Merge nested list items based on a repeating value

Question

Although poorly written, this code:

marker_array = [['hard','2','soft'],['heavy','2','light'],['rock','2','feather'],['fast','3'], ['turtle','4','wet']]
marker_array_DS = []

for i in range(len(marker_array)):
    if marker_array[i-1][1] != marker_array[i][1]:            
        marker_array_DS.append(marker_array[i])

print marker_array_DS

Returns:

[['hard', '2', 'soft'], ['fast', '3'], ['turtle', '4', 'wet']]

It accomplishes part of the task which is to create a new list containing all nested lists except those that have duplicate values in index [1]. But what I really need is to concatenate the matching index values from the removed lists creating a list like this:

[['hard heavy rock', '2', 'soft light feather'], ['fast', '3'], ['turtle', '4', 'wet']]

The values in index [1] must not be concatenated. I kind of managed to do the concatenation part using a tip from another post:

newlist = [i + n for i, n in zip(list_a, list_b]

But I am struggling with figuring out the way to produce the desired result. The "marker_array" list will be already sorted in ascending order before being passed to this code. All like-values in index [1] position will be contiguous. Some nested lists may not have any values beyond [0] and [1] as illustrated above.

Is it on purpose that one list only has two elements instead of three ? — SylvainD
– SylvainD, Commented Jun 24, 2013 at 22:50
Yes. The intention is to illustrate that some nested list may be shorter. My real lists will have up to five elements, but not fewer than four. Basically, the value at index position [4] is a text note that may or may not be there. — I_Ridanovic
– I_Ridanovic, Commented Jun 25, 2013 at 5:22

Jon Clements · Accepted Answer · 2013-06-24 22:52:54Z

2

Quick stab at it... use itertools.groupby to do the grouping for you, but do it over a generator that converts the 2 element list into a 3 element.

from itertools import groupby
from operator import itemgetter

marker_array = [['hard','2','soft'],['heavy','2','light'],['rock','2','feather'],['fast','3'], ['turtle','4','wet']]  

def my_group(iterable):
    temp = ((el + [''])[:3] for el in marker_array)
    for k, g in groupby(temp, key=itemgetter(1)):
        fst, snd = map(' '.join, zip(*map(itemgetter(0, 2), g)))
        yield filter(None, [fst, k, snd])

print list(my_group(marker_array))

answered Jun 24, 2013 at 22:52

Jon Clements

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Elazar · Accepted Answer · 2020-11-10 20:11:37Z

1

from collections import defaultdict
d1 = defaultdict(list)
d2 = defaultdict(list)
for pxa in marker_array:
    d1[pxa[1]].extend(pxa[:1])
    d2[pxa[1]].extend(pxa[2:])

res = [[' '.join(d1[x]), x, ' '.join(d2[x])] for x in sorted(d1)]

If you really need 2-tuples (which I think is unlikely):

for p in res:
    if not p[-1]:
        p.pop()

edited Nov 10, 2020 at 20:11

answered Jun 24, 2013 at 22:56

Elazar

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7 Comments

7stud Over a year ago

Ooooh. Nice...but you can't rely on the order of 'for x in d1'. So you should write: for x in sorted(d1.keys()).

Elazar Over a year ago

@7stud I can't see the requirement for ordering in the question.

7stud Over a year ago

The "marker_array" list will be already sorted in ascending order before being passed to this code. It's pretty easy to surmise from that statement that the results should be in sorted order.

Jon Clements Over a year ago

Worth noting that p, x, *a style unpacking will only work in 3.x

Elazar Over a year ago

@JonClements so I guess pxa[2:] is better? I'll change it.

|

Jean-Bernard Pellerin · Accepted Answer · 2013-06-24 22:58:33Z

0

marker_array = [['hard','2','soft'],['heavy','2','light'],['rock','2','feather'],['fast','3'], ['turtle','4','wet']]
marker_array_DS = []
marker_array_hit = []

for i in range(len(marker_array)):
    if marker_array[i][1] not in marker_array_hit:
        marker_array_hit.append(marker_array[i][1])

for i in marker_array_hit:
    lists = [item for item in marker_array if item[1] == i]
    temp = []
    first_part = ' '.join([str(item[0]) for item in lists])
    temp.append(first_part)
    temp.append(i)
    second_part = ' '.join([str(item[2]) for item in lists if len(item) > 2])
    if second_part != '':
        temp.append(second_part);
    marker_array_DS.append(temp)

print marker_array_DS

I learned python for this because I'm a shameless rep whore

answered Jun 24, 2013 at 22:58

Jean-Bernard Pellerin

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1 Comment

I_Ridanovic Over a year ago

Wow, these are all great solutions. I need to dig in and understand what each does. They all do it correctly.

7stud · Accepted Answer · 2013-06-24 22:59:56Z

0

marker_array = [
    ['hard','2','soft'],
    ['heavy','2','light'],
    ['rock','2','feather'],
    ['fast','3'], 
    ['turtle','4','wet'],
]

data = {}

for arr in marker_array:
    if len(arr) == 2:
        arr.append('')

    (first, index, last) = arr
    firsts, lasts = data.setdefault(index, [[],[]])
    firsts.append(first)
    lasts.append(last)


results = []

for key in sorted(data.keys()):
    current = [
        " ".join(data[key][0]),
        key,
        " ".join(data[key][1])
    ]

    if current[-1] == '':
        current = current[:-1]

    results.append(current)



print results

--output:--
[['hard heavy rock', '2', 'soft light feather'], ['fast', '3'], ['turtle', '4', 'wet']]

answered Jun 24, 2013 at 22:59

7stud

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Comments

Jakub Wasilewski · Accepted Answer · 2013-06-24 23:43:02Z

A different solution based on itertools.groupby:

from itertools import groupby

# normalizes the list of markers so all markers have 3 elements
def normalized(markers):
    for marker in markers:
        yield marker + [""] * (3 - len(marker))

def concatenated(markers):
  # use groupby to iterator over lists of markers sharing the same key
  for key, markers_in_category in groupby(normalized(markers), lambda m: m[1]):
    # get separate lists of left and right words
    lefts, rights = zip(*[(m[0],m[2]) for m in markers_in_category])
    # remove empty strings from both lists
    lefts, rights = filter(bool, lefts), filter(bool, rights)
    # yield the concatenated entry for this key (also removing the empty string at the end, if necessary)
    yield filter(bool, [" ".join(lefts), key, " ".join(rights)])

The generator concatenated(markers) will yield the results. This code correctly handles the ['fast', '3'] case and doesn't return an additional third element in such cases.

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