# Sampling Algorithms

Published January 17, 2021

Recently I played around with some reinforcement learning algorithms and came across the problem of sampling from replay buffers. I thought it is a fun exercise to write a blog post about.

There are two popular strategies to sample from a dataset in reinforcement learning: uniformly, where each data point has the same probability, and prioritised, where each data point has a different probability of getting sampled, i.e. weighted by training error.1

The aim of this post is to give an outline how to implement these sampling strategies as efficient algorithms.

# Uniform Distribution

The default strategy in stochastic gradient descent is sampling batches uniformly. All data points have the same probability.

$P(x_i) = \frac{1}{n} \; \text{ where i \in \{0, 1, \dots, n-1\}}$

Here $x_i$ is a data point from our dataset $\mathcal{D} = \{x_0, x_1, \dots, x_{n-1}\}$.

The goal is to sample $k$ items from $\mathcal{D}$. We can either choose to sample without or with replacement. What’s the difference?

Sampling without replacement means samples are unique and cannot be sampled twice per batch. On the flip-side, sampling with replacement means items can appear twice in one batch. Implementing the latter is trivial. Simply pick $k$ integers in $\left[0; n\right)$ and you are done.

Sampling without replacement is more interesting. How can we implement this efficiently?

## Sampling without Replacement

Let’s think of a physical analogy — picking cards. If you were asked to pick $k$ random cards out of a poker deck without replacement, you might come up with the following strategy: 1. Pick a random card out of one stack of cards, 2. Put that card on a different stack, 3. Repeat.

If we replace stacks with arrays in this description, we already have a quite convincing algorithm, although a few details are missing. Here is one way to do it:

Much like with a stack of cards, we keep two arrays. The first array stores the result (result array), the second array stores the integers which we can still choose from (open array). Initially, there are zero integers in the result array and $n$ integers in the open array. On each iteration we then randomly pick one integer in the open array, remove it and put it into the result array. After doing this $k$ times, we have a batch of unique samples.

To make this more efficient, we can store both arrays in one “super” array which we divide into two at index $r$. The result array goes from index $0$ to index $r$, while the open array goes from index $r$ to $n-1$ (inclusive). When we want to transfer the sample from open to result array, we simply swap the picked index $i$ with the integer at index $r$ and increase $r$ by one. Figure 1 visualises this algorithm.

Let’s implement this algorithm using C++. As we need to have an array of size $n$ in memory, it makes sense to reuse that througout multiple samples. Therefore we create a class for the sampler.

Now that we have our class, we can implement the algorithm. On each iteration $i$, we sample $j \sim U\{r, N\}$ and swap the $r$-th with $j$-th integer as previously discussed.

The per-sample runtime of this algorithm is $O(1)$. The memory requirement is $O(n)$. In contrast to other algorithms, the runtime does not depend on $n$. More analysis on the runtime can be found in Figure 9.

# Unnormalised Categorical Distribution

As mentioned in the introduction, there are other ways to sample from a dataset besides uniform sampling. In the following, I will be focusing on the categorical distribution where each data point has its own probability of getting sampled.

$P(x_i) = p_i$

Most of the time, however, we do not have probabilities. Instead we are given weights $w_i$ which indicate relative importance of a sample. For example, a set of weights might be $\{1, 3, 8, 1, 3, 2, 1, 4\}$. In this case, the third data point has the highest importance. As a valid probability distribution requires that $\sum_i p_i = 1$, we need to normalise the weights before we can sample.

$P(x_i) = \frac{w_i}{\sum_j w_j}$

Returning to the example from above, we would have a probability distribution of $\{\frac{1}{23}, \frac{3}{23}, \frac{8}{23}, \frac{1}{23}, \frac{3}{23}, \frac{2}{23}, \frac{1}{23}, \frac{4}{23}\}$. For larger datasets, sampling from this distribution is tricky. Summing over all elements has linear complexity. Moreover, we cannot cache the sum, because in reinforcement learning, for example, weights are changing after each iteration. Is there a way to improve over linear complexity?

## Segment Tree

The answer is yes. We can make use of a binary tree structure to efficiently compute the sum over all elements. The idea is to arrange all data points as leafs of a tree. The key of each leaf is the weight $w_i$ of its data point. To maintain a sum over all weights, the key of a higher-level node is defined as the sum of its child nodes.

$\mathrm{key}\left(i\right) = \sum_{j \in \mathrm{children}\left(i\right)} \mathrm{key}\left(j\right)$

This is known as a segment tree or sum heap. Take a look at Figure 4 for an illustration of this data structure.

Given a segment tree of the weights, it is straightforward to sample from a categorical distribution. Take a random number $u \in \mathbb{R}$ between $0$ and $1$ sampled from a uniform distribution $u \sim U\{0, 1\}$. Let $v = \hat{w} \cdot u$ where $\hat{w}$ is the value of the root node. Start at the root node and descent left if $v < \mathrm{key}\left(\mathrm{left\,node}\right)$, otherwise set $v = v - \mathrm{key}\left(\mathrm{left\,node}\right)$ and descent right. Once you reach a leaf node, you have your sample.

Essentially, we partition $\left[0, \hat{w}\right)$ into $n$ intervals with the size $w_i$, sample from $\left[0, \hat{w}\right)$ and then find the interval in which that number falls.

To sample without replacement, simply set the weight to zero after each sample and update the tree. This has $\log n$ computational cost. The only drawback is that you need to store the old weights and restore them afterwards, if you want to sample from the same distribution again.

We start implementing this algorithm by designing a class that can hold the tree. To make things easier, we store the tree using an array representation where the two children of node $j$ are at position $2j$ and $2j + 1$. This requires that $n$ is a power of $2$, resulting in an array that has size $2n$.

Next, we implement some basic functions to maintain the sum tree. Heapify walks up the tree and restores the sum condition. SetWeight and GetWeight are for users who wish to update the distribution.

After having all the setup done, we can focus on implementing the sampling algorithm as described earlier.

Show code for sampling with replacement [+]

This segment tree-based algorithm has runtime complexity $O(\log n)$ per sample. Its memory complexity is $O(n + k)$.

# Runtime Analysis

You might wonder how well these algorithms perform for bigger datasets in practice. To find that out, I ran some performance tests. Take a look at Figure 9.

The log-complexity of categorical sampling w.r.t. to $n$ is noticeable in plot (1). There is some performance trade-off when using sampling without replacement. It takes around $1.5$ms to sample a batch of size $1024$ from a dataset with $100$ million data points (which by today’s machine learning standards is a reasonably large dataset). Categorical sampling is around $25$ to $30$ times slower compared to uniform sampling for $N = 64000$, as seen in plot (4).

# Effect on Training Neural Nets

We discussed sampling without replacement throughout this post. It might not be obvious what the effect on training a neural net is. From a theoretical standpoint, it reduces the variance of an estimator. Take for example a mean estimator $Y = \frac{1}{k}\sum_{i=1}^k X_i$ with $X_i \sim \mathcal{D}$. Using sampling without replacement, the variance of this estimator reduces to2

$\hat{\sigma}^2 = \frac{N-k}{N-1} \cdot \sigma^2$

where $\sigma^2$ is the variance of the same estimator but with replacement sampling.

In domains where $N$ is small, this can make a difference. If $N$ is large, however, the factor $\frac{N-k}{N-1}$ gets closer and closer to $1$. For example, with a batch size of $b = 1\%$ of the dataset size, we get a factor of $N \cdot \frac{1 - b}{N-1} \approx 1 - b = 99 \%$. Sampling without replacement reduces the variance by $1\%$ in this case.

Although this applies only to mean estimators, the analogy might give you some intuition about the effect on neural network training of sampling without replacement.

# Conclusion

In this post, I presented algorithms to sample from a dataset uniformly or weighted. These algorithms find application in reinforcement learning for prioritised replay buffers 1 or in supervised learning when sampling batches. The effect of sampling without replacement is larger for small datasets.

Further research into the effect of prioritising training examples based on learning error would be interesting. The algorithms presented here could be used as a basis for ideas in that space.

If you want to read more about sampling algorithms, take a look at Tim Vieira’s blog: https://timvieira.github.io/blog/post/2019/09/16/algorithms-for-sampling-without-replacement/