Convert a List of Tuples into Dictionary with Grouped Dict Values in Python-CodePudding

Lets say I have a list of tuples like the following

l: list[tuple] = [
    ('bobs trucking', 'ID1'),
    ('bobs trucking', 'ID2'),
    ('bobs trucking', 'ID3'),
    ('bobs groceries', 'ID6'),
    ('bobs groceries', 'ID7'),
    ('bobs groceries', 'ID8'),
    ('bobs groceries', 'ID9')
]

I'd like to group the repeated first tuple elements (e.g. 'bobs trucking') into a dictionary with values that have the grouped IDs concatenated like a CSV but for every n values for each item in a list

# An example of how I'd accomplish this (but need a way to do it for every n elements)
l2: list[str] = ['ID1','ID2','ID3']
csv_delim: str = ','.join(l2)

Every unique key has a list where each list item is a CSV string of size n=3

# Desired ultimate result
bobs_stuff: dict = {
    'bobs trucking': ['ID1,ID2,ID3'],
    'bobs groceries': ['ID6,ID7,ID8','ID9']
}

Note: I'd like to do this without Pandas.

CodePudding user response：

Iterate over your list of tuples and populate a dictionary with its contents as follows:

list_of_tuples = [
    ('bobs trucking', 'ID1'),
    ('bobs trucking', 'ID2'),
    ('bobs trucking', 'ID3'),
    ('bobs groceries', 'ID6'),
    ('bobs groceries', 'ID7'),
    ('bobs groceries', 'ID8'),
    ('bobs groceries', 'ID9')
]

bobs_stuff = {}

N = 3
td = {}

for k, v in list_of_tuples:
    td.setdefault(k, []).append(v)

for k, v in td.items():
    for i in range(0, len(td[k]), N):
        bobs_stuff.setdefault(k, []).append(','.join(v[i:i N]))

print(bobs_stuff)

Output:

{'bobs trucking': ['ID1,ID2,ID3'], 'bobs groceries': ['ID6,ID7,ID8', 'ID9']}

CodePudding user response：

Try:

def chunker(seq, size):
    return (seq[pos : pos   size] for pos in range(0, len(seq), size))


l: list[tuple] = [
    ("bobs trucking", "ID1"),
    ("bobs trucking", "ID2"),
    ("bobs trucking", "ID3"),
    ("bobs groceries", "ID6"),
    ("bobs groceries", "ID7"),
    ("bobs groceries", "ID8"),
    ("bobs groceries", "ID9"),
]

n = 3

out = {}
for k, v in l:
    out.setdefault(k, []).append(v)

for v in out.values():
    v[:] = [",".join(chunk) for chunk in chunker(v, n)]

print(out)

Prints:

{
 "bobs trucking": ["ID1,ID2,ID3"], 
 "bobs groceries": ["ID6,ID7,ID8", "ID9"]
}