This is the third post of the series about Napoleon X Challenge.
Post 1
is the data analysis part, where we discovered how to put stuff together.Post 2
is the time series disambiguation part. We were able to partially find which crypto-asset corresponds to which time series.Post 3
explains how we obtained very good prediction accuracy using what we learned in Post 1
and a simple linear regression model.
As a reminder of the dataset and the prediction task:
What we learned in Post 1
is:
In Post 2
, we understood what is the meaning of md
and were able to identify the name of the crypto-asset (e.g. ethereum, litecoin, …).
Post 2 doesn’t really help, but confirms that we were right about asset matching.
In post 1, we saw that assets are correlated, or more specifically, they are impacted by a global trend.
Therefore, they more or less go in the same direction.
We exploited this fact to fill the test sheet in post 1
, filling the value with the average given the current week-day.
Here, we will try to be more accurate by exploiting assets of clusters.
As a cluster exists for three weeks, and a new set of clusters is issued every week, it means that for a given week, we have three generation of clusters at the same time. The following figure illustrate the process with:
4
clusters generated at week w_i
,3
clusters generated at week w_i+1
,4
clusters generated at week w_i+2
.Which means that if we want to infer something about returns for a given day occurring in week w_i+2
, we have information about 11
clusters.
We have clusters and we have assets. Fortunately, the overlap between train and test assets is relatively large.
If we take as an example week 12
, there are 8
clusters:
4
train clusters4
test clustersAnd for week 11
, there are 4
clusters, and 8
for week 10
.
Therefore, we have 16
clusters to help us predicting returns for the test clusters.
Now, if we look at assets, among these 16
clusters, there are 120
unique assets.
For the 4
test clusters, there are only 70
.
Among these 70
, 69
are shared with the train clusters.
Because the number of clusters is much smaller than the number of assets, it is impossible to recover the exact value of each asset. However, we can identity the group trend, which is useful enough.
We can represent assets as a binary matrix, where each row represents a cluster, and each column an asset.
For training, we get the binary matrix:
For testing, we get the binary matrix:
We have a set of \(c\) clusters, each made of a given set of assets, represented by the matrix \(M \in \{0, 1\}^{c \times a}\). The average return of the cluster is \(\mathbf{y} \in \mathcal{R}^{c}\).
We want to find \(\mathbf{w} \in \mathbb{R}^{a}\) such as:
\[\arg\min_Y \|\hat{M} \mathbf{w} - \mathbf{y}\|^2\]where \(\hat{M_{i, j}} = \frac{M_{i,j}}{\sum_k M_{i, k}}\) is the matrix \(M\) row-normalized. This normalization is necessary because \(\mathbf{y}\) is the average return. Therefore, the more assets there are, the less each individually contributes.
Coding what we presented above is easy.
sklearn
and numpy
are needed to not reinvent the wheel.
The only difficulty is to recover predicted values for the different weeks without getting wrong on the index.
Otherwise, when the binary matrices are isolated and average returns of train clusters extracted, the game is easy:
W = M.sum(axis=1) # M is the train binary matrix
W1 = M1.sum(axis=1) # M1 is the test binary matrix
# We normalize both, because what we measure is the average return
M = (M.T / W).T
M1 = (M1.T / W1).T
for i in range(7): # For each day of the week
md = LinearRegression()
md.fit(M, Y_train[:, i]) # Y_train is the average return for each train cluster
y_test_i = md.predict(M1) # This is the estimated average return for the test clusters.
# + Other operations to store `y_test_i` in the correct location of the submission sheet.
The linear regression is possible and effective if:
These are hypothesis, and are not always fulfilled. Because we have 1463 train clusters and 627 test clusters, the probability that the first hypothesis is not verified is low. However, for the second, it happens that the overlap is too small to be accurate.
In that case, we remember that crypto-assets are affected by the main global trend.
Therefore, we go back with prediction made in post 1
using the average when we do not have enough information about assets.
In our code, we have reshaped the dataset in a more convenient form.
Instead of having rows with one asset and one day of data,
we have DX
where each row represents one asset over 21 days in its cluster, with columns:
21 x 24 - 1
, with 24 hours filled with 0.
)md
and bc
time
, which corresponds to the starting week of the clusterassignment
, which is the global asset ID
, which is independent of time.With this format, it is easier to extract average return for each day.
# Initialization
dic_mean_list = {}
for tx in range(0, 216):
for i in range(7):
dic_mean_list[tx*7+i] = []
for tx in range(0, 216):
S = set(DX[DX.time == tx].cluster)
S = list(filter(lambda x: x <= 1463, S))
for c in S:
for i in range(21):
y = data_y.loc[c*21 + i].values[0] # Y Train sheet
dic_mean_list[tx*7+i].append(y)
We just have to find where are located extreme values, and replace them with the mean return of train clusters.
vlim = 0.1 # Threshold for correction
for i in np.where((data_y_pred.values > vlim) | (data_y_pred.values < -vlim))[0]:
sample_id = data_y_pred.index[i] # Get the sample ID
cls = sample_id//21 # Extract corresponding cluster
day = sample_id % 21 # Extract day
# DX is a dataframe where for each cluster, we have a `time` column corresponding to the starting week.
t0 = int(DX[DX.cluster == cls].time.values[0] * 7 + day)
vr = np.mean(dic_mean_list[t0]) # Averaging
data_y_pred.loc[sample_id] = vr
In our case, we found 250
extreme values, which is very small compared to the 13133
sample ID.
However, because of their amplitude, they would affect the evaluation score.
Now that we have filled the blanks, we can check if the train and test sets have a similar returns distribution.
When looking at it, we still get a Cauchy distribution, with a negligible difference.
We reduced our MSE from 0.0061
to 0.0045
, and moved from rank 7
to rank 1
.
1st rank, without any complicated algorithms. Just pure logic and basic things.
Enjoy !
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