making index with python

Question

the following is my question:

data = ['a1_1 00000001\n', 'a1_2 00000002\n', 'a1_3 00000003\n',
        'b1_1 00000004\n', 'b1_2 00000005\n', 'b1_3 00000006']

candidate = ['a', 'b']

for i in candidate:
    for j in data:
        if i in j.split()[0]:
            print i, j.split()[1]

a 00000001
a 00000002
a 00000003
b 00000004
b 00000005
b 00000006

But what I want to do is to make the result like the following:

a 00000001, 00000002, 00000003
b 00000004, 00000005, 00000006

How do I solve this problem? Thanks in advance!

Answer 1

data = ['a1_1 00000001\n', 'a1_2 00000002\n', 'a1_3 00000003\n', 'b1_1 00000004\n', 'b1_2 00000005\n', 'b1_3 00000006']

candidate = ['a', 'b']


for i in candidate:
    print i, ', '.join(v for k,v in (a.split() for a in data) if k.startswith(i))

prints:

a 00000001, 00000002, 00000003
b 00000004, 00000005, 00000006

This will, however, go through data several times (once for each candidate). To do it once only, you will need an intermediary dictionary:

d = {}
for a in data:
    d.setdefault(a[0], []).append(a.split()[1])
for k,v in sorted(d.iteritems()):
    print k, ', '.join(v)

Answer 2

One way is to store the intermediate result in a variable and then print it out outside of your loop.

E.g. something like (not tested):

candidate_print = {}
for i in candidate:
    for j in data:
        if i in j.split()[0]:
            if not i in candidate_print:
                candidate_print[i] = []
            candidate_print[i] += [j.split()[1]]

# Now print it out
for i in candidate:
    print i, ", ".join(candidate_print[i])