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musicinformationretrieval-com
Commit 19bb5d91 authored by Steve Tjoa's avatar Steve Tjoa
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lcs; dtw

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dtw.html
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dtw.ipynb
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......@@ -2,16 +2,14 @@
"cells": [
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": false
},
"execution_count": 74,
"metadata": {},
"outputs": [],
"source": [
"%matplotlib inline\n",
"import seaborn\n",
"import numpy, scipy, scipy.spatial, matplotlib.pyplot as plt\n",
"plt.rcParams['figure.figsize'] = (14, 2)"
"plt.rcParams['figure.figsize'] = (14, 3)"
]
},
{
......@@ -32,16 +30,20 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"(Credit for some of the ideas in this notebook come from [FastDTW](https://github.com/slaypni/fastdtw/blob/master/fastdtw.py).)"
"In MIR, we often want to compare two sequences of different lengths. For example, we may want to compute a similarity measure between two versions of the same song. These two signals, $x$ and $y$, may have similar sequences of chord progressions and instrumentations, but there may be timing deviations between the two. Even if we were to express the two audio signals using the same feature space (e.g. chroma or MFCCs), we cannot simply sum their pairwise distances because the signals have different lengths.\n",
"\n",
"As another example, you might want to align two different performances of the same musical work, e.g. so you can hop from one performance to another at any moment in the work. This problem is known as **music synchronization** (FMP, p. 115)."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Dynamic time warping (DTW) is an algorithm used to align two sequences of similar content but possibly different lengths. For example, you might want to align two different performances of the same musical work. These two signals, $x$ and $y$, may have similar sequences of chord progressions and instrumentations, but there may be timing deviations between the two.\n",
"**Dynamic time warping (DTW)** ([Wikipedia](https://en.wikipedia.org/wiki/Dynamic_time_warping); FMP, p. 131) is an algorithm used to align two sequences of similar content but possibly different lengths. \n",
"\n",
"Given two sequences, $x[n], n \\in \\{0, ..., N_x - 1\\}$, and $y[n], n \\in \\{0, ..., N_y - 1\\}$, DTW produces a set of index pairs $\\{ (i, j) ... \\}$ such that $x[i]$ and $y[j]$ are similar."
"Given two sequences, $x[n], n \\in \\{0, ..., N_x - 1\\}$, and $y[n], n \\in \\{0, ..., N_y - 1\\}$, DTW produces a set of index coordinate pairs $\\{ (i, j) ... \\}$ such that $x[i]$ and $y[j]$ are similar.\n",
"\n",
"We will use the same approach described in the notebooks [Dynamic Programming](dp.html) and [Longest Common Subsequence](lcs.html)."
]
},
{
......@@ -60,50 +62,36 @@
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"8 9\n"
]
}
],
"execution_count": 123,
"metadata": {},
"outputs": [],
"source": [
"x = scipy.array([3, 3, 1, 4, 6, 1, 5, 5])\n",
"y = scipy.array([4, 2, 1, 3, 3, 4, 1, 4, 5])\n",
"Nx = len(x)\n",
"Ny = len(y)\n",
"print Nx, Ny"
"x = [0, 4, 4, 0, -4, -4, 0]\n",
"y = [1, 3, 4, 3, 1, -1, -2, -1, 0]\n",
"nx = len(x)\n",
"ny = len(y)"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {
"collapsed": false
},
"execution_count": 127,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[<matplotlib.lines.Line2D at 0x110817ad0>,\n",
" <matplotlib.lines.Line2D at 0x1140bced0>]"
"<matplotlib.legend.Legend at 0x11ce5d910>"
]
},
"execution_count": 3,
"execution_count": 127,
"metadata": {},
"output_type": "execute_result"
},
{
"data": {
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aUAiys21MmWIjEFB44AE/48f7YnacqORBf6fMQTl6mwQvuljbZ3J8SZcBe5tM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\nHCp/+5uPvn0DVdo2Ylm3Rus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bAPD06IXjoZ/gnPoASrsLN9R7FYUNXx1gw9cHsZoNPHpnHPF9gqcJ/3y09ksQ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\nREREtN4/u6cPBCJ5QIcAAAAASUVORK5CYII=\n",
"text/plain": [
"<matplotlib.figure.Figure at 0x110817790>"
"<matplotlib.figure.Figure at 0x11cd40410>"
]
},
"metadata": {},
......@@ -111,7 +99,9 @@
}
],
"source": [
"plt.plot(range(Nx), x, 'b', range(Ny), y, 'r')"
"plt.plot(x)\n",
"plt.plot(y, c='r')\n",
"plt.legend(('x', 'y'))"
]
},
{
......@@ -125,261 +115,247 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"But for now, we will add an empty dimension to the array because DTW expects a vector at each time index -- in this simple example, a \"one-dimensional vector\", i.e. a scalar."
"## Distance Metric"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"DTW requires the use of a distance metric between corresponding observations of `x` and `y`. One common choice is the **Euclidean distance** ([Wikipedia](https://en.wikipedia.org/wiki/Euclidean_distance); FMP, p. 454):"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": false
},
"execution_count": 128,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(8, 1)\n",
"(9, 1)\n"
]
"data": {
"text/plain": [
"5.0"
]
},
"execution_count": 128,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"x = scipy.expand_dims(x, axis=1)\n",
"y = scipy.expand_dims(y, axis=1)\n",
"print x.shape\n",
"print y.shape"
"scipy.spatial.distance.euclidean(0, [3, 4])"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {
"collapsed": false
},
"execution_count": 129,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[3]\n",
" [3]\n",
" [1]\n",
" [4]\n",
" [6]\n",
" [1]\n",
" [5]\n",
" [5]]\n"
]
"data": {
"text/plain": [
"13.0"
]
},
"execution_count": 129,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"print x"
"scipy.spatial.distance.euclidean([0, 0], [5, 12])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Step 1: Compute pairwise distances"
"Another choice might be the **cosine distance** ([Wikipedia](https://en.wikipedia.org/wiki/Cosine_similarity); FMP, p. 376) which can be interpreted as the (normalized) angle between two vectors:"
]
},
{
"cell_type": "markdown",
"cell_type": "code",
"execution_count": 134,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"0.0"
]
},
"execution_count": 134,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"Compute a matrix of pairwise distances between elements of $x$ and $y$."
"scipy.spatial.distance.cosine([1, 0], [100, 0])"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {
"collapsed": false
},
"execution_count": 132,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"(8, 9)\n"
]
"data": {
"text/plain": [
"1.0"
]
},
"execution_count": 132,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"S = scipy.spatial.distance_matrix(x, y)\n",
"print S.shape"
"scipy.spatial.distance.cosine([1, 0, 0], [0, 0, -1])"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": false
},
"execution_count": 133,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 1. 1. 2. 0. 0. 1. 2. 1. 2.]\n",
" [ 1. 1. 2. 0. 0. 1. 2. 1. 2.]\n",
" [ 3. 1. 0. 2. 2. 3. 0. 3. 4.]\n",
" [ 0. 2. 3. 1. 1. 0. 3. 0. 1.]\n",
" [ 2. 4. 5. 3. 3. 2. 5. 2. 1.]\n",
" [ 3. 1. 0. 2. 2. 3. 0. 3. 4.]\n",
" [ 1. 3. 4. 2. 2. 1. 4. 1. 0.]\n",
" [ 1. 3. 4. 2. 2. 1. 4. 1. 0.]]\n"
]
"data": {
"text/plain": [
"2.0"
]
},
"execution_count": 133,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"print S"
"scipy.spatial.distance.cosine([1, 0], [-1, 0])"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Element $S[i, j]$ indicates the distance between $x[i]$ and $y[j]$."
"## Step 1: Cost Table Construction"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The time complexity of this operation is $O(N_x N_y)$. The space complexity is $O(N_x N_y)$."
"As described in the notebooks [Dynamic Programming](dp.html) and [Longest Common Subsequence](lcs.html), we will use dynamic programming to solve this problem. First, we create a table which stores the solutions to all subproblems. Then, we will use this table to solve each larger subproblem until the problem is solved for the full original inputs."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## Step 2: Compute cumulative distances"
"The basic idea of DTW is to find a path of index coordinate pairs the sum of distances along the path $P$ is minimized:\n",
"\n",
"$$ \\min \\sum_{(i, j) \\in P} d(x[i], y[j]) $$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The basic idea of DTW is to find a path from the top left to the bottom right of matrix $S$ such that the sum of distances along the path is minimized. After all, you want to find pairs of time indices $(i, j)$ such that $x[i]$ and $y[j]$ is always low."
"The path constraint is that, at $(i, j)$, the valid steps are $(i+1, j)$, $(i, j+1)$, and $(i+1, j+1)$. In other words, the alignment always moves forward in time for at least one of the signals. It never goes forward in time for one signal and backward in time for the other signal."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We will compute a matrix of cumulative path distances, $D$, such that for any pair of time indices $(i, j)$, $D[i, j]$ is the sum of the best path from $(0, 0)$ to $(i, j)$."
"Here is the optimal substructure. Suppose that the best alignment contains index pair `(i, j)`, i.e., `x[i]` and `y[j]` are part of the optimal DTW path. Then, we prepend to the optimal path \n",
"\n",
"$$ \\mathrm{argmin} \\ \\{ d(x[i-1], y[j]), d(x[i], y[j-1]), d(x[i-1], j-1]) \\} $$"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"First, let's compute the sum of the best path from $(0, 0)$ to $(N_x-1, 0)$."
"We create a table where cell `(i, j)` stores the optimum cost of `dtw(x[:i], y[:j])`, i.e. the optimum cost from `(0, 0)` to `(i, j)`. First, we solve for the boundary cases, i.e. when either one of the two sequences is empty. Then we populate the table from the top left to the bottom right."
]
},
{
"cell_type": "code",
"execution_count": 8,
"execution_count": 135,
"metadata": {
"collapsed": false
"collapsed": true
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 1. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 2. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 5. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 5. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 7. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 10. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 11. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 12. 0. 0. 0. 0. 0. 0. 0. 0.]]\n"
]
}
],
"source": [
"D = scipy.zeros_like(S)\n",
"D[0, 0] = S[0, 0]\n",
"for i in range(1, Nx):\n",
" D[i, 0] = D[i-1, 0] + S[i, 0]\n",
"print D"
]
},
{
"cell_type": "markdown",
"metadata": {},
"outputs": [],
"source": [
"Next, compute the sum of the best path from $(0, 0)$ to $(0, N_y)$."
"def dtw_table(x, y):\n",
" nx = len(x)\n",
" ny = len(y)\n",
" table = [[0 for _ in range(ny+1)] for _ in range(nx+1)]\n",
" \n",
" # Compute left column separately, i.e. j=0.\n",
" for i in range(1, nx+1):\n",
" d = scipy.spatial.distance.euclidean(x[i-1], 0)\n",
" table[i][0] = d + table[i-1][0]\n",
" \n",
" # Compute top row separately, i.e. i=0.\n",
" for j in range(1, ny+1):\n",
" d = scipy.spatial.distance.euclidean(0, y[j-1])\n",
" table[0][j] = d + table[0][j-1]\n",
" \n",
" # Fill in the rest.\n",
" for i in range(1, nx+1):\n",
" for j in range(1, ny+1):\n",
" d = scipy.spatial.distance.euclidean(x[i-1], y[j-1])\n",
" table[i][j] = d + min(table[i-1][j], table[i][j-1], table[i-1][j-1])\n",
" return table"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 1. 2. 4. 4. 4. 5. 7. 8. 10.]\n",
" [ 2. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 5. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 5. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 7. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 10. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 11. 0. 0. 0. 0. 0. 0. 0. 0.]\n",
" [ 12. 0. 0. 0. 0. 0. 0. 0. 0.]]\n"
]
}
],
"execution_count": 136,
"metadata": {},
"outputs": [],
"source": [
"for j in range(1, len(y)):\n",
" D[0, j] = D[0, j-1] + S[0, j]\n",
"print D"
"table = dtw_table(x, y)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Finally, compute the sums of the best paths to any other pair of coordinates, $(i, j)$.\n",
"\n",
"The path constraint is that, at $(i, j)$, the valid steps are $(i+1, j)$, $(i, j+1)$, and $(i+1, j+1)$. In other words, the alignment always moves forward in time for at least one of the signals. It never goes backward in time. (You can adapt this to your application.)"
"Let's visualize this table:"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {
"collapsed": false
},
"execution_count": 142,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[[ 1. 2. 4. 4. 4. 5. 7. 8. 10.]\n",
" [ 2. 2. 4. 4. 4. 5. 7. 8. 10.]\n",
" [ 5. 3. 2. 4. 6. 7. 5. 8. 12.]\n",
" [ 5. 5. 5. 3. 4. 4. 7. 5. 6.]\n",
" [ 7. 9. 10. 6. 6. 6. 9. 7. 6.]\n",
" [ 10. 8. 8. 8. 8. 9. 6. 9. 10.]\n",
" [ 11. 11. 12. 10. 10. 9. 10. 7. 7.]\n",
" [ 12. 14. 15. 12. 12. 10. 13. 8. 7.]]\n"
" 1 3 4 3 1 -1 -2 -1 0\n",
" +----------------------------------------\n",
" | 0 1 4 8 11 12 13 15 16 16\n",
" 0 | 0 1 4 8 11 12 13 15 16 16\n",
" 4 | 4 3 2 2 3 6 11 17 20 20\n",
" 4 | 8 6 3 2 3 6 11 17 22 24\n",
" 0 | 8 7 6 6 5 4 5 7 8 8\n",
" -4 | 12 12 13 14 12 9 7 7 10 12\n",
" -4 | 16 17 19 21 19 14 10 9 10 14\n",
" 0 | 16 17 20 23 22 15 11 11 10 10\n"
]
}
],
"source": [
"for i in range(1, Nx):\n",
" for j in range(1, Ny):\n",
" D[i, j] = min(D[i-1, j-1], D[i-1, j], D[i, j-1]) + S[i, j]\n",
"print D"
"print ' ', ''.join('%4d' % n for n in y)\n",
"print ' +' + '----' * (ny+1)\n",
"for i, row in enumerate(table):\n",
" if i == 0:\n",
" z0 = ''\n",
" else:\n",
" z0 = x[i-1]\n",
" print ('%4s |' % z0) + ''.join('%4d' % z for z in row)"
]
},
{
......@@ -393,63 +369,14 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"## Step 3: Path Finding"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We will start at the end, $(N_x - 1, N_y - 1)$, and backtrace to the beginning, $(0, 0)$.\n",
"\n",
"For each pair of time indices, $(i, j)$, we will store the index pair in the path that preceded it."
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"backsteps = [[None for j in range(Ny)] for i in range(Nx)]\n",
"for i in range(1, Nx):\n",
" backsteps[i][0] = (i-1, 0)"
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"for j in range(1, Ny):\n",
" backsteps[0][j] = (0, j-1)"
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"for i in range(1, Nx):\n",
" for j in range(1, Ny):\n",
" candidate_steps = ((i-1, j-1), (i-1, j), (i, j-1),)\n",
" candidate_distances = [D[m, n] for (m, n) in candidate_steps]\n",
" backsteps[i][j] = candidate_steps[numpy.argmin(candidate_distances)]"
"## Step 2: Backtracking"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The time complexity of this operation is $O(N_x N_y)$. The space complexity is $O(N_x N_y)$."
"To assemble the best path, we use **backtracking** (FMP, p. 139). We will start at the end, $(N_x - 1, N_y - 1)$, and backtrack to the beginning, $(0, 0)$."
]
},
{
......@@ -461,120 +388,62 @@
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {
"collapsed": false
},
"outputs": [
{
"data": {
"text/plain": [
"[(0, 0), (1, 1), (2, 2), (3, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)]"
]
},
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"xi, yi = Nx-1, Ny-1\n",
"path = [(xi, yi)]\n",
"while xi > 0 or yi > 0:\n",
" xi, yi = backsteps[xi][yi]\n",
" path.insert(0, (xi, yi))\n",
"path"
]
},
{
"cell_type": "markdown",
"execution_count": 138,
"metadata": {
"collapsed": true
},
"source": [
"## Complete DTW algorithm"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here is a function that combines all the steps above. (Note that this is a naive version of the DTW algorithm. There are many optimizations that can reduce the time and space complexity below $O(N_x N_y)$.)"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {
"collapsed": false
},
"outputs": [],
"source": [
"def dtw(x, y):\n",
" Nx = len(x)\n",
" Ny = len(y)\n",
"\n",
" # Step 1: compute pairwise distances.\n",
" S = scipy.spatial.distance_matrix(x, y) \n",
" \n",
" # Step 2: compute cumulative distances.\n",
" D = scipy.zeros_like(S)\n",
" D[0, 0] = S[0, 0]\n",
" for i in range(1, Nx):\n",
" D[i, 0] = D[i-1, 0] + S[i, 0]\n",
" for j in range(1, len(y)):\n",
" D[0, j] = D[0, j-1] + S[0, j]\n",
" for i in range(1, Nx):\n",
" for j in range(1, Ny):\n",
" D[i, j] = min(D[i-1, j-1], D[i-1, j], D[i, j-1]) + S[i, j]\n",
"\n",
" # Step 3: find optimal path.\n",
" backsteps = [[None for j in range(Ny)] for i in range(Nx)]\n",
" for i in range(1, Nx):\n",
" backsteps[i][0] = (i-1, 0)\n",
" for j in range(1, Ny):\n",
" backsteps[0][j] = (0, j-1)\n",
" for i in range(1, Nx):\n",
" for j in range(1, Ny):\n",
" candidate_steps = ((i-1, j-1), (i-1, j), (i, j-1),)\n",
" candidate_distances = [D[m, n] for (m, n) in candidate_steps]\n",
" backsteps[i][j] = candidate_steps[numpy.argmin(candidate_distances)]\n",
" \n",
" xi, yi = Nx-1, Ny-1\n",
" path = [(xi, yi)]\n",
" while xi > 0 or yi > 0:\n",
" xi, yi = backsteps[xi][yi]\n",
" path.insert(0, (xi, yi))\n",
"def dtw(x, y, table):\n",
" i = len(x)\n",
" j = len(y)\n",
" path = [(i, j)]\n",
" while i > 0 or j > 0:\n",
" minval = numpy.inf\n",
" if table[i-1][j] < minval:\n",
" minval = table[i-1][j]\n",
" step = (i-1, j)\n",
" if table[i][j-1] < minval:\n",
" minval = table[i][j-1]\n",
" step = (i, j-1)\n",
" if table[i-1][j-1] < minval:\n",
" minval = table[i-1][j-1]\n",
" step = (i-1, j-1)\n",
" path.insert(0, step)\n",
" i, j = step\n",
" return path"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let's try it out:"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {
"collapsed": false
},
"execution_count": 139,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[(0, 0), (1, 1), (2, 2), (3, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8)]"
"[(0, 0),\n",
" (1, 0),\n",
" (1, 1),\n",
" (2, 2),\n",
" (2, 3),\n",
" (3, 3),\n",
" (3, 4),\n",
" (4, 5),\n",
" (4, 6),\n",
" (5, 7),\n",
" (6, 7),\n",
" (7, 8),\n",
" (7, 9)]"
]
},
"execution_count": 16,
"execution_count": 139,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"path = dtw(x, y)\n",
"path = dtw(x, y, table)\n",
"path"
]
},
......@@ -582,86 +451,61 @@
"cell_type": "markdown",
"metadata": {},
"source": [
"Take a look at the two aligned sequences:"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {
"collapsed": false
},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[3, 3, 1, 4, 4, 6, 1, 5, 5]\n",
"[4, 2, 1, 3, 3, 4, 1, 4, 5]\n"
]
}
],
"source": [
"print [x[i][0] for (i, j) in path]\n",
"print [y[j][0] for (i, j) in path]"
"The time complexity of this operation is $O(N_x + N_y)$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Sanity check: compute the total distance of this alignment:"
"As a sanity check, compute the total distance of this alignment:"
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {
"collapsed": false
},
"execution_count": 144,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"7"
"10"
]
},
"execution_count": 18,
"execution_count": 144,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"sum(abs(x[i][0] - y[j][0]) for (i, j) in path)"
"sum(abs(x[i-1] - y[j-1]) for (i, j) in path if i >= 0 and j >= 0)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Indeed, that is the same as the cumulative distance computed earlier, $D[N_x - 1, N_y - 1]$:"
"Indeed, that is the same as the cumulative distance of the optimal path computed earlier:"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {
"collapsed": false
},
"execution_count": 146,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"7.0"
"10.0"
]
},
"execution_count": 19,
"execution_count": 146,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"D[-1, -1]"
"table[-1][-1]"
]
},
{
......@@ -688,7 +532,7 @@
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython2",
"version": "2.7.10"
"version": "2.7.13"
}
},
"nbformat": 4,
......
index.html
View file @ 19bb5d91
......@@ -11940,6 +11940,7 @@ div#notebook {
<div class="text_cell_render border-box-sizing rendered_html">
<ol>
<li><a href="dp.html">Dynamic Programming</a> (<a href="dp.ipynb">ipynb</a>)</li>
<li><a href="lcs.html">Longest Common Subsequence</a> (<a href="lcs.ipynb">ipynb</a>)</li>
<li><a href="dtw.html">Dynamic Time Warping</a> (<a href="dtw.ipynb">ipynb</a>)</li>
</ol>
......
index.ipynb
View file @ 19bb5d91
......@@ -137,6 +137,7 @@
"metadata": {},
"source": [
"1. [Dynamic Programming](dp.html) ([ipynb](dp.ipynb))\n",
"1. [Longest Common Subsequence](lcs.html) ([ipynb](lcs.ipynb))\n",
"1. [Dynamic Time Warping](dtw.html) ([ipynb](dtw.ipynb))"
]
},
......
lcs.html 0 → 100644
View file @ 19bb5d91
This source diff could not be displayed because it is too large. You can view the blob instead.
lcs.ipynb 0 → 100644
View file @ 19bb5d91
{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"[&larr; Back to Index](index.html)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# Longest Common Subsequence"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"To motivate dynamic time warping, let's look at a classic dynamic programming problem: find the **longest common subsequence (LCS)** of two strings ([Wikipedia](https://en.wikipedia.org/wiki/Longest_common_subsequence_problem)). A subsequence is not required to maintain consecutive positions in the original strings, but they must retain their order. Examples:\n",
"\n",
" lcs('cake', 'baker') -> 'ake'\n",
" lcs('cake', 'cape') -> 'cae'"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We can solve this using recursion. We must find the optimal substructure, i.e. decompose the problem into simpler subproblems. \n",
"\n",
"Let `x` and `y` be two strings. Let `z` be the true LCS of `x` and `y`. \n",
"\n",
"If the first characters of `x` and `y` were the same, then that character must also be the first character of the LCS, `z`. In other words, if `x[0] == y[0]`, then `z[0]` must equal `x[0]` (which equals `y[0]`). Therefore, append `x[0]` to `lcs(x[1:], y[1:])`.\n",
"\n",
"If the first characters of `x` and `y` differ, then solve for both `lcs(x, y[1:])` and `lcs(x[1:], y)`, and keep the result which is longer."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here is the recursive solution:"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"def lcs(x, y):\n",
" if x == \"\" or y == \"\":\n",
" return \"\"\n",
" if x[0] == y[0]:\n",
" return x[0] + lcs(x[1:], y[1:])\n",
" else:\n",
" z1 = lcs(x[1:], y)\n",
" z2 = lcs(x, y[1:])\n",
" return z1 if len(z1) > len(z2) else z2"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Test:"
]
},
{
"cell_type": "code",
"execution_count": 2,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"pairs = [\n",
" ('cake', 'baker'),\n",
" ('cake', 'cape'),\n",
" ('catcga', 'gtaccgtca'),\n",
" ('zxzxzxmnxzmnxmznmzxnzm', 'nmnzxmxzmnzmx'),\n",
" ('dfkjdjkfdjkjfdkfdkfjd', 'dkfjdjkfjdkjfkdjfkjdkfjdkfj'),\n",
"]"
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"ake\n",
"cae\n",
"ctca\n",
"zxmxzmnzmx\n",
"dfjdjkfdjkjfdkfdkfj\n"
]
}
],
"source": [
"for x, y in pairs:\n",
" print lcs(x, y)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"The time complexity of the above recursive method is $O(2^{n_x+n_y})$. That is slow because we might compute the solution to the same subproblem multiple times. "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### Memoization"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"We can do better through memoization, i.e. storing solutions to previous subproblems in a table."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Here, we create a table where cell `(i, j)` stores the length `lcs(x[:i], y[:j])`. When either `i` or `j` is equal to zero, i.e. an empty string, we already know that the LCS is the empty string. Therefore, we can initialize the table to be equal to zero in all cells. Then we populate the table from the top left to the bottom right."
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"def lcs_table(x, y):\n",
" nx = len(x)\n",
" ny = len(y)\n",
" \n",
" # Initialize a table.\n",
" table = [[0 for _ in range(ny+1)] for _ in range(nx+1)]\n",
" \n",
" # Fill the table.\n",
" for i in range(1, nx+1):\n",
" for j in range(1, ny+1):\n",
" if x[i-1] == y[j-1]:\n",
" table[i][j] = 1 + table[i-1][j-1]\n",
" else:\n",
" table[i][j] = max(table[i-1][j], table[i][j-1])\n",
" return table "
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Let's visualize this table:"
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[[0, 0, 0, 0, 0, 0],\n",
" [0, 0, 0, 0, 0, 0],\n",
" [0, 0, 1, 1, 1, 1],\n",
" [0, 0, 1, 2, 2, 2],\n",
" [0, 0, 1, 2, 3, 3]]"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"x = 'cake'\n",
"y = 'baker'\n",
"table = lcs_table(x, y)\n",
"table"
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" b a k e r\n",
" 0 0 0 0 0 0\n",
"c 0 0 0 0 0 0\n",
"a 0 0 1 1 1 1\n",
"k 0 0 1 2 2 2\n",
"e 0 0 1 2 3 3\n"
]
}
],
"source": [
"xa = ' ' + x\n",
"ya = ' ' + y\n",
"print ' '.join(ya)\n",
"for i, row in enumerate(table):\n",
" print xa[i], ' '.join(str(z) for z in row)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Finally, we backtrack, i.e. read the table from the bottom right to the top left:"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {
"collapsed": true
},
"outputs": [],
"source": [
"def lcs(x, y, table, i=None, j=None):\n",
" if i is None:\n",
" i = len(x)\n",
" if j is None:\n",
" j = len(y)\n",
" if table[i][j] == 0:\n",
" return \"\"\n",
" elif x[i-1] == y[j-1]:\n",
" return lcs(x, y, table, i-1, j-1) + x[i-1]\n",
" elif table[i][j-1] > table[i-1][j]:\n",
" return lcs(x, y, table, i, j-1)\n",
" else:\n",
" return lcs(x, y, table, i-1, j)"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"ake\n",
"cae\n",
"ctca\n",
"zxmxzmnzmx\n",
"dfjdjkfdjkjfdkfdkfj\n"
]
}
],
"source": [
"for x, y in pairs:\n",
" table = lcs_table(x, y)\n",
" print lcs(x, y, table)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"Table construction has time complexity $O(mn)$, and backtracking is $O(m+n)$. Therefore, the overall running time is $O(mn)$."
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"[&larr; Back to Index](index.html)"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 2",
"language": "python",
"name": "python2"
},
"language_info": {
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