def make_entry(hit,padded=True):
'''Turns a record into a fasta entry. The 'SEQ' field is set to a default,
if not already present: 'NAME' reflects subject source
information about the hit as well as the specified group, while 'SEQ' is
set to 'SSEQ', padded on the left with *fillchar* (default '-') to fit the
query location unless *padded* is set to False.
'''
hit = hit.copy() ; hit.open()
hit.setdefault('SEQ',_fillchar*(hit.QSTART-1)*padded + hit.pop('SSEQ'))
return fasta.seq_entry(hit)
def resolve_query_overlap(standalones,nests,overlap):
'''Expects a list of standalone fragments and a list of nests. Nests are
expected to have undergone the subject overlap truncation scheme (see
*stratify*). The return value is a list of fasta entries. If the function
detects no query overlap between any pair of fragments -- including those
in nests -- the fragments are "assembled" (non-technical term) in order
of query-ordinates into a single fasta entry, which is the only element of
the returned list.
'''
standalones = list(standalones)
nests = map(list,nests)
if not (standalones or any(nests)):
raise Error('Tried to resolve query overlap on an empty set of records!')
# assign names before reordering
for j,hit in enumerate(standalones): setname(hit,'standalone[{}]'.format(j))
for i,nest in enumerate(nests,1):
for j,hit in enumerate(nest): setname(hit,'nest{}[{}]'.format(i,j))
# and then reorder by query ordinate
recs = sorted(_it.chain(standalones,*nests),key=_attrget('QSTART'))
if any(q_overlap(x,y)>=overlap for x,y in _it.izip(recs,recs[1:]))\
or len(recs)==1: return _it.imap(make_entry,recs)
prev = None
with _cont.closing(_sIO()) as seq:
for hit in recs:
seq.write('-'*(hit.QSTART-1-( prev and prev.QEND or 0 )))
if prev is None: st,end = hit._SSTART,hit._SEND
else: st,end = min(st,hit._SSTART),max(end,hit._SEND);
seq.write(hit.SSEQ)
prev = hit
result = fasta.seq_entry({'SEQ': seq.getvalue(),
'NAME': _name_fmt.format(_GRP='all',SSEQID=recs[0].SSEQID,
SSTART=min(h._SSTART for h in recs),SEND=max(h._SEND for h in recs))})
return [result]