Further simplifying and shortening a simple grouping algorithm

Question

This is the input list

input_list = ['string','string','1','string','2','string','string','3','string','5']

This is the output required

    [['string', 'string', '1'], ['string', '2'], ['string', 'string', '3'], 
     ['string', '5']]

Basically you're grouping the list data string and (numeric data) into a list of lists, each list containing the first numeric data and previous string data encountered

A simple algorithm which can accomplish this

output_list = []
inner_list = []
for data in input_list:
    inner_list.append(data)
    if data.isdigit():
        output_list.append(inner_list)
        inner_list = []

print output_list

I was wondering if this could be made even simpler (accomplished in lesser number of lines)?

you can save 1 line `inner_list.append(data)` (can be done unconditionally) and after the positive check `if data.isdigit():` you don't need the `else`. Elimination of two lines — Ronald, Aug 01 '15 at 09:06
@Ronald Yes, edited :) I meant could this be accomplished in a much more simpler way without creating all these `inner_list` temp variables — wolfgang, Aug 01 '15 at 09:10

score 3 · Answer 1 · answered Aug 01 '15 at 09:17

3

One liner with reduce. Do not use.

reduce(lambda n, e: n[:-1]+[n[-1]+[e]] + ([[]] if e.isdigit() else []), input_list, [[]])[:-1]

answered Aug 01 '15 at 09:17

Maltysen

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too computationally expensive? – wolfgang Aug 01 '15 at 09:29

Mazdak · Answer 2 · 2015-08-01T09:23:37.123

3

You can first find the index of digit elements then use zip to create the desire slice indices and slice you list :

>>> ind=[0]+[i for i,j in enumerate(input_list,1) if j.isdigit()]
>>> [input_list[i:j] for i,j in zip(ind,ind[1:])]
[['string', 'string', '1'], ['string', '2'], ['string', 'string', '3'], ['string', '5']]

edited Aug 01 '15 at 09:23

answered Aug 01 '15 at 09:17

Mazdak

105,000
18
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mhawke · Answer 3 · 2015-08-01T09:42:48.457

Even though this is a single line solution, it is not simpler, it's more complex:

input_list = ['string','string','1','string','2','string','string','3','string','5']
result = [s.split() for s in ' '.join([s+'\n' if s.isdigit() else s for s in input_list]).splitlines()]
print result

Output:

[['string', 'string', '1'], ['string', '2'], ['string', 'string', '3'], ['string', '5']]

Is this blob of complex code what you want?

(Note the bug: if one of the strings should already contain a \n. You could choose another character if you know for sure that it won't be in one of the strings).

score 1 · Answer 4 · answered Aug 01 '15 at 09:11

If you want a (slightly evil) reduce()-based version:

def reduce_string_prefix(cur, item):
    cur[-1].append(item)
    if item.isdigit():
        cur.append([])
    return cur

input_list = ['string', 'string', '1', 'string', '2', 'string', 'string', '3', 'string', '5']
output_list = reduce(reduce_string_prefix, input_list, [[]])[:-2]

I'm not so sure if this is actually easier to read, though ;-).

Further simplifying and shortening a simple grouping algorithm

4 Answers4