# Matlab: different colormaps for subplots

I often want different subplots in one Matlab figure to have different colormaps. However, colormap is a figure property, so it’s not trivial, except that it is… with these utilities:

This works for everything except colorbars: http://www.mathworks.com/matlabcentral/fileexchange/7943-freezecolors—unfreezecolors

Post-2010, Matlab refreshes colorbars with each subplot, so you’ll need this to freeze colorbars: http://www.mathworks.com/matlabcentral/fileexchange/24371-colormap-and-colorbar-utilities–feb-2014-

# Scaling up hierarchical clustering

There are lots of caveats with hierarchical clustering, and it’s often used to draw unjustifiable conclusions. But I’ll save that discussion for later, and it’s at least occasionally useful. 🙂 So far, I’ve mainly used it to reorder the columns/rows of a covariance or correlation matrix. For example, I recently generated synthetic sparse precision matrices, and I wanted to make sure that the corresponding covariance/correlation matrices were as I expected.

However, linkage/dendrogram in both Matlab and SciPy are really slow. In particular, the linkage algorithms they use are O(n^3). So instead we can use fastcluster, which is O(n^2). All I had to do was replace this:

Z = sch.linkage(P, method="average", metric="correlation")

with this:

Z = fastcluster.linkage(P, method="average", metric="correlation")

The second problem is that dendrogram does lots of unnecessary work when all I want is the reordered indices. SciPy actually implements it recursively, so it gives this error: “RuntimeError: maximum recursion depth exceeded”. So the second change is to replace this:

dendr = sch.dendrogram(Z,no_plot=True,distance_sort=True)ix = dendr["leaves"]

with this (credit to a good denizen of StackOverflow):

n = len(Z) + 1cache = dict()
for k in range(len(Z)):    c1, c2 = int(Z[k]), int(Z[k])    c1 = [c1] if c1 < n else cache.pop(c1)    c2 = [c2] if c2 < n else cache.pop(c2)    cache[n+k] = c1 + c2ix = cache[2*len(Z)]

Then it’s as simple as:

Pclust = P[ix,:]Pclust = Pclust[:,ix]pyplot.imshow(Pclust, interpolation="nearest")

“Three unblinded mice” from Andrew Gelman’s blog

I really like the phrase “pre-scientific intuitions”. Also this: “OK, sure, sure, everybody knows about statistics and p-values and all that, but my impression is that researchers see these methods as a way to prove that an effect is real. That is, statistics is seen, not as a way to model variation, but as a way to remove uncertainty.”

# Converting between decimal and unary in Matlab

Decimal to unary:

[1:4]'==3   % returns [0 0 1 0 ]

Unary to decimal:

max([0 0 1 0] .* (1:4))   % returns 3

# Fisher information, score function, oh my!

Ever wondered why $\text{Var}_{\theta}(\ell'(\theta)) = I_n(\theta)$ and $\text{Var}_{\theta}(\hat{\theta}) \approx \frac{1}{I_n(\theta)}$?

This is an excellent intuitive explanation:

http://math.stackexchange.com/a/265933/105416

.

# Confidence Intervals from Pivots

This is an example of using a pivot to find a confidence interval. $X_1,...,X_n \sim \text{Uniform}(0,\theta).$

1. Find a pivot:

Let $Q=X_{(n)}/\theta$.

2. Find its distribution: $P(Q \le t)= P(X_i \le t\theta)^n = t^n$.

3. Find an expression involving an upper and lower bound on the pivot: $P(a \le Q \le b) = b^n-a^n$ This implies $P(a \le Q \le 1) = 1-a^n$.

4. Substitute the expression for the pivot from Step 1, and set the RHS to $1-\alpha$. $P(a \le X_{(n)}/\theta \le 1)=1-a^n$ $P(1/a \ge \theta/X_{(n)} \ge 1) = 1-a^n$ $P( X_{(n)} \le \theta \le \frac{X_{(n)}}{a} ) = 1-a^n$

Let $1-\alpha = 1-a^n$. Then $a=\alpha^{1/n}$. $P(X_{(n)} \le \theta \le \frac{X_{(n)}}{\alpha^{1/n}})=1-\alpha$

This gives us $[X_{(n)},\frac{X_{(n)}}{\alpha^{1/n}}]$ as a $1-\alpha$ CI for $\theta$.

# Hello world $\int_0^1 f(x) \, \mathrm{d} x$