Splitting PDB by delimiter

csplit  --prefix=chain --suffix="%1d".pdb 3FDL_Repair.pdb "/^TER/+1" "{*}"

Renaming Files

#!/bin/bash
for file in foo[0-9]*.png; do
  # strip the prefix ("foo") off the file name
  postfile=${file#foo}
  # strip the postfix (".png") off the file name
  number=${postfile%.png}
  # subtract 1 from the resulting number
  i=$((number-1))
  # copy to a new name in a new folder
  cp ${file} ../newframes/$(printf foo%08d.png $i)
done

http://stackoverflow.com/questions/55754/how-to-zero-pad-numbers-in-file-names-wit-a-bash-script

Sum up columns

awk '{print $1}' * | paste -sd+ | bc

String manipulation with bash

String manipulation with bash

for each in *.txt; do echo ${each/.txt/.pdb};done

${string:position}

paste -d '\n' file1 file2




Replace the 4th character with another letter.

echo "abcdefghijklmnop" | sed 's/./X/4'

Using Rpy2 with Anaconda and Ipython notebook

Typically in ipython notebook, the only command required to print a plot inline is the following:

=====================
%load_ext rmagic
=====================

=====================
%%R 
x<-c(1,2,3)
y<-c(2,3,4)
plot(x,y)

=====================

While working on AWS, my plots would not show up inline. I found out that this is because rpy2 only works for python 2.7.5. 

The easy fix is 

conda install python=2.7.5=2

Parameters for Negative Binomial

The negative binomial is a useful distribution for modeling the probability that a successful event occurs after r failures in q trials.

To put this into context:

Let's say that Joe is selling candy bars in his local neighborhood. He must sell 5 candy bars to meet his quota. The chance that a household will buy his candy is 0.4. What is the probability that he sells his 5th candy bar to the 10th household.

The parameters extracted from this problem are as follows:

\[r=5=\textrm{number of failures }
\\k=5=\textrm{number of successes (aka size)}
\\p=0.4=\textrm{probability of success}
\\k+r=10=q=\textrm{total number of trials}\]


The PDF for the negative binomial follows as
\[{{r+k-1} \choose k-1} (p)^k (1-p)^r\]

The answer that follows is 0.100329, which means that Joe has about a 10% chance of selling his 5th candy bar at the 10th house. Or rephrased, it also means that Joe has 10% chance of failing 5 times and succeeding 5 times at the 10th house. 

In R, you can calculate the CDF of the first 10 failures with the R code below.

Note that by the 10th house, we will have 5 failures and 5 successes.

pnbinom(1:10,5,0.4)

 [1] 0.0409600 0.0962560 0.1736704 0.2665677 0.3668967 0.4672258 0.5618218

 [8] 0.6469582 0.7207430 0.7827223

and the PDF

dnbinom(1:10,5,0.4)

 [1] 0.03072000 0.05529600 0.07741440 0.09289728 0.10032906 0.10032906

 [7] 0.09459597 0.08513638 0.07378486 0.06197928

Parameters for R

dnbinom(q, size, prob)
q - the number of failures
size - the number of successes
prob - the probability of success

Another quick example:

Joe decides it's time for a new job. Instead of selling candy bars he now sells notebooks. If he sells three notebooks, then he will meet his quota. However, the probability that he will sell a notebook is now 0.09. What is the probability that he sells his third notebook at the 10th house.

dnbinom(7,3,0.09)
0.013561876