Jewish New Year Distribution

I was talking to somebody the other day who was pleased because his birthday (September 9) landed on the Jewish New Year. He made the dubious claim that this happened something like 17% of the time.

For a quick check, I used the JewishNewYear[] function built in to Mathematica’s Calendar package. It is good from 1900 through 2100.


Needs["Calendar`"]

In[]:= Length[Select[
Map[JewishNewYear[#] &,
Range[1900, 2100]], #[[2]] == 9 && #[[3]] == 9 &]]

Out[]= 7

7/Length[Range[1900, 2100]] == .0348, so over this 201 year time span, his birthday would have landed on Rosh Hashana a bit under 3.5% of the time.

Curious about how the holiday distributed, I generated a table of frequencies (all displayed as though they were 2010 dates so we can use the built-in DateListPlot[] function).


In[]:= dateFreq =
Map[{#[[1]], Length[#]} &,
Gather[Sort[
Map[{2010, JewishNewYear[#][[2]], JewishNewYear[#][[3]]} &,
Range[1900, 2100]], AbsoluteTime[#1] <= AbsoluteTime[#2] &]]]; In[]:= DateListPlot[dateFreq, Filling -> Axis,
FillingStyle -> {Automatic, {Black, AbsoluteThickness[10]}},
GridLines -> None, Frame -> False, Axes -> True, AxesStyle -> Black,
AxesOrigin -> {Automatic, 0}, PlotRange -> All,
PlotLabel -> "Rosh Hashana Distribution"]

Note the nested use of Length[], Gather[], Sort[]. That is often a powerful combination when using Mathematica to analyze the frequencies of events or patterns in data.

In this case we do not have a uniform distribution — it clearly fades at the edges, but it’s not normally distributed either. Figuring out whether or not there is a meaningful pattern here is beyond my statistical skills.

A thoroughly unscientific study of computers in the New York City court system

I am on jury duty today. Civil court, as it turns out. I last served in 2003, when I sat on one of the “murder and mayhem” grand juries (grand juries in New York specialize in drugs, family, violent, or white collar crime). On the violent crime grand jury we heard about a dozen cases, none of which I am allowed to disclose, though I will say that one of the accused had tried to kill himself by pressing a running power drill against his head until, I think, he broke the bit, then slashed his wrists, jumped out of an eighth story window, and survived. I proposed that we waive any indictment in exchange for this superman’s agreeing to be dropped on Afghanistan, but that power is evidently not available to grand juries.

Anyway, since my last stint they have installed Wifi in the court building.

I’m seeing 47 unambiguous PCs on the network and 12 Macs, with another 56 I’m not sure about. All the official machines seem to be PCs. I’ve noticed one machine with a public directory featuring a bootleg collection of hiphop recordings.

The man two seats to my right has fallen asleep and is snoring. Everybody else is jealous.

reversing the x-axis in a Mathematica graph

Sometimes you have data in which the dependent variable decreases as the independent variable increases, and for reasons of clarity in illustrating it, you want to flip the x-axis and show an increasing line (or vice versa). For example, sometimes you have

and you want to display it as

Unfortunately, Mathematica’s default behavior is to show both axes growing as you go up and to the right, and there is no simple option to change that. You can achieve this however by transforming your data by hand so that it is displaying the desired way, and then setting the “Ticks” variable of the plotting function to display the Ticks in reverse order. You can’t just say “reverse,” you have to create a list of the new tick labels and where you want them to go. This is generally done with a call to the Table[] command.

I more commonly flip the X axis than the Y, and given data in {{x1,y1},{x2,y2} . . . {xn,yn}} format, the following code makes it very easy to do.

xFlipper[listofXYpairs_] :=
Map[{Last[listofXYpairs][[1]] + First[listofXYpairs][[1]],
0} + {-1, 1}*# &, listofXYpairs]

Options[xFlippedTicks] = {numPoints -> 10, digits -> "All"};

xFlippedTicks[listofXYpairs_, OptionsPattern[]] :=
Table[{Last[listofXYpairs][[1]] + First[listofXYpairs][[1]] - i,
If[OptionValue[digits] == "All", i,
Round[i, 10^-OptionValue[digits] // N]]}, {i,
First[listofXYpairs][[1]], Last[listofXYpairs][[1]], (
Last[listofXYpairs][[1]] - First[listofXYpairs][[1]])/
OptionValue[numPoints]-1}]

The numpoints variable sets how many points to show on the X axis. The digits variable can reduce the number of the digits shown in the tickmarks if needed to make the graph more useful or attractive.

The first graph above was produced with

ListPlot[returnPairs, PlotLabel -> "Expected IRR vs. Upfront cost",
AxesLabel -> {"Amt Paid", "IRR"}, Joined -> True, Mesh -> True]

Given these two functions, we can transform it to the second graph above with

ListPlot[xFlipper[returnPairs], PlotLabel -> "Expected IRR vs. Upfront cost",
Ticks -> {xFlippedTicks[returnPairs, numPoints -> 5], Automatic},
AxesLabel -> {"Amt Paid", "IRR"}, Joined -> True, Mesh -> True]

(The “automatic” in the Tick specification tells Mathematica to continue to handle the Y axis automatically.)