Generating all ordered substrings of a string

[girish]

Hi,
I want to generate all non-empty substrings of a string of length >=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd', 'c'],
['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

I've tried the following but i cant prevent duplicates and i'm missing
some substrings:

>>colocn = 'abcd'
k = 4
for i in range(k-1):

for j in range(1,k):
rule1 = [colocn[i:i+j],colocn[:i]+colocn[i+j:]]
rule2 = [colocn[:i]+colocn[i+j:],colocn[i:i+j]]
rules.append(ru le1)
rules.append(ru le2)

>>rules

[['a', 'bcd'], ['bcd', 'a'], ['ab', 'cd'], ['cd', 'ab'], ['abc', 'd'],
['d', 'abc'], ['b', 'acd'], ['acd', 'b'], ['bc', 'ad'], ['ad', 'bc'],
['bcd', 'a'], ['a', 'bcd'], ['c', 'abd'], ['abd', 'c'], ['cd', 'ab'],
['ab', 'cd'], ['cd', 'ab'], ['ab', 'cd']]
Any ideas??

TIA,
girish

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[Bruno Desthuilliers]

gi****@it.usyd. edu.au a écrit :

Hi,
I want to generate all non-empty substrings of a string of length >=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd', 'c'],
['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

I've tried the following but i cant prevent duplicates and i'm missing
some substrings:

>>>>colocn = 'abcd'
k = 4
for i in range(k-1):

for j in range(1,k):
rule1 = [colocn[i:i+j],colocn[:i]+colocn[i+j:]]
rule2 = [colocn[:i]+colocn[i+j:],colocn[i:i+j]]
rules.append(ru le1)
rules.append(ru le2)

>>>>rules

[['a', 'bcd'], ['bcd', 'a'], ['ab', 'cd'], ['cd', 'ab'], ['abc', 'd'],
['d', 'abc'], ['b', 'acd'], ['acd', 'b'], ['bc', 'ad'], ['ad', 'bc'],
['bcd', 'a'], ['a', 'bcd'], ['c', 'abd'], ['abd', 'c'], ['cd', 'ab'],
['ab', 'cd'], ['cd', 'ab'], ['ab', 'cd']]
Any ideas??

Algorithmic problem.

First, you need to get all permutations. This is a known algorithm, that
you'll find examples of on the net. Then for each permutation, list
possibles 'pair-splits'.

Here's a (probably sub-optimal, but it's getting late here) possible
implementation in functional style:

def rotate(lst):
yield lst
max = len(lst)
for i in range(1, max):
yield lst[i:] + lst[:-(max-i)]

def permute(lst):
if len(lst) 2:
for rotated in rotate(lst):
head, tail = rotated[0], rotated[1:]
for permuted in permute(tail):
yield [head] + permuted
elif len(lst) == 2:
yield lst
yield lst[::-1]
else:
yield lst

def splits(lst):
for i in range(1, len(lst)):
yield lst[0:i], lst[i:]

def allsplits(lst):
for permuted in permute(lst):
for pair in splits(permuted ):
yield pair

def listsubstrings( thestr):
format = lambda pair: (''.join(pair[0]), ''.join(pair[1]))
return [format(list(pai r)) for pair in allsplits(list( thestr))]
print listsubstrings( "abcd")

[girish]

Quoting Bruno Desthuilliers <bd************ *****@free.quel quepart.fr>:

gi****@it.usyd. edu.au a écrit :

Hi,
I want to generate all non-empty substrings of a string of length
=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd',

'c'],

['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

I've tried the following but i cant prevent duplicates and i'm

missing

some substrings:

>>>colocn = 'abcd'
k = 4
for i in range(k-1):

for j in range(1,k):
rule1 = [colocn[i:i+j],colocn[:i]+colocn[i+j:]]
rule2 = [colocn[:i]+colocn[i+j:],colocn[i:i+j]]
rules.append(ru le1)
rules.append(ru le2)

>>>rules

[['a', 'bcd'], ['bcd', 'a'], ['ab', 'cd'], ['cd', 'ab'], ['abc', 'd'],
['d', 'abc'], ['b', 'acd'], ['acd', 'b'], ['bc', 'ad'], ['ad', 'bc'],
['bcd', 'a'], ['a', 'bcd'], ['c', 'abd'], ['abd', 'c'], ['cd', 'ab'],
['ab', 'cd'], ['cd', 'ab'], ['ab', 'cd']]
Any ideas??

Algorithmic problem.

First, you need to get all permutations. This is a known algorithm, that
you'll find examples of on the net. Then for each permutation, list
possibles 'pair-splits'.

Here's a (probably sub-optimal, but it's getting late here) possible
implementation in functional style:

def rotate(lst):
yield lst
max = len(lst)
for i in range(1, max):
yield lst[i:] + lst[:-(max-i)]

def permute(lst):
if len(lst) 2:
for rotated in rotate(lst):
head, tail = rotated[0], rotated[1:]
for permuted in permute(tail):
yield [head] + permuted
elif len(lst) == 2:
yield lst
yield lst[::-1]
else:
yield lst

def splits(lst):
for i in range(1, len(lst)):
yield lst[0:i], lst[i:]

def allsplits(lst):
for permuted in permute(lst):
for pair in splits(permuted ):
yield pair

def listsubstrings( thestr):
format = lambda pair: (''.join(pair[0]), ''.join(pair[1]))
return [format(list(pai r)) for pair in allsplits(list( thestr))]
print listsubstrings( "abcd")

Thanks Bruno. I wanted to avoid permutations as it would take more time,
but i guess will have to deal with them now :((
Also this one gives each rule twice...i've to search for some double
counting...

>













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[girish]

Quoting Bruno Desthuilliers <bd************ *****@free.quel quepart.fr>:

gi****@it.usyd. edu.au a écrit :

Hi,
I want to generate all non-empty substrings of a string of length
=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd',

'c'],

['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

I've tried the following but i cant prevent duplicates and i'm

missing

some substrings:

>>>colocn = 'abcd'
k = 4
for i in range(k-1):

for j in range(1,k):
rule1 = [colocn[i:i+j],colocn[:i]+colocn[i+j:]]
rule2 = [colocn[:i]+colocn[i+j:],colocn[i:i+j]]
rules.append(ru le1)
rules.append(ru le2)

>>>rules

[['a', 'bcd'], ['bcd', 'a'], ['ab', 'cd'], ['cd', 'ab'], ['abc', 'd'],
['d', 'abc'], ['b', 'acd'], ['acd', 'b'], ['bc', 'ad'], ['ad', 'bc'],
['bcd', 'a'], ['a', 'bcd'], ['c', 'abd'], ['abd', 'c'], ['cd', 'ab'],
['ab', 'cd'], ['cd', 'ab'], ['ab', 'cd']]
Any ideas??

Algorithmic problem.

First, you need to get all permutations. This is a known algorithm, that
you'll find examples of on the net. Then for each permutation, list
possibles 'pair-splits'.

Here's a (probably sub-optimal, but it's getting late here) possible
implementation in functional style:

def rotate(lst):
yield lst
max = len(lst)
for i in range(1, max):
yield lst[i:] + lst[:-(max-i)]

def permute(lst):
if len(lst) 2:
for rotated in rotate(lst):
head, tail = rotated[0], rotated[1:]
for permuted in permute(tail):
yield [head] + permuted
elif len(lst) == 2:
yield lst
yield lst[::-1]
else:
yield lst

def splits(lst):
for i in range(1, len(lst)):
yield lst[0:i], lst[i:]

def allsplits(lst):
for permuted in permute(lst):
for pair in splits(permuted ):
yield pair

def listsubstrings( thestr):
format = lambda pair: (''.join(pair[0]), ''.join(pair[1]))
return [format(list(pai r)) for pair in allsplits(list( thestr))]
print listsubstrings( "abcd")

thanks Bruno....wanted to avoid permute function but i guess i've no no
option :((...
also there is some double counting in this one (all rules outputted
twice)...i've to find out where...

>












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[girish]

Quoting gi****@it.usyd. edu.au:

Quoting Bruno Desthuilliers <bd************ *****@free.quel quepart.fr>:

gi****@it.usyd. edu.au a écrit :

Hi,
I want to generate all non-empty substrings of a string of length
>=2.
Also,
each substring is to be paired with 'string - substring' part and

vice

versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'],

['c',

'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'],

['abc',

'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd',

'c'],

['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]
>
I've tried the following but i cant prevent duplicates and i'm

missing

some substrings:
>
>
>>>>colocn = 'abcd'
>>>>k = 4
>>>>for i in range(k-1):
>
for j in range(1,k):
rule1 = [colocn[i:i+j],colocn[:i]+colocn[i+j:]]
rule2 = [colocn[:i]+colocn[i+j:],colocn[i:i+j]]
rules.append(ru le1)
rules.append(ru le2)
>
>>>>rules
>
[['a', 'bcd'], ['bcd', 'a'], ['ab', 'cd'], ['cd', 'ab'], ['abc',

'd'],

['d', 'abc'], ['b', 'acd'], ['acd', 'b'], ['bc', 'ad'], ['ad',

'bc'],

['bcd', 'a'], ['a', 'bcd'], ['c', 'abd'], ['abd', 'c'], ['cd',

'ab'],

['ab', 'cd'], ['cd', 'ab'], ['ab', 'cd']]
>
>
Any ideas??

Algorithmic problem.

First, you need to get all permutations. This is a known algorithm,

that

you'll find examples of on the net. Then for each permutation, list
possibles 'pair-splits'.

Here's a (probably sub-optimal, but it's getting late here) possible
implementation in functional style:

def rotate(lst):
yield lst
max = len(lst)
for i in range(1, max):
yield lst[i:] + lst[:-(max-i)]

def permute(lst):
if len(lst) 2:
for rotated in rotate(lst):
head, tail = rotated[0], rotated[1:]
for permuted in permute(tail):
yield [head] + permuted
elif len(lst) == 2:
yield lst
yield lst[::-1]
else:
yield lst

def splits(lst):
for i in range(1, len(lst)):
yield lst[0:i], lst[i:]

def allsplits(lst):
for permuted in permute(lst):
for pair in splits(permuted ):
yield pair

def listsubstrings( thestr):
format = lambda pair: (''.join(pair[0]), ''.join(pair[1]))
return [format(list(pai r)) for pair in allsplits(list( thestr))]
print listsubstrings( "abcd")

thanks Bruno....wanted to avoid permute function but i guess i've no no
option :((...
also there is some double counting in this one (all rules outputted
twice)...i've to find out where...

A Recursive Attempt:
def gen(s):
sList = [s[:i]+s[i+1:] for i in range(len(s))]
substringList = sList
s = sList
while len(s[0]) != 1:
substrings = []
for string in s:
sList = [string[:i]+string[i+1:] for i in range(len(strin g))]
substrings.exte nd(sList)
s = set(substrings)
s = list(s)
substringList.e xtend(s)
print substringList
return substringList

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[Tal Einat]

gi****@it.usyd. edu.au wrote:

Hi,
I want to generate all non-empty substrings of a string of length >=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd', 'c'],
['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

In your last example you have ['ac','bd'], but neither 'ac' nor 'bd' is
a _substring_ of 'abcd'.

If you want to compute all possible (non-empty) sub-groups of a group
(a group of characters, in your case), that's really quite a common
algorthmic problem and you should be able to Google for a solution.

Once you have all possible subgroups, just make your (weird) pairs,
remove doubles (either by using a set or by sorting and removing
identical neighboring objects), and you're done.

If you're looking for a more efficient solution, specialized for your
specific problem, you'll have to explain more precisely what you're
trying to do, as well as why existing solutions aren't good enough.

- Tal

[Thorsten Kampe]

* gi****@it.usyd. edu.au (2006-07-11 10:20 +0000)

I want to generate all non-empty substrings of a string of length >=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd', 'c'],
['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

No, you don't want to generate all substrings, you want to generate
all partions of a given set with length 2:

filter(lambda x: len(x) == 2, part(['abcd']))

I've written an utility that generates all possible partions of a set;
the "pairing" as you call it, is trivial, so you can do it yourself

def part(seq):
import copy
partset = [[]]
for item in seq:
newpartset = []
for partition in partset:
for index in range(len(parti tion)):
newpartset.appe nd(copy.deepcop y(partition))
newpartset[-1][index].append(item)
partition.appen d([item])
newpartset.appe nd(partition)
partset = newpartset
return partset

[Gerard Flanagan]

gi****@it.usyd. edu.au wrote:

Hi,
I want to generate all non-empty substrings of a string of length >=2.
Also,
each substring is to be paired with 'string - substring' part and vice
versa.
Thus, ['abc'] gives me [['a', 'bc'], ['bc', 'a'], ['ab', 'c'], ['c',
'ab'], ['b', 'ac'], ['ac', 'b']] etc.
Similarly, 'abcd' should give me [['a', 'bcd'], ['bcd', 'a'], ['abc',
'd'], ['d', 'abc'], ['b', 'acd'], ['acd', 'b'],['c', 'abd'], ['abd', 'c'],
['ab', 'cd'], ['cd', 'ab'], ['bc', 'ad'], ['ad', 'bc'], ['ac',
'bd'],['bd','ac']]

from a previous post
(http://groups.google.com/group/comp....f5b578bb00e61b)

def nkRange(n,k):
m = n - k + 1
indexer = range(0, k)
vector = range(1, k+1)
last = range(m, n+1)
yield vector
while vector != last:
high_value = -1
high_index = -1
for i in indexer:
val = vector[i]
if val high_value and val < m + i:
high_value = val
high_index = i
for j in range(k - high_index):
vector[j+high_index] = high_value + j + 1
yield vector

def kSubsets(s, k):
for vector in nkRange(len(s), k):
yield ''.join( s[i-1] for i in vector )

print list( kSubsets('abcd' ,2) )

['ab', 'ac', 'ad', 'bc', 'bd', 'cd']

[Thorsten Kampe]

* Thorsten Kampe (2006-07-12 19:11 +0000)

filter(lambda x: len(x) == 2, part(['abcd']))

That's "filter(lam bda x: len(x) == 2, part('abcd'))" of course...