I'm trying to read and understand the paper Attention is all you need and in it, there is a picture:

enter image description here

I don't know what positional encoding is. by listening to some youtube videos I've found out that it is an embedding having both meaning and position of a word in it and has something to do with $sin(x)$ or $cos(x)$

but I couldn't understand what exactly it is and how exactly it is doing that. so I'm here for some help. thanks in advance.


4 Answers 4


Here is an awesome recent Youtube video that covers position embeddings in great depth, with beautiful animations:

Visual Guide to Transformer Neural Networks - (Part 1) Position Embeddings

Taking excerpts from the video, let us try understanding the “sin” part of the formula to compute the position embeddings:

enter image description here

Here “pos” refers to the position of the “word” in the sequence. P0 refers to the position embedding of the first word; “d” means the size of the word/token embedding. In this example d=5. Finally, “i” refers to each of the 5 individual dimensions of the embedding (i.e. 0, 1,2,3,4)

While “d” is fixed, “pos” and “i” vary. Let us try understanding the later two.


enter image description here

If we plot a sin curve and vary “pos” (on the x-axis), you will land up with different position values on the y-axis. Therefore, words with different positions will have different position embeddings values.

There is a problem though. Since “sin” curve repeat in intervals, you can see in the figure above that P0 and P6 have the same position embedding values, despite being at two very different positions. This is where the ‘i’ part in the equation comes into play.


enter image description here

If you vary “i” in the equation above, you will get a bunch of curves with varying frequencies. Reading off the position embedding values against different frequencies, lands up giving different values at different embedding dimensions for P0 and P6.


Thanks to @starriet for the correction. "i" is not the index of the element "within" each vector, but the "sequence" of the vector. It is used for making alternate even and odd sequences (2i and 2i+1).

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    $\begingroup$ The video link is awesome! It clearly explains the concept. $\endgroup$
    – Azhar Khan
    Commented Feb 10, 2022 at 13:34
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    $\begingroup$ Is there no collision induced by adding the word and position vectors? Or do we just no care? $\endgroup$ Commented Oct 5, 2022 at 17:34
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    $\begingroup$ @RyderBrooks I had exactly the same question. My hypothesis is that the model will naturally learn word embeddings that ensure that the positional modification does not induce collisions (since collisions would surely impair performance, and hence be trained out) but I would love someone with more knowledge to clarify that! $\endgroup$
    – atkins
    Commented Oct 29, 2022 at 19:50
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    $\begingroup$ @RyderBrooks The first FAQ in this article addresses this point a bit, and suggests a couple of references: kazemnejad.com/blog/… $\endgroup$
    – atkins
    Commented Oct 29, 2022 at 19:55
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    $\begingroup$ This is not correct. i itself is not the index of the element. It's just used for making alternate even and odd sequences (2i and 2i+1). The linked video is also incorrect. Actually, the original paper("attention is all you need") has a slight error on this notation, I guess that's why many people got it wrong. $\endgroup$ Commented Nov 23, 2022 at 3:13

For example, for word $w$ at position $pos \in [0, L-1]$ in the input sequence $\boldsymbol{w}=(w_0,\cdots, w_{L-1})$, with 4-dimensional embedding $e_{w}$, and $d_{model}=4$, the operation would be $$\begin{align*}e_{w}' &= e_{w} + \left[sin\left(\frac{pos}{10000^{0}}\right), cos\left(\frac{pos}{10000^{0}}\right),sin\left(\frac{pos}{10000^{2/4}}\right),cos\left(\frac{pos}{10000^{2/4}}\right)\right]\\ &=e_{w} + \left[sin\left(pos\right), cos\left(pos\right),sin\left(\frac{pos}{100}\right),cos\left(\frac{pos}{100}\right)\right]\\ \end{align*}$$

where the formula for positional encoding is as follows $$\text{PE}(pos,2i)=sin\left(\frac{pos}{10000^{2i/d_{model}}}\right),$$ $$\text{PE}(pos,2i+1)=cos\left(\frac{pos}{10000^{2i/d_{model}}}\right).$$ with $d_{model}=512$ (thus $i \in [0, 255]$) in the original paper.

This technique is used because there is no notion of word order (1st word, 2nd word, ..) in the proposed architecture. All words of input sequence are fed to the network with no special order or position; in contrast, in RNN architecture, $n$-th word is fed at step $n$, and in ConvNet, it is fed to specific input indices. Therefore, proposed model has no idea how the words are ordered. Consequently, a position-dependent signal is added to each word-embedding to help the model incorporate the order of words. Based on experiments, this addition not only avoids destroying the embedding information but also adds the vital position information.

This blog by Kazemnejad explains that the specific choice of ($sin$, $cos$) pair helps the model in learning patterns that rely on relative positions. As an example, consider a pattern like

if 'are' comes after 'they', then 'playing' is more likely than 'play'

which relies on relative position "$pos(\text{are}) - pos(\text{they})$" being 1, independent of absolute positions $pos(\text{are})$ and $pos(\text{they})$. To learn this pattern, any positional encoding should make it easy for the model to arrive at an encoding for "they are" that (a) is different from "are they" (considers relative position), and (b) is independent of where "they are" occurs in a given sequence (ignores absolute positions), which is what $\text{PE}$ manages to achieve.

This article by Jay Alammar explains the paper with excellent visualizations. The example on positional encoding calculates $\text{PE}(.)$ the same, with the only difference that it puts $sin$ in the first half of embedding dimensions (as opposed to even indices) and $cos$ in the second half (as opposed to odd indices). As pointed out by ShaohuaLi, this difference does not matter since vector operations would be invariant to the permutation of dimensions.

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    $\begingroup$ You also have this excellent article purely focused on the positional embedding : kazemnejad.com/blog/… $\endgroup$ Commented Feb 5, 2020 at 21:51
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    $\begingroup$ Is the 10000 in the denominator related to this comment by Jay Alammar in his post: "Let’s assume that our model knows 10,000 unique English words (our model’s 'output vocabulary')? $\endgroup$
    – tallamjr
    Commented Dec 3, 2020 at 15:53
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    $\begingroup$ Which half of the positional encoding is sin and cos doesn't matter. The dot product is the same after shuffling the embedding dimensions :-) $\endgroup$
    – Shaohua Li
    Commented Jan 11, 2021 at 9:15
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    $\begingroup$ @crizCraig In Transformer we do not have a unit with index 0 for the first word and so on. In other words, we can feed n words to n units in arbitrary order which is finally manifested as exchanging rows/columns in word-word attention matrix. $\endgroup$
    – Esmailian
    Commented Jul 22, 2021 at 21:56
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    $\begingroup$ @starriet If a positional encoding is added to a feature vector, the dot product between two such sums can be decomposed to two types of interactions: 1. dot product between two different positional encodings, and 2. dot product between a positional encoding and a feature vector. It should be apparent that the Type 1 dot product is shuffle-invariant w.r.t. the PE dimension. Type 2 dot product is not shuffle-invariant, but the PE channels have totally different semantics as the feature vector channels, therefore shuffling PE dimensions, the model is still equivalent to the original model. $\endgroup$
    – Shaohua Li
    Commented Nov 23, 2022 at 14:27

Positional encoding is a re-representation of the values of a word and its position in a sentence (given that is not the same to be at the beginning that at the end or middle).

But you have to take into account that sentences could be of any length, so saying '"X" word is the third in the sentence' does not make sense if there are different length sentences: 3rd in a 3-word-sentence is completely different to 3rd in a 20-word-sentence.

What a positional encoder does is to get help of the cyclic nature of $sin(x)$ and $cos(x)$ functions to return information of the position of a word in a sentence.

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    $\begingroup$ thank you. could you elaborate on how this positional encoder does this with $sin$ and $cos$? $\endgroup$
    – Peyman
    Commented Apr 28, 2019 at 16:56

To add to other answers, OpenAI's ref implementation calculates it in natural log-space (to improve precision, I think). They did not come up with the encoding.

Here is the PE lookup table generation rewritten in C with a nested for loop:

int d_model = 512, max_len = 5000;
double pe[max_len][d_model];

for (int i = 0; i < max_len; i++) {
   for (int k = 0; k < d_model; k = k + 2) {
      double div_term = exp(k * -log(10000.0) / d_model);
      pe[i][k] = sin(i * div_term);
      pe[i][k + 1] = cos(i * div_term);

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