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Pairwise ranking using scikit-learn LinearSVC
Implementation of pairwise ranking using scikit-learn LinearSVC
Reference: "Large Margin Rank Boundaries for Ordinal Regression", R. Herbrich,
T. Graepel, K. Obermayer.
Authors: Fabian Pedregosa <>
Alexandre Gramfort <>
import itertools
import numpy as np
from sklearn import svm, linear_model, cross_validation
def transform_pairwise(X, y):
"""Transforms data into pairs with balanced labels for ranking
Transforms a n-class ranking problem into a two-class classification
problem. Subclasses implementing particular strategies for choosing
pairs should override this method.
In this method, all pairs are choosen, except for those that have the
same target value. The output is an array of balanced classes, i.e.
there are the same number of -1 as +1
X : array, shape (n_samples, n_features)
The data
y : array, shape (n_samples,) or (n_samples, 2)
Target labels. If it's a 2D array, the second column represents
the grouping of samples, i.e., samples with different groups will
not be considered.
X_trans : array, shape (k, n_feaures)
Data as pairs
y_trans : array, shape (k,)
Output class labels, where classes have values {-1, +1}
X_new = []
y_new = []
y = np.asarray(y)
if y.ndim == 1:
y = np.c_[y, np.ones(y.shape[0])]
comb = itertools.combinations(range(X.shape[0]), 2)
for k, (i, j) in enumerate(comb):
if y[i, 0] == y[j, 0] or y[i, 1] != y[j, 1]:
# skip if same target or different group
X_new.append(X[i] - X[j])
y_new.append(np.sign(y[i, 0] - y[j, 0]))
# output balanced classes
if y_new[-1] != (-1) ** k:
y_new[-1] = - y_new[-1]
X_new[-1] = - X_new[-1]
return np.asarray(X_new), np.asarray(y_new).ravel()
class RankSVM(svm.LinearSVC):
"""Performs pairwise ranking with an underlying LinearSVC model
Input should be a n-class ranking problem, this object will convert it
into a two-class classification problem, a setting known as
`pairwise ranking`.
See object :ref:`svm.LinearSVC` for a full description of parameters.
def fit(self, X, y):
Fit a pairwise ranking model.
X : array, shape (n_samples, n_features)
y : array, shape (n_samples,) or (n_samples, 2)
X_trans, y_trans = transform_pairwise(X, y)
super(RankSVM, self).fit(X_trans, y_trans)
return self
def predict(self, X):
Predict an ordering on X. For a list of n samples, this method
returns a list from 0 to n-1 with the relative order of the rows of X.
X : array, shape (n_samples, n_features)
ord : array, shape (n_samples,)
Returns a list of integers representing the relative order of
the rows in X.
if hasattr(self, 'coef_'):
np.argsort(, self.coef_.T))
raise ValueError("Must call fit() prior to predict()")
def score(self, X, y):
Because we transformed into a pairwise problem, chance level is at 0.5
X_trans, y_trans = transform_pairwise(X, y)
return np.mean(super(RankSVM, self).predict(X_trans) == y_trans)
if __name__ == '__main__':
# as showcase, we will create some non-linear data
# and print the performance of ranking vs linear regression
n_samples, n_features = 300, 5
true_coef = np.random.randn(n_features)
X = np.random.randn(n_samples, n_features)
noise = np.random.randn(n_samples) / np.linalg.norm(true_coef)
y =, true_coef)
y = np.arctan(y) # add non-linearities
y += .1 * noise # add noise
Y = np.c_[y, np.mod(np.arange(n_samples), 5)] # add query fake id
cv = cross_validation.KFold(n_samples, 5)
train, test = iter(cv).next()
# make a simple plot out of it
import pylab as pl
pl.scatter(, true_coef), y)
pl.title('Data to be learned')
pl.xlabel('<X, coef>')
# print the performance of ranking
rank_svm = RankSVM().fit(X[train], Y[train])
print 'Performance of ranking ', rank_svm.score(X[test], Y[test])
# and that of linear regression
ridge = linear_model.RidgeCV(fit_intercept=True)[train], y[train])
X_test_trans, y_test_trans = transform_pairwise(X[test], y[test])
score = np.mean(np.sign(, ridge.coef_)) == y_test_trans)
print 'Performance of linear regression ', score

kigawas commented Apr 8, 2016

Hey, just found a bug.
np.argsort(, self.coef_.T)) should be return np.argsort(, self.coef_.T).ravel())

suzanv commented Apr 16, 2016

Thank you for this code!
I have one question/comment: you transform the test data with the same function as the training data. In this function, you don't create pairs of two items that have the same target value. I wonder: wouldn't it be more correct to assume that in the test data, you don't know the target values? I would expect that for the test data, we create pairs of all items regardless of their target value. Or perhaps I misunderstood something? Thanks for answering my question!

@suzanv I was wondering about this as well.

If you think about this being used in production/an application, the predict() method returns the indices of the data that's passed to it in ordinal fashion, and therefore (I think) you can only submit observations that are in the same group. In other words, although the model can be fit using data from multiple groups (multiple query_id's), when using predict() in production, you should only submit data to it that's of the same group (...the same query_id).

@suzanv (and @agramfort) do you have the same understanding as me?

If so, then, since all observations should have a unique ordinal position, there won't be any case where there is more than one observation that has the same target_value (when using predict() in production), and so in response to @suzanv comment, there should be a parameter added to fit() that allows you to create pairs for all observations, and predict() can use this paramter.

Curious to know what you all think.

suzanv commented Apr 18, 2016

@brendanf22 Thank you for your reply!

Yes, I agree that the test data should also be grouped (e.g. per query id).
About the target values, I indeed think that there should be a parameter that allows you to create pairs for all items in the test data, and not using the target labels. However, I tried this by making a copy of the function transform_pairwise function in which I replaced
if y[i, 0] == y[j, 0] or y[i, 1] != y[j, 1]:
if y[i, 1] != y[j, 1]:
(so, removed the requirement that target values should be different)
And the rankingSVM performance for my data drops from 0.78 to 0.60. Can you replicate this? Why does this happen?

Another recommendation is to add a function that outputs a ranking for the test data based on the predicted pairwise preferences (e.g. based on greedy ordering ), because many evaluation measures use the resulting ranking (e.g. precision@n, MAP, nDCG), and not so much the accuracy of pairwise predictions (-1 or 1)

suzanv commented Apr 18, 2016

Sorry, I have another question/comment:
About cross validation: it seems to me that you randomly create 5 partitions of the data, without keeping the items that belong to the same group (query) together in one partition. Do I understand that correctly?

For correct cross validation in a real ranking scenario, you should divide the groups (queries) in partitions, not the individual items (documents). You train on one set of queries, and evaluate on another set of queries. Instead of having a subset of the documents for a given query in each partition.

Hi, can you tell an example of the dataset that should be used here

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