split list of tuples in lists of list of tuples

Question

I have a list of tuples like:

[(1,a),(2,b), (1, e), (3, b), (2,c), (4,d), (1, b), (0,b), (6, a), (8, e)]

I want to split it into list of lists at every "b"

[[(1,a),(2,b)], [(1, e), (3, b)], [(2,c), (4,d), (1, b)], [(0,b)], [(6, a), (8, e)]]

is there any pythonic way to do this?

I'd claim that a straight-forward `for` loop here is pythonic as the intention might not be clear if you mangle this into some comprehension-generator-thingy. Explicit is better then implicit. — dornhege, Mar 20 '14 at 10:52

Jayanth Koushik · Answer 1 · 2014-03-20T11:00:09.560

3

You can use yield.

def get_groups(lst):
    t = []
    for i in lst:
        t.append(i)
        if i[1] == 'b':
            yield t
            t = []
    if t:
        yield t

my_list = [(1,a),(2,b), (1, e), (3, b), (2,c), (1, b), (0,b)]
groups = list(get_groups(my_list))

edited Mar 20 '14 at 11:00

answered Mar 20 '14 at 10:54

Jayanth Koushik

9,476
1
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1

Does not work if the last element does not end with `b`. You should check if `t` contains elements and yield it in this case at the end of your function. – sloth Mar 20 '14 at 10:57
1

What should do if the last one doesnt end in 'b', include it or not? – Raul Guiu Mar 20 '14 at 11:10
It should be included – Abhishek Thakur Mar 20 '14 at 11:16
1

That is not nice. Changing the question afterwards. – Raul Guiu Mar 20 '14 at 11:21
@Avempace I added one more tuple to make it clear that even if it doesnt end in "b", it should return the full list as a sublist – Abhishek Thakur Mar 20 '14 at 11:31
Thanks! Changed the answer for that. I was experimenting anyway. I would not do that way in production as Janyanth said – Raul Guiu Mar 20 '14 at 11:37
The code fails for huge lists with no "b". So I accepted the answer by @thefourtheye although I like generators – Abhishek Thakur Mar 21 '14 at 09:33
@AbhishekThakur: we both have the same solution except that this uses a generator. Why would this fail for 'huge lists'? – Jayanth Koushik Mar 21 '14 at 09:36
Sorry. Its not huge lists. It produces repetitions and the generated list becomes "huge" – Abhishek Thakur Mar 21 '14 at 09:38

score 1 · Accepted Answer · answered Mar 20 '14 at 10:52

my_list = [(1, "a"),(2, "b"), (1, "e"), (3, "b"), (2, "c"), (1, "b"), (0, "b")]
result, temp = [], []

for item in my_list:
    temp.append(item)
    if item[1] == 'b':
        result.append(temp)
        temp = []

if len(temp) > 0:
    result.append(temp)

print result
# [[(1, 'a'), (2, 'b')], [(1, 'e'), (3, 'b')], [(2, 'c'), (1, 'b')], [(0, 'b')]]

Raul Guiu · Answer 3 · 2014-03-20T11:33:20.710

1

My solution, directly in the python console:

>>> l = [(1,'a'),(2,'b'), (1, 'e'), (3, 'b'), (2,'c'), (1, 'b'), (0,'b')]
>>> b = [-1] + [x for x, y in enumerate(l) if y[1] == 'b' or x == len(l)-1]
>>> u = zip(b,b[1:])
>>> m = [l[x[0]+1:x[1]+1] for x in u]
>>> m
[[(1, 'a'), (2, 'b')], [(1, 'e'), (3, 'b')], [(2, 'c'), (1, 'b')], [(0, 'b')]]

b is the indices of the tuples with 'b', starting by -1.

[-1, 1, 3, 5, 6]

u is tuples of indices of the sublist that we will create:

[(-1, 1), (1, 3), (3, 5), (5, 6)]

And for the case not ending in a tuple with 'b':

[(1, 'a'), (2, 'b'), (1, 'e'), (3, 'b'), (2, 'c'), (1, 'b'), (0, 'b'), (6, 'a'), (8, 'e')]

gives:

[[(1, 'a'), (2, 'b')], [(1, 'e'), (3, 'b')], [(2, 'c'), (1, 'b')], [(0, 'b')], [(6, 'a'), (8, 'e')]]

edited Mar 20 '14 at 11:33

answered Mar 20 '14 at 11:08

Raul Guiu

2,374
22
37

This is less intuitive *and* less efficient than a simple for loop. – Jayanth Koushik Mar 20 '14 at 11:11
The questions says "is there any pythonic way to do this?". Only showing other alternatives. – Raul Guiu Mar 20 '14 at 11:12
Intuitiveness is a big part of 'pythonic'. – Jayanth Koushik Mar 20 '14 at 11:13
1

This fails if last item doesn't contain `'b'`. – Ashwini Chaudhary Mar 20 '14 at 11:15
What should it give? I tried, and split in blocks ending by the tuple with the b. Are you assuming something not specify in the question? – Raul Guiu Mar 20 '14 at 11:17
fixed now, by adding x == len(l)-1 – Raul Guiu Mar 20 '14 at 11:33

score 1 · Answer 4 · answered Mar 20 '14 at 11:57

1

What about this?

xs = [(1, 'a'), (2, 'b'), (1, 'e'), (3, 'b'), (2, 'c'), (4, 'd'), (1, 'b'), (0, 'b'), (6, 'a'), (8, 'e')]

result = [[]]
for x in xs:
  result[0].append(x)
  if x[1] == 'b':
    result.insert(0, [])
result.reverse()

answered Mar 20 '14 at 11:57

Aegis

1,749
11
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score 1 · Answer 5 · answered Mar 20 '14 at 12:01

A pure iterator solution:

def get_groups(lst):
    i = iter(lst)
    def row(x):
        while True:
            yield x
            if x[1] == 'b':
                return
            x = next(i)

    for x in i:
        yield row(x)

for r in get_groups(data):
    print list(r)

split list of tuples in lists of list of tuples

5 Answers5

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