Predict NA (missing values) with machine learning

Question

I have a huge data set and want to predict (not replace) missing values with a machine learning algorithm like svm or random forest in python.

My data set looks like this:

ID i0   i1    i2    i3    i4   i5     j0    j1   j2   j3    j4    j5    

0  0.19 -0.02 -0.20 0.07 -0.06 -0.06  -0.06 1.48 0.33 -0.46 -0.37 -0.11
1 -0.61 -0.19 -0.10 -0.1 -0.21  0.63   NA    NA   NA   NA    NA    NA
2 -0.31 -0.14 -0.64 -0.5 -0.20 -0.30  -0.08 1.56 -0.2 -0.33  0.81 -0.03
.
.

What I want to do:
On the basis of ID 0 and 2 I want to train the values of j0 to j5 with i0 to i5. Subsequent there should be a prediction of the NA's from j0-j5 for ID 1.

Question:
As the data is not continuous (the time steps end at i5 and start again at j0), is it possible to use some kind of regression?

How should the X and the y for the .fit(X, y) and .predict(X) function look like in this example?

Answer 1

In your case, you're looking at at a multi-output regression problem:

• A regression problem - as opposed to classification - since you are trying to predict a value and not a class/state variable/category
• Multi-output since you are trying to predict 6 values for each data point

You can read more in the sklearn documentation about multiclass.

Here I'm going to show you how you can use sklearn.multioutput.MultiOutputRegressor with a sklearn.ensemble.RandomForestRegressor to predict your values.

Construct some dummy data

from sklearn.datasets import make_regression

X,y = make_regression(n_samples=1000, n_features=6,
                                 n_informative=3, n_targets=6,  
                                 tail_strength=0.5, noise=0.02, 
                                 shuffle=True, coef=False, random_state=0)

# Convert to a pandas dataframe like in your example
icols = ['i0','i1','i2','i3','i4','i5']
jcols = ['j0', 'j1', 'j2', 'j3', 'j4', 'j5']
df = pd.concat([pd.DataFrame(X, columns=icols),
                pd.DataFrame(y, columns=jcols)], axis=1)

# Introduce a few np.nans in there
df.loc[0, jcols] = np.nan
df.loc[10, jcols] = np.nan
df.loc[100, jcols] = np.nan

df.head()

Out:
     i0    i1    i2    i3    i4    i5     j0     j1     j2     j3     j4  \
0 -0.21 -0.18 -0.06  0.27 -0.32  0.00    NaN    NaN    NaN    NaN    NaN   
1  0.65 -2.16  0.46  1.82  0.22 -0.13  33.08  39.85   9.63  13.52  16.72   
2 -0.75 -0.52 -1.08  0.14  1.12 -1.05  -0.96 -96.02  14.37  25.19 -44.90   
3  0.01  0.62  0.20  0.53  0.35 -0.73   6.09 -12.07 -28.88  10.49   0.96   
4  0.39 -0.70 -0.55  0.10  1.65 -0.69  83.15  -3.16  93.61  57.44 -17.33   

      j5  
0    NaN  
1  17.79  
2 -77.48  
3 -35.61  
4  -2.47  

Exclude the nans initially, and split into 75% train and 25% test

The split is done in order to be able to validate our model.

notnans = df[jcols].notnull().all(axis=1)
df_notnans = df[notnans]

# Split into 75% train and 25% test
X_train, X_test, y_train, y_test = train_test_split(df_notnans[icols], df_notnans[jcols],
                                                    train_size=0.75,
                                                    random_state=4)

Use a multi output regression based on a random forest regressor

from sklearn.ensemble import RandomForestRegressor
from sklearn.multioutput import MultiOutputRegressor
from sklearn.model_selection import train_test_split

regr_multirf = MultiOutputRegressor(RandomForestRegressor(max_depth=30,
                                                          random_state=0))

# Fit on the train data
regr_multirf.fit(X_train, y_train)

# Check the prediction score
score = regr_multirf.score(X_test, y_test)
print("The prediction score on the test data is {:.2f}%".format(score*100))

Out: The prediction score on the test data is 96.76%

Predict the nan rows

df_nans = df.loc[~notnans].copy()
df_nans[jcols] = regr_multirf.predict(df_nans[icols])
df_nans

Out:

           i0        i1        i2        i3        i4        i5         j0  \
0   -0.211620 -0.177927 -0.062205  0.267484 -0.317349  0.000341 -41.254983   
10   1.138974 -1.326378  0.123960  0.982841  0.273958  0.414307  46.406351   
100 -0.682390 -1.431414 -0.328235 -0.886463  1.212363 -0.577676  94.971966   

            j1         j2         j3         j4         j5  
0   -18.197513 -31.029952 -14.749244  -5.990595  -9.296744  
10   67.915628  59.750032  15.612843  10.177314  38.226387  
100  -3.724223  65.630692  44.636895 -14.372414  11.947185