# coding: utf-8 """ ============================================= Introduction to Optimal Transport with Python ============================================= This example gives an introduction on how to use Optimal Transport in Python. """ # Author: Remi Flamary, Nicolas Courty, Aurelie Boisbunon # # License: MIT License # sphinx_gallery_thumbnail_number = 1 ############################################################################## # POT Python Optimal Transport Toolbox # ------------------------------------ # # POT installation # ``````````````````` # # * Install with pip:: # # pip install pot # * Install with conda:: # # conda install -c conda-forge pot # # Import the toolbox # ``````````````````` # import numpy as np # always need it import pylab as pl # do the plots import ot # ot import time ############################################################################## # Getting help # ````````````` # # Online documentation : ``_ # # Or inline help: # help(ot.dist) ############################################################################## # First OT Problem # ---------------- # # We will solve the Bakery/Cafés problem of transporting croissants from a # number of Bakeries to Cafés in a City (in this case Manhattan). We did a # quick google map search in Manhattan for bakeries and Cafés: # # .. image:: ../_static/images/bak.png # :align: center # :alt: bakery-cafe-manhattan # :width: 600px # :height: 280px # # We extracted from this search their positions and generated fictional # production and sale number (that both sum to the same value). # # We have acess to the position of Bakeries ``bakery_pos`` and their # respective production ``bakery_prod`` which describe the source # distribution. The Cafés where the croissants are sold are defined also by # their position ``cafe_pos`` and ``cafe_prod``, and describe the target # distribution. For fun we also provide a # map ``Imap`` that will illustrate the position of these shops in the city. # # # Now we load the data # # data = np.load('../data/manhattan.npz') bakery_pos = data['bakery_pos'] bakery_prod = data['bakery_prod'] cafe_pos = data['cafe_pos'] cafe_prod = data['cafe_prod'] Imap = data['Imap'] print('Bakery production: {}'.format(bakery_prod)) print('Cafe sale: {}'.format(cafe_prod)) print('Total croissants : {}'.format(cafe_prod.sum())) ############################################################################## # Plotting bakeries in the city # ----------------------------- # # Next we plot the position of the bakeries and cafés on the map. The size of # the circle is proportional to their production. # pl.figure(1, (7, 6)) pl.clf() pl.imshow(Imap, interpolation='bilinear') # plot the map pl.scatter(bakery_pos[:, 0], bakery_pos[:, 1], s=bakery_prod, c='r', ec='k', label='Bakeries') pl.scatter(cafe_pos[:, 0], cafe_pos[:, 1], s=cafe_prod, c='b', ec='k', label='Cafés') pl.legend() pl.title('Manhattan Bakeries and Cafés') ############################################################################## # Cost matrix # ----------- # # # We can now compute the cost matrix between the bakeries and the cafés, which # will be the transport cost matrix. This can be done using the # `ot.dist `_ function that # defaults to squared Euclidean distance but can return other things such as # cityblock (or Manhattan distance). # C = ot.dist(bakery_pos, cafe_pos) labels = [str(i) for i in range(len(bakery_prod))] f = pl.figure(2, (14, 7)) pl.clf() pl.subplot(121) pl.imshow(Imap, interpolation='bilinear') # plot the map for i in range(len(cafe_pos)): pl.text(cafe_pos[i, 0], cafe_pos[i, 1], labels[i], color='b', fontsize=14, fontweight='bold', ha='center', va='center') for i in range(len(bakery_pos)): pl.text(bakery_pos[i, 0], bakery_pos[i, 1], labels[i], color='r', fontsize=14, fontweight='bold', ha='center', va='center') pl.title('Manhattan Bakeries and Cafés') ax = pl.subplot(122) im = pl.imshow(C, cmap="coolwarm") pl.title('Cost matrix') cbar = pl.colorbar(im, ax=ax, shrink=0.5, use_gridspec=True) cbar.ax.set_ylabel("cost", rotation=-90, va="bottom") pl.xlabel('Cafés') pl.ylabel('Bakeries') pl.tight_layout() ############################################################################## # The red cells in the matrix image show the bakeries and cafés that are # further away, and thus more costly to transport from one to the other, while # the blue ones show those that are very close to each other, with respect to # the squared Euclidean distance. ############################################################################## # Solving the OT problem with `ot.emd `_ # ----------------------------------------------------------------------------------- start = time.time() ot_emd = ot.emd(bakery_prod, cafe_prod, C) time_emd = time.time() - start ############################################################################## # The function returns the transport matrix, which we can then visualize (next section). ############################################################################## # Transportation plan vizualization # ````````````````````````````````` # # A good vizualization of the OT matrix in the 2D plane is to denote the # transportation of mass between a Bakery and a Café by a line. This can easily # be done with a double ``for`` loop. # # In order to make it more interpretable one can also use the ``alpha`` # parameter of plot and set it to ``alpha=G[i,j]/G.max()``. # Plot the matrix and the map f = pl.figure(3, (14, 7)) pl.clf() pl.subplot(121) pl.imshow(Imap, interpolation='bilinear') # plot the map for i in range(len(bakery_pos)): for j in range(len(cafe_pos)): pl.plot([bakery_pos[i, 0], cafe_pos[j, 0]], [bakery_pos[i, 1], cafe_pos[j, 1]], '-k', lw=3. * ot_emd[i, j] / ot_emd.max()) for i in range(len(cafe_pos)): pl.text(cafe_pos[i, 0], cafe_pos[i, 1], labels[i], color='b', fontsize=14, fontweight='bold', ha='center', va='center') for i in range(len(bakery_pos)): pl.text(bakery_pos[i, 0], bakery_pos[i, 1], labels[i], color='r', fontsize=14, fontweight='bold', ha='center', va='center') pl.title('Manhattan Bakeries and Cafés') ax = pl.subplot(122) im = pl.imshow(ot_emd) for i in range(len(bakery_prod)): for j in range(len(cafe_prod)): text = ax.text(j, i, '{0:g}'.format(ot_emd[i, j]), ha="center", va="center", color="w") pl.title('Transport matrix') pl.xlabel('Cafés') pl.ylabel('Bakeries') pl.tight_layout() ############################################################################## # The transport matrix gives the number of croissants that can be transported # from each bakery to each café. We can see that the bakeries only need to # transport croissants to one or two cafés, the transport matrix being very # sparse. ############################################################################## # OT loss and dual variables # -------------------------- # # The resulting wasserstein loss loss is of the form: # # .. math:: # W=\sum_{i,j}\gamma_{i,j}C_{i,j} # # where :math:`\gamma` is the optimal transport matrix. # W = np.sum(ot_emd * C) print('Wasserstein loss (EMD) = {0:.2f}'.format(W)) ############################################################################## # Regularized OT with Sinkhorn # ---------------------------- # # The Sinkhorn algorithm is very simple to code. You can implement it directly # using the following pseudo-code # # .. image:: ../_static/images/sinkhorn.png # :align: center # :alt: Sinkhorn algorithm # :width: 440px # :height: 240px # # In this algorithm, :math:`\oslash` corresponds to the element-wise division. # # An alternative is to use the POT toolbox with # `ot.sinkhorn `_ # # Be careful of numerical problems. A good pre-processing for Sinkhorn is to # divide the cost matrix ``C`` by its maximum value. ############################################################################## # Algorithm # ````````` # Compute Sinkhorn transport matrix from algorithm reg = 0.1 K = np.exp(-C / C.max() / reg) nit = 100 u = np.ones((len(bakery_prod), )) for i in range(1, nit): v = cafe_prod / np.dot(K.T, u) u = bakery_prod / (np.dot(K, v)) ot_sink_algo = np.atleast_2d(u).T * (K * v.T) # Equivalent to np.dot(np.diag(u), np.dot(K, np.diag(v))) # Compute Sinkhorn transport matrix with POT ot_sinkhorn = ot.sinkhorn(bakery_prod, cafe_prod, reg=reg, M=C / C.max()) # Difference between the 2 print('Difference between algo and ot.sinkhorn = {0:.2g}'.format(np.sum(np.power(ot_sink_algo - ot_sinkhorn, 2)))) ############################################################################## # Plot the matrix and the map # ``````````````````````````` print('Min. of Sinkhorn\'s transport matrix = {0:.2g}'.format(np.min(ot_sinkhorn))) f = pl.figure(4, (13, 6)) pl.clf() pl.subplot(121) pl.imshow(Imap, interpolation='bilinear') # plot the map for i in range(len(bakery_pos)): for j in range(len(cafe_pos)): pl.plot([bakery_pos[i, 0], cafe_pos[j, 0]], [bakery_pos[i, 1], cafe_pos[j, 1]], '-k', lw=3. * ot_sinkhorn[i, j] / ot_sinkhorn.max()) for i in range(len(cafe_pos)): pl.text(cafe_pos[i, 0], cafe_pos[i, 1], labels[i], color='b', fontsize=14, fontweight='bold', ha='center', va='center') for i in range(len(bakery_pos)): pl.text(bakery_pos[i, 0], bakery_pos[i, 1], labels[i], color='r', fontsize=14, fontweight='bold', ha='center', va='center') pl.title('Manhattan Bakeries and Cafés') ax = pl.subplot(122) im = pl.imshow(ot_sinkhorn) for i in range(len(bakery_prod)): for j in range(len(cafe_prod)): text = ax.text(j, i, np.round(ot_sinkhorn[i, j], 1), ha="center", va="center", color="w") pl.title('Transport matrix') pl.xlabel('Cafés') pl.ylabel('Bakeries') pl.tight_layout() ############################################################################## # We notice right away that the matrix is not sparse at all with Sinkhorn, # each bakery delivering croissants to all 5 cafés with that solution. Also, # this solution gives a transport with fractions, which does not make sense # in the case of croissants. This was not the case with EMD. ############################################################################## # Varying the regularization parameter in Sinkhorn # ```````````````````````````````````````````````` # reg_parameter = np.logspace(-3, 0, 20) W_sinkhorn_reg = np.zeros((len(reg_parameter), )) time_sinkhorn_reg = np.zeros((len(reg_parameter), )) f = pl.figure(5, (14, 5)) pl.clf() max_ot = 100 # plot matrices with the same colorbar for k in range(len(reg_parameter)): start = time.time() ot_sinkhorn = ot.sinkhorn(bakery_prod, cafe_prod, reg=reg_parameter[k], M=C / C.max()) time_sinkhorn_reg[k] = time.time() - start if k % 4 == 0 and k > 0: # we only plot a few ax = pl.subplot(1, 5, k // 4) im = pl.imshow(ot_sinkhorn, vmin=0, vmax=max_ot) pl.title('reg={0:.2g}'.format(reg_parameter[k])) pl.xlabel('Cafés') pl.ylabel('Bakeries') # Compute the Wasserstein loss for Sinkhorn, and compare with EMD W_sinkhorn_reg[k] = np.sum(ot_sinkhorn * C) pl.tight_layout() ############################################################################## # This series of graph shows that the solution of Sinkhorn starts with something # very similar to EMD (although not sparse) for very small values of the # regularization parameter, and tends to a more uniform solution as the # regularization parameter increases. # ############################################################################## # Wasserstein loss and computational time # ``````````````````````````````````````` # # Plot the matrix and the map f = pl.figure(6, (4, 4)) pl.clf() pl.title("Comparison between Sinkhorn and EMD") pl.plot(reg_parameter, W_sinkhorn_reg, 'o', label="Sinkhorn") XLim = pl.xlim() pl.plot(XLim, [W, W], '--k', label="EMD") pl.legend() pl.xlabel("reg") pl.ylabel("Wasserstein loss") ############################################################################## # In this last graph, we show the impact of the regularization parameter on # the Wasserstein loss. We can see that higher # values of ``reg`` leads to a much higher Wasserstein loss. # # The Wasserstein loss of EMD is displayed for # comparison. The Wasserstein loss of Sinkhorn can be a little lower than that # of EMD for low values of ``reg``, but it quickly gets much higher. #