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Question

Answer 1

+4 A:

import itertools

l = ['1', 'a', 'b','---', '2','c','d','---','3','123','e','---','4']
r = []

i = iter(l)
while True:
  a = [x for x in itertools.takewhile(lambda x: x != '---', i)]
  if not a:
    break
  r.append(a)
print r

# [['1', 'a', 'b'], ['2', 'c', 'd'], ['3', '123', 'e'], ['4']]

Ignacio Vazquez-Abrams 2010-07-12 20:22:00

+1 I like this better then my similar solution (will delete it shortly) because of the efficient `iter` approach (only consume it once). But you probably mean to replace `for j in range(7)` with `while True` to handle arbitrary lengths.

ChristopheD 2010-07-12 20:35:07

@ChristopheD: Yes, fixed.

Ignacio Vazquez-Abrams 2010-07-12 20:37:38

Answer 2

+13 A:

In [17]: import itertools
# putter around 22 times
In [39]: l=['1', 'a', 'b','---', '2','c','d','---','3','123','e','---','4']

In [40]: [list(g) for k,g in itertools.groupby(l,'---'.__ne__) if k]
Out[40]: [['1', 'a', 'b'], ['2', 'c', 'd'], ['3', '123', 'e'], ['4']]

unutbu 2010-07-12 20:25:34

+1 Nice (didn't immediately think of `itertools.groupby` here but it fits the bill indeed)

ChristopheD 2010-07-12 20:37:29

Thank you for your answer. Is there a way to check x=='---' in the above line with a regular expression (something like x==re.match('-'))? Many thanks

DGT 2010-07-13 00:37:17

Yes, you could use something like `[list(g) for k,g in itertools.groupby(l,lambda x: re.match('---',x)) if not k]`. The expression `re.match(...)` returns None when `x` does not match the pattern. Thus `k` is `None` for the elements you want to keep. So I changed the condition to `if not k`.

unutbu 2010-07-13 01:16:10

great! thanks a lot.

DGT 2010-07-13 01:21:33

You can use `'---'.__ne__` instead of the lambda function

gnibbler 2010-07-13 11:13:34

@gnibbler: Oh, that's slick!

unutbu 2010-07-13 12:11:16

Answer 3

A:

It's been a while since I've done any python so my syntax is going to be way off, but a simple loop should suffice.

Keep track of the indexes in two numbers

firstList = ['1', 'a', 'b','---', '2','c','d','---','3','123','e','---','4']
listIndex = 0
itemIndex = 0
ii = 0
foreach item in firstList
  if(firstList[ii] == '---') 
    listIndex = listIndex + 1
    itemIndex = 0
    ii = ii + 1
  else secondList[listIndex][itemIndex] = firstList[ii]

mphair 2010-07-12 20:26:12

Answer 4

A:

import itertools

a = ['1', 'a', 'b','---', '2','c','d','---','3','123','e','---','4']
b = [list(x[1]) for x in itertools.groupby(a, '---'.__eq__) if not x[0]]

print b     # or print(b) in Python 3

Result is

[['1', 'a', 'b'], ['2', 'c', 'd'], ['3', '123', 'e'], ['4']]

JAB 2010-07-12 20:26:19

Answer 5

A:

Here's a solution without itertools:

def foo(input):
    output = []
    currentGroup = []
    for value in input:
        if '-' in value:  #if we should break on this element 
            currentGroup.append( value )
        elif currentGroup:
            output.append( currentGroup )
            currentGroup = []
    if currentGroup:
        output.append(currentGroup) #appends the rest if not followed by separator    
    return output

print ( foo ( ['1', 'a', 'b','---', '2','c','d','---','3','123','e','---','4'] ) )

CrazyJugglerDrummer 2010-07-12 20:37:43

Beyond a certain threshold, readability is relative. That just looks like a big mass of code to me.

Ignacio Vazquez-Abrams 2010-07-12 20:42:22

@Crazy: Using (1) `value.find('-') == -1` instead of `'-' in value` (2) camelCase (2) redundant parentheses in `(el)if` sometimes (3) extraneous spaces inside parentheses sometimes (4) statement long enough to make SO insert the ferschlugginer horizontal scroll bar (5) no space after comma sometimes makes it look like a big mess of fugly code to me.

John Machin 2010-07-12 21:16:29

You're both right. Edited code and description. :)

CrazyJugglerDrummer 2010-07-13 20:25:06

Answer 6

A:

Here's one way to do it:

lst=['1', 'a', 'b','---', '2','c','d','---','3','123','e','---','4']
indices=[-1]+[i for i,x in enumerate(lst) if x=='---']+[len(lst)]
answer=[lst[indices[i-1]+1:indices[i]] for i in xrange(1,len(indices))]
print answer

Basically, this finds the locations of the string '---' in the list and then slices the list accordingly.

MAK 2010-07-12 20:59:29

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