Group emails into TO & CC with itertools.groupby and convert it to a dictionary-CodePudding

I'd like to group emails by their domain and convert the result into a dictionary. So far I have figured out that itertools.groupby with a custom func will do that. It correctly assigns keys to each value, but when I try to create a dictionary only the last value is used when the values to be grouped are not continues.


import re
from itertools import groupby

{k: list(v) for k, v in groupby(["bar", "foo", "baz"], key=lambda x: "to" if re.search(r"^b", x) else "cc")}

This will produce {'to': ['baz'], 'cc': ['foo']} instead of {'to': ['bar', 'baz'], 'cc': ['foo']}.

How I can fix that?

CodePudding user response：

Sort the group first to get correct result (itertools.groupby groups continuous items):

import re
from itertools import groupby

out = {
    k: list(v)
    for k, v in groupby(
        sorted(
            ["awol", "bar", "foo", "baz"],
            key=lambda x: bool(re.search(r"^b", x)),
        ),
        key=lambda x: "to" if re.search(r"^b", x) else "cc",
    )
}

print(out)

Prints:

{'cc': ['awol', 'foo'], 'to': ['bar', 'baz']}

CodePudding user response：

You can use dict.setdefault OR collections.defaultdict(list) and extend in list like below.

# from collections import defaultdict
# dct = defaultdict(list)

from itertools import groupby
import re

dct = {}
for k, v in groupby(["awol", "bar", "foo", "baz"], 
                    key=lambda x: "to" if re.search(r"^b", x) else "cc"):
    dct.setdefault(k,[]).extend(list(v))

    # If you use 'dct = defaultdict(list)'. You can add item in 'list' like below
    # dct[k].extend(list(v))
print(dct)

{'cc': ['awol', 'foo'], 'to': ['bar', 'baz']}