How to print with itertools groupby?

Using multiple print statements with transpose

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

group = groupby(text)

# Transpose so we have row 0 with group names, and row 1 with grouping items
transposed = [(k, list(g)) for k, g in group]

for k in transposed:
    print(k[0], end = " ")
print()
for k in transposed:
    print(sum(1 for _ in k[1]), end=" ")

You can do it like this which post-processes the results groupby returns to make it easy to get the values needed for each row of the output:

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

groups = [(k, str(sum(1 for _ in g))) for k, g in groupby(text)]
a, b = zip(*groups)
print(' '.join(a))  # -> a b c
print(' '.join(b))  # -> 2 2 1

Here is one more alternative:

from itertools import groupby

text = ["a", "a", "b", "b", "c"]

group = groupby(text)
dictionary = {}

for k, g in group:
    dictionary[k] = sum(1 for _ in g)

keys = " ".join(list(dictionary.keys()))
values = " ".join(str(v) for v in list(dictionary.values()))

print(keys)
print(values)

I know you asked about itertools.groupby, BUT I would use a Counter for such task:

>>> text = ["a", "a", "b", "b", "c"]
>>> from collections import Counter
>>> c = Counter(text)
>>> print(c)
Counter({'a': 2, 'b': 2, 'c': 1})
>>> headers = c.keys()
>>> values = [str(val) for val in c.values()]
>>> print(' '.join(headers))
a b c
>>> print(' '.join(values))
2 2 1

Since groupby() does not lend itself to deploying tee() to make a copy, the best solution seems to be to create a tuple comprehension. Note that we're not interested in the values contained in each group g, so we'll just store the length of the tuple constructed on the fly from the group.

import itertools

text = ["a", "a", "b", "b", "c"]

group = tuple((k,len(tuple(v))) for k,v in itertools.groupby(text))

for t in group:
    print(t[0], end= " ")
print()
for t in group:
    print(t[1], end=" ")
print()
# a b c
# 2 2 1