Split words into alphabetic dictionary rxpy

Question

I am learning RxPY , so I need to write some code which can split each word by its first character. The results must look something like this:

{'a': ['a'], 't': ['the','the'], 'l': ['low','lazy']}

What I've tried.

from rx import Observable , Observer

list =['A', 'The', 'the', 'LAZY', 'Low']

o = Observable . from_ ( list )\
. filter (lambda i: i[0] == 'A' or 'Z' )\

words = o.map(lambda s: s.lower().split())

word_each = words.flat_map(lambda s: s)

ss = word_each.to_dict(lambda x: x[:1], lambda x : x)\
    .subscribe(lambda val: print(val))

So, how can I solve this problem? I am thinking about grouping each word by it's first character but I dont know how.

[CLOSED]

On SO it's *de rigueur* to mark an answer accepted if it does what you asked for. — Bill Bell, Nov 05 '17 at 19:06
Thank for all you, especially for Bill. I was busy looking through documentation of rx. — Alesha, Nov 05 '17 at 19:09
@Alesha - offered a modern solution using rxjs v6. appreciate this is an old question, however it would help others reviewing this Q&A if you care to mark as the correct answer. — arcseldon, May 06 '19 at 06:21

0TTT0 · Answer 1 · 2017-11-05T18:33:39.237

1

I know nothing about RxPy, but you can do it in one line with vanilla python using dict and list comps

  d = {el[0].lower(): [e.lower() for e in lst if e[0].lower() == el[0].lower()] for el in lst}

edited Nov 05 '17 at 18:33

answered Nov 05 '17 at 18:26

0TTT0

1,288
1
13
23

Bill Bell · Answer 2 · 2017-11-05T19:05:08.923

Using rx, if you really want to do it in a difficult way.

>>> from rx import Observable
>>> from collections import defaultdict
>>> source = Observable.from_(['A', 'The', 'the', 'LAZY', 'Low'])
>>> result = defaultdict(list)
>>> def add(value):
...     value_content = '{0}'.format(value)
...     result[value_content[0].lower()].append(value_content)
... 
>>> s = source.subscribe(on_next=lambda value: add(value), on_completed=lambda: print('Finished'))
Finished
>>> result
defaultdict(<class 'list'>, {'a': ['A'], 't': ['The', 'the'], 'l': ['LAZY', 'Low']})

Use dict(result) as shown in Chiheb Nexus' answer if you need this in the form of a dictionary.

I changed the name of your list to my_list.

>>> my_list =['A', 'The', 'the', 'LAZY', 'Low']
>>> from itertools import groupby
>>> {key: list(val) for (key, val) in  groupby(sorted(my_list), key=lambda x: x[0].lower())}
{'a': ['A'], 't': ['The', 'the'], 'l': ['LAZY', 'Low']}

doesn't match her output `{'a': ['a'], 't': ['the','the'], 'l': ['low','lazy']}`, just lower your strings — 0TTT0, Nov 05 '17 at 18:31

score 0 · Answer 3 · answered Nov 05 '17 at 18:53

0

You can use defaultdict module to accomplish the task. Here is an example:

from collections import defaultdict
my_list = ['A', 'The', 'the', 'LAZY', 'Low']

def split_words(a):
    b = defaultdict(list)
    for k in a:
        b[k[0].lower()].append(k.lower())
    return b

>>> split_words(my_list)
    defaultdict(list, {'a': ['a'], 'l': ['lazy', 'low'], 't': ['the', 'the']})
>>> dict(split_words(my_list))
    {'a': ['a'], 'l': ['lazy', 'low'], 't': ['the', 'the']}

answered Nov 05 '17 at 18:53

Chiheb Nexus

9,104
4
30
43

1

I thought you had scooped me but I don't think this is what she's asking for. – Bill Bell Nov 05 '17 at 18:55
@BillBell Yes indeed. But no one is answering to this question using RxPy, so i thought that she can use another work around if she is in really need. – Chiheb Nexus Nov 05 '17 at 18:57
I'd never heard of rx, I read through the intro and wrote an answer. – Bill Bell Nov 05 '17 at 18:59
@BillBell Well, you've answered using the right tool and your answer should be validated ... Otherwise, if it's a homework, it doesn't matter a lot, at least for me, because i never heard of RxPy too. I'm learning from your answer. – Chiheb Nexus Nov 05 '17 at 19:01
1

Read the intro, it's probably better! – Bill Bell Nov 05 '17 at 19:02

arcseldon · Answer 4 · 2019-05-05T20:25:50.923

Update May 2019 using RxPy v6

RxPy really shines when performing these types of transformations (although acknowledged you could equally apply a pragmatic solution in plain python).

Here is a straightforward solution:

from rx import Observable

list = ['A', 'The', 'the', 'LAZY', 'Low']

Observable.from_(list) \
    .map(lambda s: s.lower()) \
    .group_by(lambda s: s[0]) \
    .flat_map(lambda grp: grp.to_list()) \
    .to_dict(lambda grp: grp[0][0], lambda grp: grp) \
    .subscribe(print)

Output as requested:

# {'a': ['a'], 't': ['the', 'the'], 'l': ['lazy', 'low']}

The "algorithm" is:

Make the list of words an Observable.
Map over each word and convert to lowercase.
Group each word by its first letter.
Use flatmap to unnest as we convert each grouping to a list. At this point, the interim result looks as follows:
```
['a']
['the', 'the']
['lazy', 'low']
```
Now, use the to_dict operator to convert to a single dictionary, whose keys are the first letter of the first item in each grouping (list).
Subscribe to the Observable to execute the pipeline.

Split words into alphabetic dictionary rxpy

4 Answers4