Thursday, 16 June 2016

More molecular depiction in Vim

The reception for my previous blog post on ASCII depiction in Vim was somewhat muted even, dare I say, lukewarm. Buoyed by this response, I updated that script to handle quoted SMILES strings (as found in my Python scripts, for example). And here I provide similar depiction support for SDF files ("\a" and "\A" for small and big depictions).

Disclaimer: Due to my use of complex regular expressions, this script is not suitable for use by Andrew Dalke, nor for any SDF file provided by same. If this script is used on a SDF file provided by Andrew Dalke, it may result in the summoning of Cthulhu. You have been warned.

function! SDFToAscii(width)
  let prev = search("$$$$", "nbW") + 1
  let end = search("$$$$", "nW")
  if end == 0
    let end = line('$')
  silent execute prev.",".end."yank"
  let output = system("obabel -isdf -oascii -d -xa 2.0 -xw ".a:width, @")
  botright new
  setlocal buftype=nofile bufhidden=wipe nobuflisted noswapfile nowrap
  put =output
  setlocal nomodifiable
noremap <silent> <leader>a :call SDFToAscii(79)<CR>
noremap <silent> <leader>A :call SDFToAscii(189)<CR>

Monday, 13 June 2016

SMILES depiction in Vim

I spend a lot of time scrolling through files of SMILES in Vim, and copying and pasting SMILES into viewers. This is now a thing of the past. I bring you (from the depths of time) molecule depiction within Vim.

Using the script below (paste it into your .vimrc), I just hit "\s" with the cursor on a SMILES string, and up pops its ASCII depiction (":q" to close). Need a bigger image? "\S" is your friend. This is my first Vim script ("and I hope it's your last" sez you) so if you have any improvements, let me know.

[Update 16/06/2014]: Script updated to handle quoted SMILES strings and those with escaped backslashes.
noremap <silent> <leader>s :call SmiToAscii(79)<CR>
noremap <silent> <leader>S :call SmiToAscii(189)<CR>
function! SmiToAscii(width)
  let smiles = expand("<cWORD>")
  " Strip quotation marks and commas
  let smiles = substitute(smiles, "[\"',]", "", "g")
  " Handle escaped backslashes, e.g. in C++ strings
  let smiles = substitute(smiles, "\\\\", "\\", "g")
  botright new
  setlocal buftype=nofile bufhidden=wipe nobuflisted noswapfile nowrap
  execute '$read ! obabel -:'.shellescape(smiles, 1). ' -o ascii -d -xa 2.0 -xw '. a:width
  setlocal nomodifiable

Saturday, 21 May 2016

Open Babel presentation at MIOSS 2016

Thanks again to George Papadatos, Mark Forster, et al for the invitation to present at the 2016 reboot of MIOSS (Molecular Informatics in Open Source Software). You may remember the earlier one back in 2011.

I gave a presentation on the Open Babel project. While I have been largely quiescent over the past 2 years, there has been much happening besides. Rather than cover features, I tried to present some interesting stats or views of the overall project. For example, the background of some of the current developers, which file formats were recently added, and so forth. A few of my slides discuss industry involvement in the project - this is because the MIOSS workshop is 50% industry and 50% open source developers.

For info on other presentations from the meeting, see the Lanyrd page (I'll update here when the rest of the talks become available) or search #MIOSS2016 on Twitter.

Tuesday, 19 April 2016

Static static code analyser

I've just been refactoring/rewriting some C++ code and noticed that I've left some static functions around that are unused. "static" in this context means functions that are only accessible within the same file (it's a good idea to put static in front of every function until the point you need to call it from some other file). However, if nothing in the same file calls them, then they contain dead code. Unfortunately neither cppcheck nor MSVC's static code analyser will flag up static functions that are not called, and so I've written my own Python script to do so. It's fairly basic but worked for me.

Update (19/04): Doh! Adding --enable=all to cppcheck will find most (but not all) of these cases.

import os
import sys
import glob

def FindStaticFunctions(fname):
    fns = []
    for line in open(fname):
        tmp = line.strip()
        if not tmp.startswith("static"): continue
        if tmp.endswith(";"): continue
        broken = tmp.split()
        for i, x in enumerate(broken):
            if "(" in x:
                name = x[:x.find("(")].replace("*", "")
            elif i < len(broken)-1 and broken[i+1][0]=='(':
                fns.append(x.replace("*", ""))

    return fns

def FindMissingReferences(fname, staticfns):
    lines = open(fname).readlines()
    for staticfn in staticfns:
        refs = 0
        for line in lines:
            if line.find(staticfn) >= 0:
                refs += 1
        if refs <= 1:
            print "(%s) %s: %d refs" % (os.path.basename(fname), staticfn, refs)

if __name__ == "__main__":
    if len(sys.argv) < 2:
        sys.exit("You need to specify a directory")
    dirname = sys.argv[1]
    if not os.path.isdir(dirname):
        sys.exit("%s is not a directory name" % dirname)

    for cppfile in glob.glob(os.path.join(dirname, "*.cpp")):
        staticfns = FindStaticFunctions(cppfile)
        FindMissingReferences(cppfile, staticfns)

Wednesday, 13 January 2016

Do BEDROC results correlate with AUC?

Following up on a post over at NM where I was perhaps overly harsh on AUC, it is natural to instead consider early recognition metrics such as EF, BEDROC and so forth as potentially better measures of performance in virtual screens. Here I want to pour cold water on the idea that results for AUC and BEDROC are highly correlated and so we might as well just use the more straightfoward of the two (i.e. AUC).

In a 2010 review of current trends by Geppert, Vogt and Bajorath [1], they interpret the "What do we know and when do we know it?" paper by Nicholls [2] as presenting "evidence for a strong correlation between AUC and BEDROC [3], suggesting AUC as a sufficient measure for virtual screening performance." Now, Nicholls doesn't quite say that but I can see that Figure 8 and 9 in that paper could be interpreted in that way. For the Warren study [4] data shown, the Pearson R2 between the two is 0.901. It is unfortunate that the correlation is only shown for the averaged data (Figure 9) - it would be nice to see the Spearman correlation for Figure 8. But either way, it's clear that they are highly correlated.

So a large AUC tends to imply a large BEDROC (and v.v), and so forth. But the thing is, that's not quite what we are interested in. The real question is whether the relative ordering of methods obtained with AUC are the same as those obtained with BEDROC - if it is, then we might as well use AUC. If not, we'll have to exercise those little grey cells (I've been reading a lot of Agatha Christie recently) and figure out which is best. In other words, it could be that methods A and B both have a high AUC, and so have a high BEDROC, but does A>B in AUC space imply A>B in BEDROC space?

As it happens I have the results to hand (as one does) for AUC (*) and BEDROC(20) for the Riniker and Landrum dataset [5] (88 proteins, 28 fingerprints, 50 repetitions). This is a fingerprint-based virtual screen as opposed to a docking study, but the conclusions should be the same. The 4400 Spearman correlations between the AUC and BEDROC results are shown here as a histogram (bins of 0.05):

The median value of the Spearman correlation is 0.713. This is not that high (remember that the square is 0.51) indicating that while there is a moderate correlate between the orderings obtained by BEDROC and AUC, they are not highly correlated.

One more nail in the AUC's coffin I hope. Grab your hammer and join me!

[1] Current Trends in Ligand-Based Virtual Screening: Molecular Representations, Data Mining Methods, New Application Areas, and Performance Evaluation. Hanna Geppert, Martin Vogt, and Jürgen Bajorath. J. Chem. Inf. Model. 2010, 50, 205–216.
[2] What do we know and when do we know it? Anthony Nicholls. J. Comput. Aided Mol. Des. 2008, 22, 239–255.
[3] Evaluating Virtual Screening Methods: Good and Bad Metrics for the “Early Recognition” Problem. Jean-François Truchon and Christopher I. Bayly. J. Chem. Inf. Model. 2007, 47, 488-508.
[4] A critical assessment of docking programs and scoring functions. Warren GL, Andrews CW, Capelli AM, Clarke B, LaLonde J, Lambert MH, Lindvall M, Nevins N, Semus SF, Senger S, Tedesco G, Wall ID, Woolven JM, Peishoff CE, Head MS. J. Med. Chem. 2006, 49, 5912-5931.
[5] Open-source platform to benchmark fingerprints for ligand-based virtual screening. Sereina Riniker and Gregory A. Landrum. J. Cheminf. 2013, 5, 26.

* I'm lazy so I just used the mean rank of the actives instead of integrating ROC curves. You get an identical ordering (see Eq. 12 in BEDROC paper [3] for the relationship). And yes, I did check.

Wednesday, 2 December 2015

Coin-tossing power

I've been doing some coin tossing. Well actually it's cheminformatics, but the problem reduces to coin tossing. Given that I toss a coin 1000 times, how many heads must I observe to believe that it is a biased coin. My usual ramshackle approach would be to estimate the standard deviation of the random distribution (have my computer do some coin tossing) and then say something about being two sigmas away from the mean. I've done that just to be sure - the sigma is just under 16.
import random
import matplotlib.pyplot as plt
from collections import defaultdict
import scipy
import scipy.optimize
import numpy as np

values = [01]

ans = defaultdict(int)
for N in range(100000):
    tot = 0
    for i in range(1000):
        tot += random.choice(values)
    ans[tot] += 1

start, end = 350650
x = np.arange(start, end)
y = []
for i in range(start, end):
    if i in ans:
        if i > start:

n = len(x)
mean = 500.
sigma = 10.

def gaus(x,a,sigma):
    return a*scipy.exp(-(x-mean)**2/(2*sigma**2))

popt,pcov = scipy.optimize.curve_fit(gaus,x,y,p0=[1.0,sigma])
print popt

PMC points out that a slightly easier way of getting the same value would be to use an equation: variance = n(1-p)p where n = 1000 and p is 0.5. The standard deviation is then the square root of this (i.e. sqrt of 250), 15.8.

However, the proper way of handling the original question is a power calculation. Typically power calculations are associated with calculating the necessary sample size. Here I have a fixed sample size (n=1000) and I want to figure out, given that the alternative hypothesis is true (biased coin), what number of heads must I observe to be 99.9% sure (power=0.999) that it would be show up as significant at the 0.1% level (alpha=0.001). The y = 0.5 below refers to the proportion of heads expected under the null hypothesis (unbiased coin). Since n=1000, we can use the normal approximation to the binomial distribution.
import math
import scipy.stats

def qnorm(p):
    return scipy.stats.norm.ppf(p)

def solvequadratic(a, b, c):
    return [(-b+math.sqrt((b**2)-(4*(a*c))))/(2.*a),

if __name__ == "__main__":
    alpha = 0.001 # Significance level
    gamma = 0.999 # Power
    qgamma = qnorm(gamma)
    qalpha = qnorm(1.-(alpha/2.))
    p = (qalpha+qgamma)**2
    n = 1000
    y = 0.5
    a = 2*n+p
    b = (2*p-4*n)*y - 2*p
    # Note that if y is 0.5, then, b = -2*n-p (i.e. -a)
    c = -2*p*y + (2*n+p)*y*y
    print solvequadratic(a, b, c)
...and the answer? 642. Thanks to ALC for help with this, though all errors are my own.

Notes: (added 04/12/2015)
Both the power and the significance level are conditional probabilities:
  • The significance level (alpha) is the probability of rejecting the null hypothesis given that it is true, ie. prob(rejecting null hypothesis | null hypothesis is correct).
  • The power (gamma) is the probability of detecting a difference, if indeed the difference does exist, ie. prob(rejecting null hypothesis | null hypothesis is incorrect).

Thursday, 27 August 2015

The whole of cheminformatics best practice in 15 minutes

Here's the talk I co-presented with Rajarshi at the recent Boston ACS in a session about best practices in cheminformatics. It was entitled "So I have an SD File...What do I do next?" and went something like this...