[MRG] Batch/Lazy Log Sinkhorn Knopp on samples (#259)

* Add batch implementation of Sinkhorn

* Reformat to pep8 and modify parameter

* Fix error in batch size

* Code review and add test

* Fix accidental typo in test_empirical_sinkhorn

* Remove whitespace

* Edit config.yml

Author: 6Ulm

.circleci/config.yml

@@ -73,6 +73,7 @@ jobs:
             command: |
               cd docs;
               make html;
+            no_output_timeout: 30m
 
         # Save the outputs
         - store_artifacts:

ot/bregman.py

@@ -11,13 +11,15 @@
 #         Mokhtar Z. Alaya <mokhtarzahdi.alaya@gmail.com>
 #         Alexander Tong <alexander.tong@yale.edu>
 #         Ievgen Redko <ievgen.redko@univ-st-etienne.fr>
+#         Quang Huy Tran <quang-huy.tran@univ-ubs.fr>
 #
 # License: MIT License
 
 import warnings
 
 import numpy as np
 from scipy.optimize import fmin_l_bfgs_b
+from scipy.special import logsumexp
 
 from ot.utils import unif, dist, list_to_array
 from .backend import get_backend
@@ -1684,7 +1686,7 @@ def jcpot_barycenter(Xs, Ys, Xt, reg, metric='sqeuclidean', numItermax=100,
 
 
 def empirical_sinkhorn(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean',
-                       numIterMax=10000, stopThr=1e-9, verbose=False,
+                       numIterMax=10000, stopThr=1e-9, isLazy=False, batchSize=100, verbose=False,
                        log=False, **kwargs):
     r'''
     Solve the entropic regularization optimal transport problem and return the
@@ -1723,6 +1725,12 @@ def empirical_sinkhorn(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean',
         Max number of iterations
     stopThr : float, optional
         Stop threshol on error (>0)
+    isLazy: boolean, optional
+        If True, then only calculate the cost matrix by block and return the dual potentials only (to save memory)
+        If False, calculate full cost matrix and return outputs of sinkhorn function.
+    batchSize: int or tuple of 2 int, optional
+        Size of the batcheses used to compute the sinkhorn update without memory overhead.
+        When a tuple is provided it sets the size of the left/right batches.
     verbose : bool, optional
         Print information along iterations
     log : bool, optional
@@ -1758,24 +1766,78 @@ def empirical_sinkhorn(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean',
 
     .. [10] Chizat, L., Peyré, G., Schmitzer, B., & Vialard, F. X. (2016). Scaling algorithms for unbalanced transport problems. arXiv preprint arXiv:1607.05816.
     '''
-
+    ns, nt = X_s.shape[0], X_t.shape[0]
     if a is None:
-        a = unif(np.shape(X_s)[0])
+        a = unif(ns)
     if b is None:
-        b = unif(np.shape(X_t)[0])
+        b = unif(nt)
+
+    if isLazy:
+        if log:
+            dict_log = {"err": []}
 
-    M = dist(X_s, X_t, metric=metric)
+        log_a, log_b = np.log(a), np.log(b)
+        f, g = np.zeros(ns), np.zeros(nt)
+
+        if isinstance(batchSize, int):
+            bs, bt = batchSize, batchSize
+        elif isinstance(batchSize, tuple) and len(batchSize) == 2:
+            bs, bt = batchSize[0], batchSize[1]
+        else:
+            raise ValueError("Batch size must be in integer or a tuple of two integers")
+
+        range_s, range_t = range(0, ns, bs), range(0, nt, bt)
+
+        lse_f = np.zeros(ns)
+        lse_g = np.zeros(nt)
+
+        for i_ot in range(numIterMax):
+
+            for i in range_s:
+                M = dist(X_s[i:i + bs, :], X_t, metric=metric)
+                lse_f[i:i + bs] = logsumexp(g[None, :] - M / reg, axis=1)
+            f = log_a - lse_f
+
+            for j in range_t:
+                M = dist(X_s, X_t[j:j + bt, :], metric=metric)
+                lse_g[j:j + bt] = logsumexp(f[:, None] - M / reg, axis=0)
+            g = log_b - lse_g
+
+            if (i_ot + 1) % 10 == 0:
+                m1 = np.zeros_like(a)
+                for i in range_s:
+                    M = dist(X_s[i:i + bs, :], X_t, metric=metric)
+                    m1[i:i + bs] = np.exp(f[i:i + bs, None] + g[None, :] - M / reg).sum(1)
+                err = np.abs(m1 - a).sum()
+                if log:
+                    dict_log["err"].append(err)
+
+                if verbose and (i_ot + 1) % 100 == 0:
+                    print("Error in marginal at iteration {} = {}".format(i_ot + 1, err))
+
+                if err <= stopThr:
+                    break
+
+        if log:
+            dict_log["u"] = f
+            dict_log["v"] = g
+            return (f, g, dict_log)
+        else:
+            return (f, g)
 
-    if log:
-        pi, log = sinkhorn(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=True, **kwargs)
-        return pi, log
     else:
-        pi = sinkhorn(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=False, **kwargs)
-        return pi
+        M = dist(X_s, X_t, metric=metric)
+
+        if log:
+            pi, log = sinkhorn(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=True, **kwargs)
+            return pi, log
+        else:
+            pi = sinkhorn(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=False, **kwargs)
+            return pi
 
 
 def empirical_sinkhorn2(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean', numIterMax=10000, stopThr=1e-9,
-                        verbose=False, log=False, **kwargs):
+                        isLazy=False, batchSize=100, verbose=False, log=False, **kwargs):
     r'''
     Solve the entropic regularization optimal transport problem from empirical
     data and return the OT loss
@@ -1814,6 +1876,12 @@ def empirical_sinkhorn2(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean', num
         Max number of iterations
     stopThr : float, optional
         Stop threshol on error (>0)
+    isLazy: boolean, optional
+        If True, then only calculate the cost matrix by block and return the dual potentials only (to save memory)
+        If False, calculate full cost matrix and return outputs of sinkhorn function.
+    batchSize: int or tuple of 2 int, optional
+        Size of the batcheses used to compute the sinkhorn update without memory overhead.
+        When a tuple is provided it sets the size of the left/right batches.
     verbose : bool, optional
         Print information along iterations
     log : bool, optional
@@ -1850,21 +1918,45 @@ def empirical_sinkhorn2(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean', num
     .. [10] Chizat, L., Peyré, G., Schmitzer, B., & Vialard, F. X. (2016). Scaling algorithms for unbalanced transport problems. arXiv preprint arXiv:1607.05816.
     '''
 
+    ns, nt = X_s.shape[0], X_t.shape[0]
     if a is None:
-        a = unif(np.shape(X_s)[0])
+        a = unif(ns)
     if b is None:
-        b = unif(np.shape(X_t)[0])
+        b = unif(nt)
 
-    M = dist(X_s, X_t, metric=metric)
+    if isLazy:
+        if log:
+            f, g, dict_log = empirical_sinkhorn(X_s, X_t, reg, a, b, metric, numIterMax=numIterMax, stopThr=stopThr,
+                                                isLazy=isLazy, batchSize=batchSize, verbose=verbose, log=log)
+        else:
+            f, g = empirical_sinkhorn(X_s, X_t, reg, a, b, metric, numIterMax=numIterMax, stopThr=stopThr,
+                                      isLazy=isLazy, batchSize=batchSize, verbose=verbose, log=log)
+
+        bs = batchSize if isinstance(batchSize, int) else batchSize[0]
+        range_s = range(0, ns, bs)
+
+        loss = 0
+        for i in range_s:
+            M_block = dist(X_s[i:i + bs, :], X_t, metric=metric)
+            pi_block = np.exp(f[i:i + bs, None] + g[None, :] - M_block / reg)
+            loss += np.sum(M_block * pi_block)
+
+        if log:
+            return loss, dict_log
+        else:
+            return loss
 
-    if log:
-        sinkhorn_loss, log = sinkhorn2(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=log,
-                                       **kwargs)
-        return sinkhorn_loss, log
     else:
-        sinkhorn_loss = sinkhorn2(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=log,
-                                  **kwargs)
-        return sinkhorn_loss
+        M = dist(X_s, X_t, metric=metric)
+
+        if log:
+            sinkhorn_loss, log = sinkhorn2(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=log,
+                                           **kwargs)
+            return sinkhorn_loss, log
+        else:
+            sinkhorn_loss = sinkhorn2(a, b, M, reg, numItermax=numIterMax, stopThr=stopThr, verbose=verbose, log=log,
+                                      **kwargs)
+            return sinkhorn_loss
 
 
 def empirical_sinkhorn_divergence(X_s, X_t, reg, a=None, b=None, metric='sqeuclidean', numIterMax=10000, stopThr=1e-9,

test/test_bregman.py

@@ -2,6 +2,7 @@
 
 # Author: Remi Flamary <remi.flamary@unice.fr>
 #         Kilian Fatras <kilian.fatras@irisa.fr>
+#         Quang Huy Tran <quang-huy.tran@univ-ubs.fr>
 #
 # License: MIT License
 
@@ -329,6 +330,49 @@ def test_empirical_sinkhorn():
     np.testing.assert_allclose(loss_emp_sinkhorn, loss_sinkhorn, atol=1e-05)
 
 
+def test_lazy_empirical_sinkhorn():
+    # test sinkhorn
+    n = 100
+    a = ot.unif(n)
+    b = ot.unif(n)
+    numIterMax = 1000
+
+    X_s = np.reshape(np.arange(n), (n, 1))
+    X_t = np.reshape(np.arange(0, n), (n, 1))
+    M = ot.dist(X_s, X_t)
+    M_m = ot.dist(X_s, X_t, metric='minkowski')
+
+    f, g = ot.bregman.empirical_sinkhorn(X_s, X_t, 1, numIterMax=numIterMax, isLazy=True, batchSize=(1, 1), verbose=True)
+    G_sqe = np.exp(f[:, None] + g[None, :] - M / 1)
+    sinkhorn_sqe = ot.sinkhorn(a, b, M, 1)
+
+    f, g, log_es = ot.bregman.empirical_sinkhorn(X_s, X_t, 0.1, numIterMax=numIterMax, isLazy=True, batchSize=1, log=True)
+    G_log = np.exp(f[:, None] + g[None, :] - M / 0.1)
+    sinkhorn_log, log_s = ot.sinkhorn(a, b, M, 0.1, log=True)
+
+    f, g = ot.bregman.empirical_sinkhorn(X_s, X_t, 1, metric='minkowski', numIterMax=numIterMax, isLazy=True, batchSize=1)
+    G_m = np.exp(f[:, None] + g[None, :] - M_m / 1)
+    sinkhorn_m = ot.sinkhorn(a, b, M_m, 1)
+
+    loss_emp_sinkhorn, log = ot.bregman.empirical_sinkhorn2(X_s, X_t, 1, numIterMax=numIterMax, isLazy=True, batchSize=1, log=True)
+    loss_sinkhorn = ot.sinkhorn2(a, b, M, 1)
+
+    # check constratints
+    np.testing.assert_allclose(
+        sinkhorn_sqe.sum(1), G_sqe.sum(1), atol=1e-05)  # metric sqeuclidian
+    np.testing.assert_allclose(
+        sinkhorn_sqe.sum(0), G_sqe.sum(0), atol=1e-05)  # metric sqeuclidian
+    np.testing.assert_allclose(
+        sinkhorn_log.sum(1), G_log.sum(1), atol=1e-05)  # log
+    np.testing.assert_allclose(
+        sinkhorn_log.sum(0), G_log.sum(0), atol=1e-05)  # log
+    np.testing.assert_allclose(
+        sinkhorn_m.sum(1), G_m.sum(1), atol=1e-05)  # metric euclidian
+    np.testing.assert_allclose(
+        sinkhorn_m.sum(0), G_m.sum(0), atol=1e-05)  # metric euclidian
+    np.testing.assert_allclose(loss_emp_sinkhorn, loss_sinkhorn, atol=1e-05)
+
+
 def test_empirical_sinkhorn_divergence():
     # Test sinkhorn divergence
     n = 10