[bip] how to join files together
amrita bhu2005
amritabhu2005 at yahoo.com
Mon Apr 26 03:03:54 PDT 2010
Dear Sir,
I have four text files:
file1
1 8.40 123.54 180.07 3.78 54.98 17.88
2 8.70 121.94 182.97 4.68 54.78
4 8.21 119.84 179.17 4.13 55.48 18.48
6 7.97 124.04 178.27 3.65 55.48 17.68
file2
156 1 0 0 0 TOTAL NUMBER OF RESIDUES, NUMBER OF CHAINS, NUMBER OF SS-BRIDGES(TOTAL,INTRACHAIN,INTERCHAIN) .
8970.0 ACCESSIBLE SURFACE OF PROTEIN (ANGSTROM**2) .
116 74.4 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(J) , SAME NUMBER PER 100 RESIDUES .
8 5.1 TOTAL NUMBER OF HYDROGEN BONDS IN PARALLEL BRIDGES, SAME NUMBER PER 100 RESIDUES .
34 21.8 TOTAL NUMBER OF HYDROGEN BONDS IN ANTIPARALLEL BRIDGES, SAME NUMBER PER 100 RESIDUES .
0 0.0 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I-5), SAME NUMBER PER 100 RESIDUES .
1 0.6 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I-4), SAME NUMBER PER 100 RESIDUES .
3 1.9 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I-3), SAME NUMBER PER 100 RESIDUES .
0 0.0 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I-2), SAME NUMBER PER 100 RESIDUES .
0 0.0 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I-1), SAME NUMBER PER 100 RESIDUES .
0 0.0 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I+0), SAME NUMBER PER 100 RESIDUES .
0 0.0 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I+1), SAME NUMBER PER 100 RESIDUES .
12 7.7 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I+2), SAME NUMBER PER 100 RESIDUES .
15 9.6 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I+3), SAME NUMBER PER 100 RESIDUES .
47 30.1 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I+4), SAME NUMBER PER 100 RESIDUES .
2 1.3 TOTAL NUMBER OF HYDROGEN BONDS OF TYPE O(I)-->H-N(I+5), SAME NUMBER PER 100 RESIDUES .
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 *** HISTOGRAMS OF *** .
0 0 0 1 0 0 1 0 0 1 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 RESIDUES PER ALPHA HELIX .
1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 PARALLEL BRIDGES PER LADDER .
1 0 0 1 0 2 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ANTIPARALLEL BRIDGES PER LADDER .
0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 LADDERS PER SHEET .
# RESIDUE AA STRUCTURE BP1 BP2 ACC N-H-->O O-->H-N N-H-->O O-->H-N TCO KAPPA ALPHA PHI PSI X-CA Y-CA Z-CA
1 1 A A 0 0 91 0, 0.0 38,-0.1 0, 0.0 5,-0.0 0.000 360.0 360.0 360.0 156.1 38.1 24.6 -5.0
2 2 A A >> - 0 0 86 1,-0.1 4,-1.8 38,-0.1 3,-1.0 -0.377 360.0-116.9 -66.2 144.2 35.1 25.2 -7.3
3 3 A P H 3> S+ 0 0 81 0, 0.0 4,-2.7 0, 0.0 5,-0.2 0.843 111.7 60.4 -55.3 -36.8 34.6 28.8 -8.2
4 4 A A H 3> S+ 0 0 75 1,-0.2 4,-1.6 2,-0.2 5,-0.1 0.876 107.3 46.9 -62.2 -32.5 31.2 29.1 -6.4
5 5 A E H 3> S+ 0 0 75 1,-0.2 4,-1.6 2,-0.2 5,-0.1 0.876 107.3 46.9 -62.2 -32.5 31.2 29.1 -6.4
6 6 A A H <> S+ 0 0 23 -3,-1.0 4,-2.7 2,-0.2 -1,-0.2 0.873 110.0 52.5 -73.0 -37.9 32.9 28.3 -3.1
file3
1 ALA coil
2 ALA coil
3 ARG+ turn
4 ALA helix (helix_alpha, helix1)
5 GLU helix (helix_alpha, helix1)
6 ALA helix
file4
REM --------------- Detailed secondary structure assignment------------- 2EWR
REM 2EWR
REM |---Residue---| |--Structure--| |-Phi-| |-Psi-| |-Area-| 2EWR
ASG ALA A 1 1 C Coil 360.00 156.09 90.6 2EWR
ASG ALA A 2 2 C Coil -66.24 144.20 87.3 2EWR
ASG ARG A 3 3 H AlphaHelix -55.29 -36.84 77.3 2EWR
ASG ALA A 4 4 H AlphaHelix -62.25 -32.49 73.6 2EWR
ASG GLU A 5 5 H AlphaHelix -73.04 -37.93 22.8 2EWR
ASG ALA A 6 6 H AlphaHelix -61.85 -42.14 39.3 2EWR
file5
***********************************
* MAIN CHAIN INFORMATION PANEL *
***********************************
RES. RES. SCND HBOND BTURN RES. FRAC. RES. FRAC. PHI PSI OMEGA PRBLM
NUM. NAME STRUC HBOND BTURN ASA ASA VOL. VOL. PHI PSI OMEGA PRBLM
----------------------------------------------------------------------------------------
Chain A
1 ALA CCC C 83.5 0.42 166.1 1.03 360.0 156.1 178.5
2 ALA CHC C 6A III 93.7 0.38 137.0 0.78 -66.2 144.2 -175.4 V
3 ARG HHH H 7A,8A III 87.9 0.57 109.6 0.95 -55.3 -36.8 179.2
4 ALA HHH H 8A,9A III 63.8 0.34 135.2 1.01 -62.2 -32.5 174.5
5 GLU HHH H 3A,10A III 16.9 0.07 203.8 1.07 -73.0 -37.9 176.2
6 ALA HHH H 4A,11A 33.8 0.16 152.5 0.93 -61.9 -42.1 179.2
file1 is the chemical shift value of six backbone of alanine amino acid atom and other four file are giving its secondary structure information, now what i want is to match the residue no. from file1 with its secondary structure information:
so I want the output file to be like this:
1 8.40 123.54 180.07 3.78 54.98 17.88 | 1 A A | 1 ALA coil | ALA 1 C Coil | 1 ALA CCC C | coil
2 8.70 121.94 182.97 4.68 54.78 | 2 A A >> | 2 ALA coil | ALA 2 C Coil | 2 ALA CHC C | coil
4 8.21 119.84 179.17 4.13 55.48 18.48 | 4 A A H 3> S+ | 4 ALA helix (helix_alpha, helix1) | ALA 4 H AlphaHelix | 4 ALA HHH H | helix
6 7.97 124.04 178.27 3.65 55.48 17.68 | 6 A A H <> S+ | 6 ALA helix (helix_alpha, helix1) | ALA 6 H AlphaHelix | 6 ALA HHH H | helix
and if two or file is showing same secondary structure then it should write that secondary structure at the end otherwise it should write error at the end column.
Thanks
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