[Tutor] arrangement of datafile

[Amrita Kumari]

Hi Peter,

Thankyou very much for your kind help. I got the output like the way I
wanted (which you have also shown in your output). I really appreciate your
effort.

Thanks for your time.
Amrita


On Thu, Jan 9, 2014 at 8:41 PM, Peter Otten <__peter__ at web.de> wrote:

> Amrita Kumari wrote:
>
> > On 17th Dec. I posted one question, how to arrange datafile in a
> > particular fashion so that I can have only residue no. and chemical
> > shift value of the atom as:
> > 1  H=nil
> > 2  H=8.8500
> > 3  H=8.7530
> > 4  H=7.9100
> > 5  H=7.4450
> > ........
> > Peter has replied to this mail but since I haven't subscribe to the
> > tutor mailing list earlier hence I didn't receive the reply, I
> > apologize for my mistake, today I checked his reply and he asked me to
> > do few things:
>
> I'm sorry, I'm currently lacking the patience to tune into your problem
> again, but maybe the script that I wrote (but did not post) back then is of
> help.
>
> The data sample:
>
> $ cat residues.txt
> 1 GLY HA2=3.7850 HA3=3.9130
> 2 SER H=8.8500 HA=4.3370 N=115.7570
> 3 LYS H=8.7530 HA=4.0340 HB2=1.8080 N=123.2380
> 4 LYS H=7.9100 HA=3.8620 HB2=1.7440 HG2=1.4410 N=117.9810
> 5 LYS H=7.4450 HA=4.0770 HB2=1.7650 HG2=1.4130 N=115.4790
> 6 LEU H=7.6870 HA=4.2100 HB2=1.3860 HB3=1.6050 HG=1.5130 HD11=0.7690
> HD12=0.7690 HD13=0.7690 N=117.3260
> 7 PHE H=7.8190 HA=4.5540 HB2=3.1360 N=117.0800
> 8 PRO HD2=3.7450
> 9 GLN H=8.2350 HA=4.0120 HB2=2.1370 N=116.3660
> 10 ILE H=7.9790 HA=3.6970 HB=1.8800 HG21=0.8470 HG22=0.8470 HG23=0.8470
> HG12=1.6010 HG13=2.1670 N=119.0300
> 11 ASN H=7.9470 HA=4.3690 HB3=2.5140 N=117.8620
> 12 PHE H=8.1910 HA=4.1920 HB2=3.1560 N=121.2640
> 13 LEU H=8.1330 HA=3.8170 HB3=1.7880 HG=1.5810 HD11=0.8620 HD12=0.8620
> HD13=0.8620 N=119.1360
>
> The script:
>
> $ cat residues.py
> def process(filename):
>     residues = {}
>     with open(filename) as infile:
>         for line in infile:
>             parts = line.split()            # split line at whitespace
>             residue = int(parts.pop(0))     # convert first item to integer
>             if residue in residues:
>                 raise ValueError("duplicate residue {}".format(residue))
>             parts.pop(0)                    # discard second item
>
>             # split remaining items at "=" and put them in a dict,
>             # e. g. {"HA2": 3.7, "HA3": 3.9}
>             pairs = (pair.split("=") for pair in parts)
>             lookup = {atom: float(value) for atom, value in pairs}
>
>             # put previous lookup dict in residues dict
>             # e. g. {1: {"HA2": 3.7, "HA3": 3.9}}
>             residues[residue] = lookup
>
>     return residues
>
> def show(residues):
>     atoms = set().union(*(r.keys() for r in residues.values()))
>     residues = sorted(residues.items())
>     for atom in sorted(atoms):
>         for residue, lookup in residues:
>             print "{} {}={}".format(residue, atom, lookup.get(atom, "nil"))
>         print
>         print "-----------"
>         print
>
> if __name__ == "__main__":
>     r = process("residues.txt")
>     show(r)
>
> Note that converting the values to float can be omitted if all you want to
> do is print them. Finally the output of the script:
>
> $ python residues.py
> 1 H=nil
> 2 H=8.85
> 3 H=8.753
> 4 H=7.91
> 5 H=7.445
> 6 H=7.687
> 7 H=7.819
> 8 H=nil
> 9 H=8.235
> 10 H=7.979
> 11 H=7.947
> 12 H=8.191
> 13 H=8.133
>
> -----------
>
> 1 HA=nil
> 2 HA=4.337
> 3 HA=4.034
> 4 HA=3.862
> 5 HA=4.077
> 6 HA=4.21
> 7 HA=4.554
> 8 HA=nil
> 9 HA=4.012
> 10 HA=3.697
> 11 HA=4.369
> 12 HA=4.192
> 13 HA=3.817
>
> -----------
>
> [snip]
>
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