Rearanging Data Using Dictionary

I have a dataset and I extracted three columns contain: pdb name and chain 1 and chain 2.

I want to write a file with pdb , chains related 2 it but remove redundant combinations

I tried to use both a list and dictionary for the same purpose,

I wanted something like;

ABC A B
ABC c D

and I am getting,

ABC A B
ABC C D
ABC B A

since A, B and B, A are the same combination I want to remove the redundancy...

heres a part of my code..

pdb_name  = cols[4].strip(string.punctuation)

    chain_1= cols[5].strip(string.punctuation)
    chain_2 = cols[10].strip(string.punctuation)
    if a.has_key(pdb_name):
        a[pdb_name][chain_1] = a[pdb_name].get(chain_1, {})
        a[pdb_name][chain_1][chain_2] =   a[pdb_name][chain_1].get(chain_2, +1)
    else:
        a[pdb_name]= a.get(pdb_name, {})
        a[pdb_name][chain_1] = a[pdb_name].get(chain_1, {})
        a[pdb_name][chain_1][chain_2] =   a[pdb_name][chain_1].get(chain_2, +1)
   #     a[pdb_name][chain_1].append(chain_2)
     #   a[pdb_name] = [chain_1]
      #  a[pdb_name].append(chain_2)

###    if a.key() in list:
#        chain_1

        # if there is a second combination why its not coming ??


outfile_f = open('pdbextract.txt', 'w')    
aa = a.keys()
aa.sort()

for l in aa:

    te = a[l].keys()
    te.sort()
    for m in te:
       tf = a[l][m].keys()
       tf.sort()
       for n in tf:
           l = l.strip(string.punctuation)
           m = m.strip(string.punctuation)
           n = n.strip(string.punctuation)
           outfile_f.write("%-4s%-1s%-1s%-1s%-1s\n" % ( l, '\t ', m, '\t ', n))

python programming

• 3,272 views • link

updated 15.6 years ago by brentp

• written 15.6 years ago by Samihk •

2 answers

If I understand correctly, and you sort the keys, you'll get the same for A, B and B, A. And if you do a bit of initialization with the python collections.defaultdict, you can make the code a lot more readable. Something like this should work

# initialize a to take form a['pdb']['A']['B'] += 1 
a = collections.defaultdict(lambda \
     :collections.defaultdict(lambda: collections.defaultdict(int)))

# later
pdb_name  = cols[4].strip(string.punctuation)

chain_1= cols[5].strip(string.punctuation)
chain_2 = cols[10].strip(string.punctuation)

# so B, A is sorted to A, B
chain_a, chain_b = sorted((chain_1, chain_2))

a[pdb_name][chain_a][chain_b] += 1

Once that's done, the logic to do the printing is much simpler.

• 0 views • link

updated 6.9 years ago by Ram

• written 15.6 years ago by brentp

i just love defaultdict ... I don't think I go any project without using it at least once

• 0 views • link

written 15.6 years ago by Will

Why not just check that chain_1 <= chain_2 and if not, switch them before adding to your dict.

You can eliminate a bit of redundancy by using dict.setdefault() like so:

a.setdefault(pdb_name, {})
a[pdb_name].setdefault(chain_1, {})
a[pdb_name][chain_1].setdefault(chain_2, 1)

Also, you are stripping punctuation twice (when you read it in and before writing)...

And in your last line, is there any reason not to leave the tabs in the write string, rather than substituting them in?

outfile_f.write("%-4s\t%-1s\t%-1s\n" % ( l, m, n))

• 0 views • link

updated 6.9 years ago by Ram

• written 15.6 years ago by Jashapiro

Thanks a lot, I am a newbie so I dont know many rules or easy solution!

• 0 views • link

written 15.6 years ago by Samihk •