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.
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import pandas as pd import numpy as np from itertools import product from grader import Grader
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.
# For every month we create a grid from all shops/items combinations from that month grid = [] for block_num in sales['date_block_num'].unique(): cur_shops = sales[sales['date_block_num']==block_num]['shop_id'].unique() cur_items = sales[sales['date_block_num']==block_num]['item_id'].unique() grid.append(np.array(list(product(*[cur_shops, cur_items, [block_num]])),dtype='int32'))
#turn the grid into pandas dataframe grid = pd.DataFrame(np.vstack(grid), columns = index_cols,dtype=np.int32)
#get aggregated values for (shop_id, item_id, month) gb = sales.groupby(index_cols,as_index=False).agg({'item_cnt_day':{'target':'sum'}})
#fix column names gb.columns = [col[0] if col[-1]==''else col[-1] for col in gb.columns.values] #join aggregated data to the grid all_data = pd.merge(grid,gb,how='left',on=index_cols).fillna(0) #sort the data all_data.sort_values(['date_block_num','shop_id','item_id'],inplace=True)
/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
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# Calculate a mapping: {item_id: target_mean} item_id_target_mean = all_data.groupby('item_id').target.mean()
# In our non-regularized case we just *map* the computed means to the `item_id`'s all_data['item_target_enc'] = all_data['item_id'].map(item_id_target_mean)
# Fill NaNs all_data['item_target_enc'].fillna(0.3343, inplace=True)
''' Differently to `.target.mean()` function `transform` will return a dataframe with an index like in `all_data`. Basically this single line of code is equivalent to the first two lines from of Method 1. ''' all_data['item_target_enc'] = all_data.groupby('item_id')['target'].transform('mean')
# Fill NaNs all_data['item_target_enc'].fillna(0.3343, inplace=True)
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.
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 with shuffle=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.
# YOUR CODE GOES HERE from sklearn.model_selection import KFold kf = KFold(shuffle=False, n_splits = 5)
for train_index, test_index in kf.split(all_data): x_tr = all_data.iloc[train_index] mean = x_tr.groupby("item_id").target.mean() all_data.loc[all_data.index[test_index],'item_target_enc'] = all_data.loc[all_data.index[test_index],'item_id'].map(mean)
## fill with global mean #all_data['item_target_enc'].fillna(all_data['target'].mean(), inplace = True)
# You will need to compute correlation like that corr = np.corrcoef(all_data['target'].values, encoded_feature)[0][1] print(corr) grader.submit_tag('KFold_scheme', corr)
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.
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).
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.
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# YOUR CODE GOES HERE cumsum = all_data.groupby('item_id')['target'].cumsum() - all_data['target'] cumcnt = all_data.groupby('item_id').cumcount()
0.502524521108
Current answer for task Expanding_mean_scheme is: 0.502524521108
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