference for transition $i$.

We correct the bias introduced by prioritized replay using importance-sampling (IS) weights in the loss function. This fully compensates when $\beta = 1$. We normalize weights by $\frac{1}{\max_i w_i}$ for stability. Unbiased nature is most important towards the convergence at end of training. Therefore we increase $\beta$ towards end of training.

Binary Segment Tree

We use a binary segment tree to efficiently calculate $\sum_k^i p_k^\alpha$, the cumulative probability, which is needed to sample. We also use a binary segment tree to find $\min p_i^\alpha$, which is needed for $\frac{1}{\max_i w_i}$. We can also use a min-heap for this. Binary Segment Tree lets us calculate these in $\mathcal{O}(\log n)$ time, which is way more efficient that the naive $\mathcal{O}(n)$ approach.

This is how a binary segment tree works for sum; it is similar for minimum. Let $x_i$ be the list of $N$ values we want to represent. Let $b_{i,j}$ be the $j^{\mathop{th}}$ node of the $i^{\mathop{th}}$ row in the binary tree. That is two children of node $b_{i,j}$ are $b_{i+1,2j}$ and $b_{i+1,2j + 1}$.

The leaf nodes on row $D = \left\lceil {1 + \log_2 N} \right\rceil$ will have values of $x$. Every node keeps the sum of the two child nodes. That is, the root node keeps the sum of the entire array of values. The left and right children of the root node keep the sum of the first half of the array and the sum of the second half of the array, respectively. And so on…

Number of nodes in row $i$, This is equal to the sum of nodes in all rows above $i$. So we can use a single array $a$ to store the tree, where,

Then child nodes of $a_i$ are $a_{2i}$ and $a_{2i + 1}$. That is,

This way of maintaining binary trees is very easy to program. Note that we are indexing starting from 1.

We use the same structure to compute the minimum.

18class ReplayBuffer:


88    def __init__(self, capacity, alpha):

We use a power of $2$ for capacity because it simplifies the code and debugging

93        self.capacity = capacity


95        self.alpha = alpha

Maintain segment binary trees to take sum and find minimum over a range

98        self.priority_sum = [0 for _ in range(2 * self.capacity)]
99        self.priority_min = [float('inf') for _ in range(2 * self.capacity)]

Current max priority, $p$, to be assigned to new transitions

102        self.max_priority = 1.

Arrays for buffer

105 = {
106            'obs': np.zeros(shape=(capacity, 4, 84, 84), dtype=np.uint8),
107            'action': np.zeros(shape=capacity, dtype=np.int32),
108            'reward': np.zeros(shape=capacity, dtype=np.float32),
109            'next_obs': np.zeros(shape=(capacity, 4, 84, 84), dtype=np.uint8),
110            'done': np.zeros(shape=capacity, dtype=np.bool)
111        }

We use cyclic buffers to store data, and next_idx keeps the index of the next empty slot

114        self.next_idx = 0

Size of the buffer

117        self.size = 0

Add sample to queue

119    def add(self, obs, action, reward, next_obs, done):

Get next available slot

125        idx = self.next_idx

store in the queue

128['obs'][idx] = obs
129['action'][idx] = action
130['reward'][idx] = reward
131['next_obs'][idx] = next_obs
132['done'][idx] = done

Increment next available slot

135        self.next_idx = (idx + 1) % self.capacity

Calculate the size

137        self.size = min(self.capacity, self.size + 1)

$p_i^\alpha$, new samples get max_priority

140        priority_alpha = self.max_priority ** self.alpha

Update the two segment trees for sum and minimum

142        self._set_priority_min(idx, priority_alpha)
143        self._set_priority_sum(idx, priority_alpha)

Set priority in binary segment tree for minimum

145    def _set_priority_min(self, idx, priority_alpha):

Leaf of the binary tree

151        idx += self.capacity
152        self.priority_min[idx] = priority_alpha

Update tree, by traversing along ancestors. Continue until the root of the tree.

156        while idx >= 2:

Get the index of the parent node

158            idx //= 2

Value of the parent node is the minimum of it’s two children

160            self.priority_min[idx] = min(self.priority_min[2 * idx], self.priority_min[2 * idx + 1])

Set priority in binary segment tree for sum

162    def _set_priority_sum(self, idx, priority):

Leaf of the binary tree

168        idx += self.capacity

Set the priority at the leaf

170        self.priority_sum[idx] = priority

Update tree, by traversing along ancestors. Continue until the root of the tree.

174        while idx >= 2:

Get the index of the parent node

176            idx //= 2

Value of the parent node is the sum of it’s two children

178            self.priority_sum[idx] = self.priority_sum[2 * idx] + self.priority_sum[2 * idx + 1]

$\sum_k p_k^\alpha$

180    def _sum(self):

The root node keeps the sum of all values

186        return self.priority_sum[1]

$\min_k p_k^\alpha$

188    def _min(self):

The root node keeps the minimum of all values

194        return self.priority_min[1]

Find largest $i$ such that $\sum_{k=1}^{i} p_k^\alpha \le P$

196    def find_prefix_sum_idx(self, prefix_sum):

Start from the root

202        idx = 1
203        while idx < self.capacity:

If the sum of the left branch is higher than required sum

205            if self.priority_sum[idx * 2] > prefix_sum:

Go to left branch of the tree

207                idx = 2 * idx
208            else:

Otherwise go to right branch and reduce the sum of left branch from required sum

211                prefix_sum -= self.priority_sum[idx * 2]
212                idx = 2 * idx + 1

We are at the leaf node. Subtract the capacity by the index in the tree to get the index of actual value

216        return idx - self.capacity

Sample from buffer

218    def sample(self, batch_size, beta):

Initialize samples

224        samples = {
225            'weights': np.zeros(shape=batch_size, dtype=np.float32),
226            'indexes': np.zeros(shape=batch_size, dtype=np.int32)
227        }

Get sample indexes

230        for i in range(batch_size):
231            p = random.random() * self._sum()
232            idx = self.find_prefix_sum_idx(p)
233            samples['indexes'][i] = idx

$\min_i P(i) = \frac{\min_i p_i^\alpha}{\sum_k p_k^\alpha}$

236        prob_min = self._min() / self._sum()

$\max_i w_i = \bigg(\frac{1}{N} \frac{1}{\min_i P(i)}\bigg)^\beta$

238        max_weight = (prob_min * self.size) ** (-beta)
240        for i in range(batch_size):
241            idx = samples['indexes'][i]

$P(i) = \frac{p_i^\alpha}{\sum_k p_k^\alpha}$

243            prob = self.priority_sum[idx + self.capacity] / self._sum()

$w_i = \bigg(\frac{1}{N} \frac{1}{P(i)}\bigg)^\beta$

245            weight = (prob * self.size) ** (-beta)

Normalize by $\frac{1}{\max_i w_i}$, which also cancels off the $\frac{1}{N}$ term

248            samples['weights'][i] = weight / max_weight

Get samples data

251        for k, v in
252            samples[k] = v[samples['indexes']]
254        return samples

Update priorities

256    def update_priorities(self, indexes, priorities):
261        for idx, priority in zip(indexes, priorities):

Set current max priority

263            self.max_priority = max(self.max_priority, priority)

Calculate $p_i^\alpha$

266            priority_alpha = priority ** self.alpha

Update the trees

268            self._set_priority_min(idx, priority_alpha)
269            self._set_priority_sum(idx, priority_alpha)

Whether the buffer is full

271    def is_full(self):
275        return self.capacity == self.size