Speed-up Sinkhorn

Speed-up in 3 places:
 - the computation of pairwise distance is faster with sklearn.metrics.pairwise.euclidean_distances
 - faster computation of K = np.exp(-M / reg)
 - faster computation of the error every 10 iterations

Example with this little script:

import time
import numpy as np
import ot
rng = np.random.RandomState(0)
transport = ot.da.SinkhornTransport()
time1 = time.time()
Xs, ys, Xt = rng.randn(10000, 100), rng.randint(0, 2, size=10000), rng.randn(10000, 100)
transport.fit(Xs=Xs, Xt=Xt)
time2 = time.time()
print("OT Computation Time {:6.2f} sec".format(time2-time1))
transport = ot.da.SinkhornLpl1Transport()
transport.fit(Xs=Xs, ys=ys, Xt=Xt)
time3 = time.time()
print("OT LpL1 Computation Time {:6.2f} sec".format(time3-time2))

Before
OT Computation Time  19.93 sec
OT LpL1 Computation Time 133.43 sec

After
OT Computation Time   7.55 sec
OT LpL1 Computation Time  82.25 sec

1 files changed, 3 insertions, 1 deletions

diff --git a/ot/utils.py b/ot/utils.py
index 7dac283..5b052ac 100644
--- a/ot/utils.py
+++ b/ot/utils.py
@@ -13,6 +13,7 @@ import time
 
 import numpy as np
 from scipy.spatial.distance import cdist
+from sklearn.metrics.pairwise import euclidean_distances
 import sys
 import warnings
 try:
@@ -104,7 +105,8 @@ def dist(x1, x2=None, metric='sqeuclidean'):
     """
     if x2 is None:
         x2 = x1
-
+    if metric == "sqeuclidean":
+        return euclidean_distances(x1, x2, squared=True)
     return cdist(x1, x2, metric=metric)