Version 1.1.0
Mean encodings
In this programming assignment you will be working with 1C
dataset from the final competition. You are asked to encode item_id
in 4 different ways:
1) Via KFold scheme;
2) Via Leave-one-out scheme;
3) Via smoothing scheme;
4) Via expanding mean scheme.
You will need to submit the correlation coefficient between resulting encoding and target variable up to 4 decimal places.
General tips
- Fill NANs in the encoding with
0.3343
. - Some encoding schemes depend on sorting order, so in order to avoid confusion, please use the following code snippet to construct the data frame. This snippet also implements mean encoding without regularization.
1 | import pandas as pd |
1 | list(product('ABCD', 'xy')) |
[('A', 'x'),
('A', 'y'),
('B', 'x'),
('B', 'y'),
('C', 'x'),
('C', 'y'),
('D', 'x'),
('D', 'y')]
Read data
1 | sales = pd.read_csv('../readonly/final_project_data/sales_train.csv.gz') |
1 | sales.head(10) |
date | date_block_num | shop_id | item_id | item_price | item_cnt_day | |
---|---|---|---|---|---|---|
0 | 02.01.2013 | 0 | 59 | 22154 | 999.00 | 1.0 |
1 | 03.01.2013 | 0 | 25 | 2552 | 899.00 | 1.0 |
2 | 05.01.2013 | 0 | 25 | 2552 | 899.00 | -1.0 |
3 | 06.01.2013 | 0 | 25 | 2554 | 1709.05 | 1.0 |
4 | 15.01.2013 | 0 | 25 | 2555 | 1099.00 | 1.0 |
5 | 10.01.2013 | 0 | 25 | 2564 | 349.00 | 1.0 |
6 | 02.01.2013 | 0 | 25 | 2565 | 549.00 | 1.0 |
7 | 04.01.2013 | 0 | 25 | 2572 | 239.00 | 1.0 |
8 | 11.01.2013 | 0 | 25 | 2572 | 299.00 | 1.0 |
9 | 03.01.2013 | 0 | 25 | 2573 | 299.00 | 3.0 |
1 | sales.nunique() |
date 1034
date_block_num 34
shop_id 60
item_id 21807
item_price 19993
item_cnt_day 198
dtype: int64
1 | sales["date_block_num"].unique() |
array([ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33])
1 | sales[sales["date_block_num"] ==1].head() |
date | date_block_num | shop_id | item_id | item_price | item_cnt_day | |
---|---|---|---|---|---|---|
115690 | 21.02.2013 | 1 | 50 | 3880 | 1499.0 | 1.0 |
115691 | 14.02.2013 | 1 | 50 | 3880 | 1499.0 | 1.0 |
115692 | 21.02.2013 | 1 | 50 | 4128 | 899.0 | 1.0 |
115693 | 13.02.2013 | 1 | 50 | 4124 | 249.0 | 1.0 |
115694 | 24.02.2013 | 1 | 50 | 3880 | 1499.0 | 1.0 |
Aggregate data
Since the competition task is to make a monthly prediction, we need to aggregate the data to montly level before doing any encodings. The following code-cell serves just that purpose.
1 | index_cols = ['shop_id', 'item_id', 'date_block_num'] |
/opt/conda/lib/python3.6/site-packages/pandas/core/groupby.py:4036: FutureWarning: using a dict with renaming is deprecated and will be removed in a future version
return super(DataFrameGroupBy, self).aggregate(arg, *args, **kwargs)
1 | all_data.head(10) |
shop_id | item_id | date_block_num | target | |
---|---|---|---|---|
139255 | 0 | 19 | 0 | 0.0 |
141495 | 0 | 27 | 0 | 0.0 |
144968 | 0 | 28 | 0 | 0.0 |
142661 | 0 | 29 | 0 | 0.0 |
138947 | 0 | 32 | 0 | 6.0 |
138948 | 0 | 33 | 0 | 3.0 |
138949 | 0 | 34 | 0 | 0.0 |
139247 | 0 | 35 | 0 | 1.0 |
142672 | 0 | 40 | 0 | 0.0 |
142065 | 0 | 41 | 0 | 0.0 |
Mean encodings without regularization
After we did the techinical work, we are ready to actually mean encode the desired item_id
variable.
Here are two ways to implement mean encoding features without any regularization. You can use this code as a starting point to implement regularized techniques.
Method 1
1 | # Calculate a mapping: {item_id: target_mean} |
0.483038698862
1 | all_data[all_data["item_id"] == 19].head() |
shop_id | item_id | date_block_num | target | item_target_enc | |
---|---|---|---|---|---|
139255 | 0 | 19 | 0 | 0.0 | 0.022222 |
147370 | 1 | 19 | 0 | 0.0 | 0.022222 |
114910 | 2 | 19 | 0 | 0.0 | 0.022222 |
123025 | 3 | 19 | 0 | 0.0 | 0.022222 |
98680 | 4 | 19 | 0 | 0.0 | 0.022222 |
Method 2
1 | ''' |
0.483038698862
See the printed value? It is the correlation coefficient between the target variable and your new encoded feature. You need to compute correlation coefficient between the encodings, that you will implement and submit those to coursera.
1 | grader = Grader() |
1. KFold scheme
Explained starting at 41 sec of Regularization video.
Now it’s your turn to write the code!
You may use ‘Regularization’ video as a reference for all further tasks.
First, implement KFold scheme with five folds. Use KFold(5) from sklearn.model_selection.
- Split your data in 5 folds with
sklearn.model_selection.KFold
withshuffle=False
argument. Iterate through folds: use all but the current fold to calculate mean target for each level
item_id
, and fill the current fold.- See the Method 1 from the example implementation. In particular learn what
map
and pd.Series.map functions do. They are pretty handy in many situations.
- See the Method 1 from the example implementation. In particular learn what
1 | # YOUR CODE GOES HERE |
0.41645907128
Current answer for task KFold_scheme is: 0.41645907128
2. Leave-one-out scheme
Now, implement leave-one-out scheme. Note that if you just simply set the number of folds to the number of samples and run the code from the KFold scheme, you will probably wait for a very long time.
To implement a faster version, note, that to calculate mean target value using all the objects but one given object, you can:
- Calculate sum of the target values using all the objects.
- Then subtract the target of the given object and divide the resulting value by
n_objects - 1
.
Note that you do not need to perform 1.
for every object. And 2.
can be implemented without any for
loop.
It is the most convenient to use .transform
function as in Method 2.
1 | # YOUR CODE GOES HERE |
0.480384831129
Current answer for task Leave-one-out_scheme is: 0.480384831129
3. Smoothing
Explained starting at 4:03 of Regularization video.
Next, implement smoothing scheme with $\alpha = 100$. Use the formula from the first slide in the video and $0.3343$ as globalmean
. Note that nrows
is the number of objects that belong to a certain category (not the number of rows in the dataset).
1 | # YOUR CODE GOES HERE |
0.48181987971
Current answer for task Smoothing_scheme is: 0.48181987971
4. Expanding mean scheme
Explained starting at 5:50 of Regularization video.
Finally, implement the expanding mean scheme. It is basically already implemented for you in the video, but you can challenge yourself and try to implement it yourself. You will need cumsum
and cumcount
functions from pandas.
1 | # YOUR CODE GOES HERE |
0.502524521108
Current answer for task Expanding_mean_scheme is: 0.502524521108
Authorization & Submission
To submit assignment parts to Cousera platform, please, enter your e-mail and token into variables below. You can generate token on this programming assignment page. Note: Token expires 30 minutes after generation.
1 | STUDENT_EMAIL = "lvduzhen@gmail.com" |
You want to submit these numbers:
Task KFold_scheme: 0.41645907128
Task Leave-one-out_scheme: 0.480384831129
Task Smoothing_scheme: 0.48181987971
Task Expanding_mean_scheme: 0.502524521108
1 | grader.submit(STUDENT_EMAIL, STUDENT_TOKEN) |
Submitted to Coursera platform. See results on assignment page!
1 |