Python: create dictionary with multi list comprehension

Question

Is there a way I can use list comprehension answered in this thread to create dictionary?

listA = [
    "apple_v001",
    "apple_v002",
    "banana_v001",
    "orange_v001",
]
keywords = ["apple", "banana", "orange"]
[[item for item in listA if kw in item] for kw in keywords]
# Result: [['apple_v001', 'apple_v002'], ['banana_v001'], ['orange_v001']] #

What I'm trying to do is create a dictionary using keywords as key out of this result. So

dictA["apple"] = ['apple_v001', 'apple_v002']

and so forth. I tried to do the dict = {key, value for ...(iteration) } but always get a syntax error. I really don't know how to start, any help appreciate.

why is the tag `python-2.7` if you use `python-2.6`? – jfs Mar 05 '13 at 04:39 — jfs, Mar 05 '13 at 04:39

John La Rooy · Answer 1 · 2013-03-05T12:43:36.033

2

{kw: [item for item in listA if kw in item] for kw in keywords}

But this doesn't seem a particularly efficient way to create such a dict

For example this doesn't need the list of keywords in advance and is reasonably efficient

>>> from itertools import groupby
>>> {k:list(g) for k,g in groupby(sorted(listA), key=lambda x:x.partition('_')[0])}
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

For Python2.6 the equivalent is

dict((kw, [item for item in listA if kw in item]) for kw in keywords)

and

>>> from itertools import groupby
>>> dict((k,list(g)) for k,g in groupby(sorted(listA), key=lambda x:x.partition('_')[0]))
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

edited Mar 05 '13 at 12:43

answered Mar 05 '13 at 04:00

John La Rooy

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Very interesting. Thanks for showing me. Is this specific for Python 3? Because I'm having a syntax error using Python 2.6. – Panupat Mar 05 '13 at 04:31
@Panupat, 2.7+. I'll add a note for 2.6 to the answer – John La Rooy Mar 05 '13 at 12:43

score 2 · Answer 2 · answered Mar 05 '13 at 04:01

2

if you don't want to go for one liner solution, check this

In [58]: d
Out[58]: defaultdict(<type 'list'>, {})

In [59]: for elem in keywords:
   ....:     for item in listA:
   ....:         if item.startswith(elem):
   ....:             d[elem].append(item)
   ....:

In [60]: d
Out[60]: defaultdict(<type 'list'>, {'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']})

answered Mar 05 '13 at 04:01

avasal

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This code is so much easier to read than the one-liner above! – JesseBikman Mar 05 '13 at 04:39

score 1 · Answer 3 · edited May 23 '17 at 11:52

In the comments you mentioned that you use Python 2.6. There is no dict comprehension in Python 2.6, you could use dict() with a generator expresion instead:

d = dict((kw, [item for item in listA if kw in item]) for kw in keywords)

Here's a possibly more efficient version:

import re
from collections import defaultdict

search_word = re.compile("(%s)" % "|".join(map(re.escape, keywords))).search

d = defaultdict(list)
for item in listA:
    m = search_word(item)
    if m:
       d[m.group(1)].append(item)

If listA is always in the format given in the question:

from collections import defaultdict

keywords = set(keywords)
d = defaultdict(list)
for item in listA:
    word = item.partition("_")[0]
    if word in keywords:
       d[word].append(item)

If listA doesn't contain items that are not in keywords:

from collections import defaultdict

d = defaultdict(list)
for item in listA:
    d[item.partition('_')[0]].append(item)

score 0 · Answer 4 · answered Mar 05 '13 at 05:04

You can use a regex:

>>> import re
>>> listA = [
...     "apple_v001",
...     "apple_v002",
...     "banana_v001",
...     "orange_v001",
... ]
>>> keywords = ["apple", "banana", "orange"]
>>> s=' '.join(listA)
>>> dict([(e,re.findall(r'{}_v\d+'.format(e),s)) for e in keywords])
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

Or (post Python 2.7) a dict comprehension:

>>> {e:re.findall(r'{}_v\d+'.format(e),s) for e in keywords}
{'orange': ['orange_v001'], 'apple': ['apple_v001', 'apple_v002'], 'banana': ['banana_v001']}

Python: create dictionary with multi list comprehension

4 Answers4