# [reportlab-users] code128.py

Robin Becker robin at reportlab.com
Mon Jan 25 09:21:01 EST 2016

```On 11/01/2016 16:19, Klaas Feenstra wrote:
> Hi Robin,
>
> Maybe we should first break the list in separate list starting with START_X
> or TO_X and STOP.
> Then check each list, if last 4 position are digits, we do first a backward
> iteration until the first break. So we divide the list in 2 parts, right
> list and left list. After the backward iteration,until first break, we
> continue from the left side (and we remove the right part from the list.)
> If the list is an even number of digits, all string is converted to C in
> the first iteration run.

H Klaas,

I believe I have you ends at stop case covered. See the algorithm below.

1) I believe we only need to check the special case when we break at a non
start. If we break at a start we have to do the start and we won't see a TO_C if
we do this algorithm

2) we apply the backward algorithm to i, i-1, ..... startpos+1. Startpos cannot
be part of the digits. We compute rsavings and then compare with savings.

3) In the comparison the savings ta has to be 1 because l[i] is not STOP and we
didn't stop at a start that means we test

rsavings > savings+int(savings>=0 and (startpos and nl[-1] in starts))-1

and if this is true we fix up the values of rl, nl, startpos and i so it looks
like we did the forward algorithm starting at startpos + 1. That means we will
eliminate the dangling digit. When I run this algorithm I find it applies twice
on the expanded data set.

Can you test this and see if it works for you?

###########################################################################
from string import digits

starts = ['START_B','TO_A','TO_B']
def _trailingDigitsToC(self, l):
# Optimization: trailing digits -> set C double-digits

i = 0
nl = []
while i < len(l):
startpos = i
rl = []
savings = -1 # the TO_C costs one character
while i < len(l):
if l[i] in starts:
j = i
break
elif l[i] == '\xf1':
rl.append(l[i])
i += 1
continue
elif l[i] in digits \
and l[i+1] in digits:
rl.append(l[i] + l[i+1])
i += 2
savings += 1
continue
else:
if l[i] in digits and l[i+1]=='STOP':
rrl = []
rsavings = -1	#we need a TO_C
k = i
while k>startpos:
if l[k]=='\xf1':
rrl.append(l[i])
k -= 1
elif l[k] in digits and l[k-1] in digits:
rrl.append(l[k-1]+l[k])
rsavings += 1
k -= 2
else:
break
rrl.reverse()
if rsavings>savings+int(savings>=0 and (startpos and nl[-1] in starts))-1:
nl += l[startpos]
startpos += 1
rl = rrl
del rrl
i += 1
break
ta = not (l[i]=='STOP' or j==i)
xs = savings>=0 and (startpos and nl[-1] in starts)
if savings+int(xs) > int(ta):
if xs:
toc = nl[-1][:-1]+'C'
del nl[-1]
else:
toc = 'TO_C'
nl += [toc]+rl
if ta:
nl.append('TO'+l[j][-2:])
nl.append(l[i])
else:
nl += l[startpos:i+1]
i += 1
return nl

for x,l in enumerate((
'START_B b b b b 1 2 3 4 5 6 b b b STOP',
'START_B b b b b 1 2 3 4 5 b b b STOP',
'START_B b b b b 1 2 3 4 b b b STOP',
'START_B b b b b 1 2 3 b b b STOP',
'START_B b b b b 1 2 b b b STOP',
'START_B b b b b 1 b b b STOP',
'START_B b b b b 1 2 3 4 5 6 TO_A a a STOP',
'START_B b b b b 1 2 3 4 TO_A a a STOP',
'START_B b b b b 1 2 3 TO_A a a STOP',
'START_B b b b b 1 2 TO_A a a STOP',
'START_B b b b b 1 TO_A a a STOP',
'START_B b b b b 1 2 3 4 5 6 STOP',
'START_B b b b b 1 2 3 4 5 STOP',
'START_B b b b b 1 2 3 4 STOP',
'START_B b b b b 1 2 3 STOP',
'START_B b b b b 1 2 STOP',
'START_B b b b b 1 STOP',
'START_B \xf1 1 2 3 4 b STOP',
'START_B b TO_A a TO_B \xf1 1 2 3 4 b STOP',
'START_B b TO_A a TO_B 1 2 STOP',
'START_B b TO_A a TO_B 1 2 3 b STOP',
'START_B b TO_A a TO_B 1 2 3 4 b STOP',
'START_B b b b b 1 2 3 4 5 6 7 STOP',
'START_B b b b b 1 2 3 4 5 6 7 TO_A a a STOP',
)):
l = l.split()
lo = _trailingDigitsToC(None, l)
print '%2d: %s(%d) --> %s(%d)' % (x,' '.join(l),len(l),' '.join(lo),len(lo))
###########################################################################
--
Robin Becker
```