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I'm provided with the following dataset: Dataset. I'm meant to use sklearn to create a Support Vector Machine that can predict it.

I load A and B from my dataset into a 2 dimensional array called input_data and load the label from my dataset into an array called label.

First of all I'm scaling A and B to fit in the range of -1 to 1 using sklearn.preprocessing.MinMaxScaler.

input_data = np.array(input_data)
minmax = pre.MinMaxScaler(feature_range=(-1,1))
input_data = minmax.fit(input_data).transform(input_data)
label = np.array(label)

Then I use sklearn.model_selection.train_test_split to stratify and split my data keeping 40% for my test set and 60% for my training set.

model_selector = model_selection
X_train, X_test, y_train, y_test = model_selector.train_test_split(input_data, label, stratify=label, test_size=0.4)

After that I use sklearn.model_selection.GridSearchCV to try and come up with the optimal parameters (in my case only C).

gscv = model_selector.GridSearchCV
clf = gscv(svm.SVC(), tuned_parameters, cv=5, scoring='precision')
clf.fit(X_train, y_train)

After that I evaluate my results, but my Support Vector Machine just outputs 1 no matter what the input is, I'm not sure what exactly I'm doing wrong so after trying to figure things out by reading the documentation for a few hours I decided to ask for help, can anyone point out to me what I'm doing wrong? I'm pretty new in this area so my mistake is probably something simple I missed. Thanks in advance for any advice!

Here's an image of the dataset using a scatter plot: enter image description here

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  • $\begingroup$ Not sure about this @patrickdamery but you might want to try changing your scoring metric to ´accuracy´ from ´precision´ in your ´GridSearchCV´ function. Maybe that helps. $\endgroup$
    – Daneel Olivaw
    Mar 29, 2017 at 9:30

1 Answer 1

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What is tuned_parameters?

If I train SVC with default parameters on your dataset, it works fine with 61% accuracy, predicting both classes.

model.predict(X_test) with the model trained with your parameters outputs both 0 and 1 for me, with 98% accuracy:

model = svm.SVC(kernel = 'rbf', C=10, gamma=10)
model.fit(X_train, y_train)
print(model.predict(X_test))
print(model.score(X_test, y_test))

So the question is, how do you check what the model outputs on your test set? You may have a mistake there.

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  • $\begingroup$ The labels where already in the range of 0, 1 so I didn't think I needed to scale it with minmax, am I wrong? You can see the dataset in the link I provided. I've updated the question with a scatter plot of the dataset. If I set SVC to use the parameters: {'kernel': 'rbf', 'C': 10, 'gamma': 10} I get 95% accuracy on the training set and 92% on the test set, however I also need to present the svc using the linear kernel. Should the linear kernel not output 0 at least sometimes when using the linear kernel? $\endgroup$
    – patrickdamery
    Mar 29, 2017 at 8:04
  • $\begingroup$ See my edited answer. $\endgroup$
    – rinspy
    Mar 29, 2017 at 8:17
  • $\begingroup$ tuned_parameters is an array with an object that contains a variety of parameter information like this: tuned_parameters = [{'kernel': ['rbf'], 'gamma': [0.1, 0.5, 1, 3, 6, 10], 'C': [0.1, 0.5, 1, 3, 6, 10]},{'kernel': ['linear'], 'C': [0.1, 0.5, 1, 5, 10, 50, 100]}]. I know the rbf kernel works, but the linear one just returns 0 in my case, is this normal? I would have expected at least a few 0's even if it wasn't that accurate am I wrong? $\endgroup$
    – patrickdamery
    Mar 30, 2017 at 2:47
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    $\begingroup$ With a linear kernel, you are trying to linearly separate your data (which is not linearly separable). Since you have more 1s in your sample than 0s, I suspect the best linear model you can do is just predict 1 all the time. $\endgroup$
    – rinspy
    Mar 30, 2017 at 7:24
  • $\begingroup$ You are indeed correct, it turned out predicting only 1s was the best possible linear model! $\endgroup$
    – patrickdamery
    Mar 31, 2017 at 7:12

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