Given two permutations A and B of L different elements, L is even, let's call these permutations "symmetric" (for a lack of a better term), if there exist n and m, m > n such as (in python notation):
- A[n:m] == B[L-m:L-n]
- B[n:m] == A[L-m:L-n]
- all other elements are in place
Informally, consider
A = 0 1 2 3 4 5 6 7
Take any slice of it, for example 1 2. It starts at the second index and its length is 2. Now take a slice symmetric to it: it ends at the penultimate index and is 2 chars long too, so it's 5 6. Swapping these slices gives
B = 0 5 6 3 4 1 2 7
Now, A and B are "symmetric" in the above sense (n=1, m=3). On the other hand
A = 0 1 2 3 4 5 6 7
B = 1 0 2 3 4 5 7 6
are not "symmetric" (no n,m with above properties exist).
How can I write an algorithm in python that finds if two given permutations (=lists) are "symmetric" and if yes, find the n and m? For simplicity, let's consider only even L (because the odd case can be trivially reduced to the even one by eliminating the middle fixed element) and assume correct inputs (set(A)==set(B), len(set(A))==len(A)).
(I have no problem bruteforcing all possible symmetries, but looking for something smarter and faster than that).
Fun fact: the number of symmetric permutations for the given L is a Triangular number.
I use this code to test out your answers.
Bounty update: many excellent answers here. #Jared Goguen's solution appears to be the fastest.
Final timings:
testing 0123456789 L= 10
test_alexis ok in 15.4252s
test_evgeny_kluev_A ok in 30.3875s
test_evgeny_kluev_B ok in 27.1382s
test_evgeny_kluev_C ok in 14.8131s
test_ian ok in 26.8318s
test_jared_goguen ok in 10.0999s
test_jason_herbburn ok in 21.3870s
test_tom_karzes ok in 27.9769s
Here is the working solution for the question:
def isSymmetric(A, B):
L = len(A) #assume equivalent to len(B), modifying this would be as simple as checking if len(A) != len(B), return []
la = L//2 # half-list length
Al = A[:la]
Ar = A[la:]
Bl = B[:la]
Br = B[la:]
for i in range(la):
lai = la - i #just to reduce the number of computation we need to perform
for j in range(1, lai + 1):
k = lai - j #same here, reduce computation
if Al[i] != Br[k] or Ar[k] != Bl[i]: #the key for efficient computation is here: do not proceed unnecessarily
continue
n = i #written only for the sake of clarity. i is n, and we can use i directly
m = i + j
if A[n:m] == B[L-m:L-n] and B[n:m] == A[L-m:L-n]: #possibly symmetric
if A[0:n] == B[0:n] and A[m:L-m] == B[m:L-m] and A[L-n:] == B[L-n:]:
return [n, m]
return []
As you have mentioned, though the idea looks simple, but it is actually quite a tricky one. Once we see the patterns, however, the implementation is straight-forward.
The central idea of the solution is this single line:
if Al[i] != Br[k] or Ar[k] != Bl[i]: #the key for efficient computation is here: do not proceed unnecessarily
All other lines are just either direct code translation from the problem statement or optimization made for more efficient computation.
There are few steps involved in order to find the solution:
Firstly, we need to split the each both list Aand list B into two half-lists (called Al, Ar, Bl, and Br). Each half-list would contain half of the members of the original lists:
Al = A[:la]
Ar = A[la:]
Bl = B[:la]
Br = B[la:]
Secondly, to make the evaluation efficient, the goal here is to find what I would call pivot index to decide whether a position in the list (index) is worth evaluated or not to check if the lists are symmetric. This pivot index is the central idea to find an efficient solution. So I would try to elaborate it quite a bit:
Consider the left half part of the A list, suppose you have a member like this:
Al = [al1, al2, al3, al4, al5, al6]
We can imagine that there is a corresponding index list for the mentioned list like this
Al = [al1, al2, al3, al4, al5, al6]
iAl = [0, 1, 2, 3, 4, 5 ] #corresponding index list, added for explanation purpose
(Note: the reason why I mention of imagining a corresponding index list is for ease of explanation purposes)
Likewise, we can imagine that the other three lists may have similar index lists. Let's name them iAr, iBl, and iBr respectively and they are all having identical members with iAl.
It is the index of the lists which would really matter for us to look into - in order to solve the problem.
Here is what I mean: suppose we have two parameters:
index (let's give a variable name i to it, and I would use symbol ^ for current i)
length (let's give a variable name j to it, and I would use symbol == to visually represent its length value)
for each evaluation of the index element in iAl - then each evaluation would mean:
Given an index value i and length value of j in iAl, do
something to determine if it is worth to check for symmetric
qualifications starting from that index and with that length
(Hence the name pivot index come).
Now, let's take example of one evaluation when i = 0 and j = 1. The evaluation can be illustrated as follow:
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 0
== <-- now this has length (j) of 1
In order for those index i and length j to be worth evaluated further, then the counterpart iBr must have the same item value with the same length but on different index (let's name it index k)
iBr = [0, 1, 2, 3, 4, 5]
^ <-- must compare the value in this index to what is pointed by iAl
== <-- must evaluate with the same length = 1
For example, for the above case, this is a possible "symmetric" permutation just for the two lists Al-Br (we will consider the other two lists Ar-Bl later):
Al = [0, x, x, x, x, x] #x means don't care for now
Br = [x, x, x, x, x, 0]
At this moment, it is good to note that
It won't worth evaluating further if even the above condition is not
true
And this is where you get the algorithm to be more efficient; that is, by selectively evaluating only the few possible cases among all possible cases. And how to find the few possible cases?
By trying to find relationship between indexes and lengths of the
four lists. That is, for a given index i and length j in a
list (say Al), what must be the index k in the counterpart
list (in the case is Br). Length for the counterpart list need not
be found because it is the same as in the list (that is j).
Having know that, let's now proceed further to see if we can see more patterns in the evaluation process.
Consider now the effect of length (j). For example, if we are to evaluate from index 0, but the length is 2 then the counterpart list would need to have different index k evaluated than when the length is 1
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 0
===== <-- now this has length (j) of 2
iBr = [0, 1, 2, 3, 4, 5]
^ <-- must compare the value in this index to what is pointed by iAl
===== <-- must evaluate with the same length = 2
Or, for the illustration above, what really matters fox i = 0 and y = 2 is something like this:
# when i = 0 and y = 2
Al = [0, y, x, x, x, x] #x means don't care for now
Br = [x, x, x, x, 0, y] #y means to be checked later
Take a look that the above pattern is a bit different from when i = 0 and y = 1 - the index position for 0 value in the example is shifted:
# when i = 0 and y = 1, k = 5
Al = [0, x, x, x, x, x] #x means don't care for now
Br = [x, x, x, x, x, 0]
# when i = 0 and y = 2, k = 4
Al = [0, y, x, x, x, x] #x means don't care for now
Br = [x, x, x, x, 0, y] #y means to be checked later
Thus, length shifts where the index of the counterpart list must be checked. In the first case, when i = 0 and y = 1, then the k = 5. But in the second case, when i = 0 and y = 1, then the k = 4. Thus we found the pivot indexes relationship when we change the length j for a fixed index i (in this case being 0) unto the counterpart list index k.
Now, consider the effects of index i with fixed length j for counterpart list index k. For example, let's fix the length as y = 4, then for index i = 0, we have:
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 0
========== <-- now this has length (j) of 4
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 1
========== <-- now this has length (j) of 4
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 2
========== <-- now this has length (j) of 4
#And no more needed
In the above example, it can be seen that we need to evaluate 3 possibilities for the given i and j, but if the index i is changed to 1 with the same length j = 4:
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 1
========== <-- now this has length (j) of 4
iAl = [0, 1, 2, 3, 4, 5]
^ <-- now evaluate this index (i) = 2
========== <-- now this has length (j) of 4
Note that we only need to evaluate 2 possibilities. Thus the increase of index i decreases the number of possible cases to be evaluated!
With all the above patterns found, we almost found all the basis we need to make the algorithm works. But to complete that, we need to find the relationship between indexes which appear in Al-Br pair for a given [i, j] => [k, j] with the indexes in Ar-Bl pair for the same [i, j].
Now, we can actually see that they are simply mirroring the relationship we found in Al-Br pair!
(IMHO, this is really beautiful! and thus I think term "symmetric" permutation is not far from truth)
For example, if we have the following Al-Br pair evaluated with i = 0 and y = 2
Al = [0, y, x, x, x, x] #x means don't care for now
Br = [x, x, x, x, 0, y] #y means to be checked later
Then, to make it symmetric, we must have the corresponding Ar-Bl:
Ar = [x, x, x, x, 3, y] #x means don't care for now
Bl = [3, y, x, x, x, x] #y means to be checked later
The indexing of Al-Br pair is mirroring (or, is symmetric to) the indexing of Ar-Bl pair!
Therefore, combining all the pattern we found above, we now could find the pivot indexes for evaluating Al, Ar, Bl, and Br.
We only need to check the values of the lists in the pivot index
first. If the values of the lists in the pivot indexes of Al, Ar, Bl, and Br
matches in the evaluation then and only then we need to check
for symmetric criteria (thus making the computation efficient!)
Putting up all the knowledge above into code, the following is the resulting for-loop Python code to check for symmetricity:
for i in range(len(Al)): #for every index in the list
lai = la - i #just simplification
for j in range(1, lai + 1): #get the length from 1 to la - i + 1
k = lai - j #get the mirror index
if Al[i] != Br[k] or Ar[k] != Bl[i]: #if the value in the pivot indexes do not match
continue #skip, no need to evaluate
#at this point onwards, then the values in the pivot indexes match
n = i #assign n
m = i + j #assign m
#test if the first two conditions for symmetric are passed
if A[n:m] == B[L-m:L-n] and B[n:m] == A[L-m:L-n]: #possibly symmetric
#if it passes, test the third condition for symmetric, the rests of the elements must stay in its place
if A[0:n] == B[0:n] and A[m:L-m] == B[m:L-m] and A[L-n:] == B[L-n:]:
return [n, m] #if all three conditions are passed, symmetric lists are found! return [n, m] immediately!
#passing this but not outside of the loop means
#any of the 3 conditions to find symmetry are failed
#though values in the pivot indexes match, simply continue
return [] #nothing can be found - asymmetric lists
And there go you with the symmetric test!
(OK, this is quite a challenge and it takes quite a while for me to figure out how.)
I rewrote the code without some of the complexity (and errors).
def test_o_o(a, b):
L = len(a)
H = L//2
n, m = 0, H-1
# find the first difference in the left-side
while n < H:
if a[n] != b[n]: break
n += 1
else: return
# find the last difference in the left-side
while m > -1:
if a[m] != b[m]: break
m -= 1
else: return
# for slicing, we want end_index+1
m += 1
# compare each slice for equality
# order: beginning, block 1, block 2, middle, end
if (a[0:n] == b[0:n] and \
a[n:m] == b[L-m:L-n] and \
b[n:m] == a[L-m:L-n] and \
a[m:L-m] == b[m:L-m] and \
a[L-n:L] == b[L-n:L]):
return n, m
The implementation is both elegant and efficient.
The break into else: return structures ensure that the function returns at the soonest possible point. They also validate that n and m have been set to valid values, but this does not appear to be necessary when explicitly checking the slices. These lines can be removed with no noticeable impact on the timing.
Explicitly comparing the slices will also short-circuit as soon as one evaluates to False.
Originally, I checked whether a permutation existed by transforming b into a:
b = b[:]
b[n:m], b[L-m:L-n] = b[L-m:L-n], b[n:m]
if a == b:
return n, m
But this is slower than explicitly comparing the slices. Let me know if the algorithm doesn't speak for itself and I can offer further explanation (maybe even proof) as to why it works and is minimal.
I tried to implement 3 different algorithms for this task. All of them have O(N) time complexity and require O(1) additional space. Interesting fact: all other answers (known so far) implement 2 of these algorithms (though they not always keep optimal asymptotic time/space complexity). Here is high-level description for each algorithm:
Algorithm A
Compare the lists, group "non-equal" intervals, make sure there are exactly two such intervals (with special case when intervals meet in the middle).
Check if "non-equal" intervals are positioned symmetrically, and their contents is also "symmetrical".
Algorithm B
Compare first halves of the lists to guess where are "intervals to be exchanged".
Check if contents of these intervals is "symmetrical". And make sure the lists are equal outside of these intervals.
Algorithm C
Compare first halves of the lists to find first mismatched element.
Find this mismatched element of first list in second one. This hints the position of "intervals to be exchanged".
Check if contents of these intervals is "symmetrical". And make sure the lists are equal outside of these intervals.
There are two alternative implementations for step 1 of each algorithm: (1) using itertools, and (2) using plain loops (or list comprehensions). itertools are efficient for long lists but relatively slow on short lists.
Here is algorithm C with first step implemented using itertools. It looks simpler than other two algorithms (at the end of this post). And it is pretty fast, even for short lists:
import itertools as it
import operator as op
def test_C(a, b):
length = len(a)
half = length // 2
mismatches = it.imap(op.ne, a, b[:half]) # compare half-lists
try:
n = next(it.compress(it.count(), mismatches))
nr = length - n
mr = a.index(b[n], half, nr)
m = length - mr
except StopIteration: return None
except ValueError: return None
if a[n:m] == b[mr:nr] and b[n:m] == a[mr:nr] \
and a[m:mr] == b[m:mr] and a[nr:] == b[nr:]:
return (n, m)
This could be done using mostly itertools:
def test_A(a, b):
equals = it.imap(op.eq, a, b) # compare lists
e1, e2 = it.tee(equals)
l = it.chain(e1, [True])
r = it.chain([True], e2)
borders = it.imap(op.ne, l, r) # delimit equal/non-equal intervals
ranges = list(it.islice(it.compress(it.count(), borders), 5))
if len(ranges) == 4:
n1, m1 = ranges[0], ranges[1]
n2, m2 = ranges[2], ranges[3]
elif len(ranges) == 2:
n1, m1 = ranges[0], len(a) // 2
n2, m2 = len(a) // 2, ranges[1]
else:
return None
if n1 == len(a) - m2 and m1 == len(a) - n2 \
and a[n1:m1] == b[n2:m2] and b[n1:m1] == a[n2:m2]:
return (n1, m1)
High-level description of this algorithm is already provided in OP comments by #j_random_hacker. Here are some details:
Start with comparing the lists:
A 0 1 2 3 4 5 6 7
B 0 5 6 3 4 1 2 7
= E N N E E N N E
Then find borders between equal/non-equal intervals:
= E N N E E N N E
B _ * _ * _ * _ *
Then determine ranges for non-equal elements:
B _ * _ * _ * _ *
[1 : 3] [5 : 7]
Then check if there are exactly 2 ranges (with special case when both ranges meet in the middle), the ranges themselves are symmetrical, and their contents too.
Other alternative is to use itertools to process only half of each list. This allows slightly simpler (and slightly faster) algorithm because there is no need to handle a special case:
def test_B(a, b):
equals = it.imap(op.eq, a, b[:len(a) // 2]) # compare half-lists
e1, e2 = it.tee(equals)
l = it.chain(e1, [True])
r = it.chain([True], e2)
borders = it.imap(op.ne, l, r) # delimit equal/non-equal intervals
ranges = list(it.islice(it.compress(it.count(), borders), 2))
if len(ranges) != 2:
return None
n, m = ranges[0], ranges[1]
nr, mr = len(a) - n, len(a) - m
if a[n:m] == b[mr:nr] and b[n:m] == a[mr:nr] \
and a[m:mr] == b[m:mr] and a[nr:] == b[nr:]:
return (n, m)
This does the right thing:
Br = B[L//2:]+B[:L//2]
same_full = [a==b for (a,b) in zip(A, Br)]
same_part = [a+b for (a,b) in zip(same_full[L//2:], same_full[:L//2])]
for n, vn in enumerate(same_part):
if vn != 2:
continue
m = n
for vm in same_part[n+1:]:
if vm != 2:
break
m+=1
if m>n:
print("n=", n, "m=", m+1)
I'm pretty sure you could do the counting a bit bettter, but... meh
I believe the following pseudocode should work:
Find the first element i for which A[i] != B[i], set n = i. If no such element, return success. If n >= L/2, return fail.
Find the first element i > n for which A[i] == B[i], set m = i. If no such element or m > L/2, set m = L/2.
Check so A[0:n] == B[0:n], A[n:m] == B[L-m:L-n], B[n:m] == A[L-m:L-n], A[m:L-m] == B[m:L-m] and A[L-n:L] == B[L-n:L]. If all are true, return success. Else, return fail.
Complexity is O(n) which should be the lowest possible as one always needs to compare all elements in the lists.
I build a map of where the characters are in list B, then use that to determine the implied subranges in list A. Once I have the subranges, I can sanity check some of the info, and compare the slices.
If A[i] == x, then where does x appear in B? Call that position p.
I know i, the start of the left subrange.
I know L (= len(A)), so I know L-i, the end of the right subrange.
If I know p, then I know the implied start of the right subrange, assuming that B[p] and A[i] are the start of a symmetric pair of ranges. Thus, the OP's L - m would be p if the lists were symmetric.
Setting L-m == p gives me m, so I have all four end points.
Sanity tests are:
n and m are in left half of list(s)
n <= m (note: OP did not prohibit n == m)
L-n is in right half of list (computed)
L-m is in right half (this is a good check for quick fail)
If all those check out, compare A[left] == B[right] and B[left] == A[right]. Return left if true.
def find_symmetry(a:list, b:list) -> slice or None:
assert len(a) == len(b)
assert set(a) == set(b)
assert len(set(a)) == len(a)
length = len(a)
assert length % 2 == 0
half = length // 2
b_loc = {bi:n for n,bi in enumerate(b)}
for n,ai in enumerate(a[:half]):
L_n = length - 1 - n # L - n
L_m = b_loc[ai] # L - m (speculative)
if L_m < half: # Sanity: bail if on wrong side
continue
m = b_loc[a[L_n]] # If A[n] starts range, A[m] ends it.
if m < n or m > half: # Sanity: bail if backwards or wrong side
continue
left = slice(n, m+1)
right = slice(L_m, L_n+1)
if a[left] == b[right] and \
b[left] == a[right]:
return left
return None
res = find_symmetry(
[ 10, 11, 12, 13, 14, 15, 16, 17, ],
[ 10, 15, 16, 13, 14, 11, 12, 17, ])
assert res == slice(1,3)
res = find_symmetry(
[ 0, 1, 2, 3, 4, 5, 6, 7, ],
[ 1, 0, 2, 3, 4, 5, 7, 6, ])
assert res is None
res = find_symmetry("abcdefghijklmn", "nbcdefghijklma")
assert res == slice(0,1)
res = find_symmetry("abcdefghijklmn", "abjklfghicdmen")
assert res == slice(3,4)
res = find_symmetry("abcdefghijklmn", "ancjkfghidelmb")
assert res == slice(3,5)
res = find_symmetry("abcdefghijklmn", "bcdefgaijklmnh")
assert res is None
res = find_symmetry("012345", "013245")
assert res == slice(2,3)
Here's an O(N) solution which passes the test code:
def sym_check(a, b):
cnt = len(a)
ml = [a[i] == b[i] for i in range(cnt)]
sl = [i for i in range(cnt) if (i == 0 or ml[i-1]) and not ml[i]]
el = [i+1 for i in range(cnt) if not ml[i] and (i == cnt-1 or ml[i+1])]
assert(len(sl) == len(el))
range_cnt = len(sl)
if range_cnt == 1:
start1 = sl[0]
end2 = el[0]
if (end2 - start1) % 2 != 0:
return None
end1 = (start1 + end2) // 2
start2 = end1
elif range_cnt == 2:
start1, start2 = sl
end1, end2 = el
else:
return None
if end1 - start1 != end2 - start2:
return None
if start1 != cnt - end2:
return None
if a[start1:end1] != b[start2:end2]:
return None
if b[start1:end1] != a[start2:end2]:
return None
return start1, end1
I only tested it with Python 2, but I believe it will also work with Python 3.
It identifies the ranges where the two lists differ. It looks for two such ranges (if there is only one such range, it tries to divide it in half). It then checks that both ranges are the same length and in the proper positions relative to each other. If so, then it checks that the elements in the ranges match.
Yet another version:
def compare(a, b):
i_zip = list(enumerate(zip(a, b)))
llen = len(a)
hp = llen // 2
def find_index(i_zip):
for i, (x, y) in i_zip:
if x != y:
return i
return i_zip[0][0]
# n and m are determined by the unmoved items:
n = find_index(i_zip[:hp])
p = find_index(i_zip[hp:])
m = llen - p
q = llen - n
# Symmetric?
if a[:n] + a[p:q] + a[m:p] + a[n:m] + a[q:] != b:
return None
return n, m
This solution is based on:
All validly permuted list pairs A, B adhering to the symmetry requirement will have the structure:
A = P1 + P2 + P3 + P4 + P5
B = P1 + P4 + P3 + P2 + P5
^n ^m ^hp ^p ^q <- indexes
,len(P1) == len(P5) and len(P2) == len(P4)
Therefore the 3 last lines of the above function will determine the correct solution provided the indexes n, m are correctly determined. (p & q are just mirror indexes of m & n)
Finding n is a matter of determining when items of A and B start to diverge. Next the same method is applied to finding p starting from midpoint hp. m is just mirror index of p. All involved indexes are found and the solution emerges.
Make a list (ds) of indices where the first halves of the two lists differ.
A possible n is the first such index, the last such index is m - 1.
Check if valid symmetry. len(ds) == m - n makes sure there aren't any gaps.
import itertools as it
import operator as op
def test(a, b):
sz = len(a)
ds = list(it.compress(it.count(), map(op.ne, a[:sz//2], b[:sz//2])))
n,m = ds[0], ds[-1]+1
if a[n:m] == b[sz-m:sz-n] and b[n:m] == a[sz-m:sz-n] and len(ds) == m - n:
return n,m
else:
return None
Here's a simple solution that passes my tests, and yours:
Compare the inputs, looking for a subsequence that does not match.
Transform A by transposing the mismatched subsequence according to the rules. Does the result match B?
The algorithm is O(N); there are no embedded loops, explicit or implicit.
In step 1, I need to detect the case where the swapped substrings are adjacent. This can only happen in the middle of the string, but I found it easier to just look out for the first element of the moved piece (firstval). Step 2 is simpler (and hence less error-prone) than explicitly checking all the constraints.
def compare(A, B):
same = True
for i, (a, b) in enumerate(zip(A,B)):
if same and a != b: # Found the start of a presumed transposition
same = False
n = i
firstval = a # First element of the transposed piece
elif (not same) and (a == b or b == firstval): # end of the transposition
m = i
break
# Construct the transposed string, compare it to B
origin = A[n:m]
if n == 0: # swap begins at the edge
dest = A[-m:]
B_expect = dest + A[m:-m] + origin
else:
dest = A[-m:-n]
B_expect = A[:n] + dest + A[m:-m] + origin + A[-n:]
return bool(B_expect == B)
Sample use:
>>> compare("01234567", "45670123")
True
Bonus: I believe the name for this relationship would be "symmetric block transposition". A block transposition swaps two subsequences, taking ABCDE to ADCBE. (See definition 4 here; I actually found this by googling "ADCBE"). I've added "symmetric" to the name to describe the length conditions.
If I have an input string and an array:
s = "to_be_or_not_to_be"
pos = [15, 2, 8]
I am trying to find the longest common prefix between the consecutive elements of the array pos referencing the original s. I am trying to get the following output:
longest = [3,1]
The way I obtained this is by computing the longest common prefix of the following pairs:
s[15:] which is _be and s[2:] which is _be_or_not_to_be giving 3 ( _be )
s[2:] which is _be_or_not_to_be and s[8:] which is _not_to_be giving 1 ( _ )
However, if s is huge, I don't want to create multiple copies when I do something like s[x:]. After hours of searching, I found the function buffer that maintains only one copy of the input string but I wasn't sure what is the most efficient way to utilize it here in this context. Any suggestions on how to achieve this?
Here is a method without buffer which doesn't copy, as it only looks at one character at a time:
from itertools import islice, izip
s = "to_be_or_not_to_be"
pos = [15, 2, 8]
length = len(s)
for start1, start2 in izip(pos, islice(pos, 1, None)):
pref = 0
for pos1, pos2 in izip(xrange(start1, length), xrange(start2, length)):
if s[pos1] == s[pos2]:
pref += 1
else:
break
print pref
# prints 3 1
I use islice, izip, and xrange in case you're talking about potentially very long strings.
I also couldn't resist this "One Liner" which doesn't even require any indexing:
[next((i for i, (a, b) in
enumerate(izip(islice(s, start1, None), islice(s, start2, None)))
if a != b),
length - max((start1, start2)))
for start1, start2 in izip(pos, islice(pos, 1, None))]
One final method, using os.path.commonprefix:
[len(commonprefix((buffer(s, n), buffer(s, m)))) for n, m in zip(pos, pos[1:])]
>>> import os
>>> os.path.commonprefix([s[i:] for i in pos])
'_'
Let Python to manage memory for you. Don't optimize prematurely.
To get the exact output you could do (as #agf suggested):
print [len(commonprefix([buffer(s, i) for i in adj_indexes]))
for adj_indexes in zip(pos, pos[1:])]
# -> [3, 1]
I think your worrying about copies is unfounded. See below:
>>> s = "how long is a piece of string...?"
>>> t = s[12:]
>>> print t
a piece of string...?
>>> id(t[0])
23295440
>>> id(s[12])
23295440
>>> id(t[2:20]) == id(s[14:32])
True
Unless you're copying the slices and leaving references to the copies hanging around, I wouldn't think it could cause any problem.
edit: There are technical details with string interning and stuff that I'm not really clear on myself. But I'm sure that a string slice is not always a copy:
>>> x = 'google.com'
>>> y = x[:]
>>> x is y
True
I guess the answer I'm trying to give is to just let python manage its memory itself, to begin with, you can look at memory buffers and views later if needed. And if this is already a real problem occurring for you, update your question with details of what the actual problem is.
One way of doing using buffer this is give below. However, there could be much faster ways.
s = "to_be_or_not_to_be"
pos = [15, 2, 8]
lcp = []
length = len(pos) - 1
for index in range(0, length):
pre = buffer(s, pos[index])
cur = buffer(s, pos[index+1], pos[index+1]+len(pre))
count = 0
shorter, longer = min(pre, cur), max(pre, cur)
for i, c in enumerate(shorter):
if c != longer[i]:
break
else:
count += 1
lcp.append(count)
print
print lcp
This is a part of my homework assignment and im close to the final answer but not quite yet. I need to write a function that counts odd numbers in a list.
Create a recursive function count_odd(l) which takes as its only argument a list of integers. The function will return a count of the number of list elements that are odd, i.e., not evenly divisible by 2.\
>>> print count_odd([])
0
>>> print count_odd([1, 3, 5])
3
>>> print count_odd([2, 4, 6])
0
>>> print count_odd([0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144])
8
Here is what i have so far:
#- recursive function count_odd -#
def count_odd(l):
"""returns a count of the odd integers in l.
PRE: l is a list of integers.
POST: l is unchanged."""
count_odd=0
while count_odd<len(l):
if l[count_odd]%2==0:
count_odd=count_odd
else:
l[count_odd]%2!=0
count_odd=count_odd+1
return count_odd
#- test harness
print count_odd([])
print count_odd([1, 3, 5])
print count_odd([2, 4, 6])
print count_odd([0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144])
Can u help explain what im missing. The first two test harness works fine but i cant get the final two. Thanks!
Since this is homework, consider this pseudo-code that just counts a list:
function count (LIST)
if LIST has more items
// recursive case.
// Add one for the current item we are counting,
// and call count() again to process the *remaining* items.
remaining = everything in LIST except the first item
return 1 + count(remaining)
else
// base case -- what "ends" the recursion
// If an item is removed each time, the list will eventually be empty.
return 0
This is very similar to what the homework is asking for, but it needs to be translate to Python and you must work out the correct recursive case logic.
Happy coding.
def count_odd(L):
return (L[0]%2) + count_odd(L[1:]) if L else 0
Are slices ok? Doesn't feel recursive to me, but I guess the whole thing is kind of against usual idioms (i.e. - recursion of this sort in Python):
def countOdd(l):
if l == list(): return 0 # base case, empty list means we're done
return l[0] % 2 + countOdd(l[1:]) # add 1 (or don't) depending on odd/even of element 0. recurse on the rest
x%2 is 1 for odds, 0 for evens. If you are uncomfortable with it or just don't understand it, use the following in place of the last line above:
thisElement = l[0]
restOfList = l[1:]
if thisElement % 2 == 0: currentElementOdd = 0
else: currentElementOdd = 1
return currentElementOdd + countOdd(restOfList)
PS - this is pretty recursive, see what your teacher says if you turn this in =P
>>> def countOdd(l):
... return fold(lambda x,y: x+(y&1),l,0)
...
>>> def fold(f,l,a):
... if l == list(): return a
... return fold(f,l[1:],f(a,l[0]))
All of the prior answers are subdividing the problem into subproblems of size 1 and size n-1. Several people noted that the recursive stack might easily blow out. This solution should keep the recursive stack size at O(log n):
def count_odd(series):
l = len(series) >> 1
if l < 1:
return series[0] & 1 if series else 0
else:
return count_odd(series[:l]) + count_odd(series[l:])
The goal of recursion is to divide the problem into smaller pieces, and apply the solution to the smaller pieces. In this case, we can check if the first number of the list (l[0]) is odd, then call the function again (this is the "recursion") with the rest of the list (l[1:]), adding our current result to the result of the recursion.
def count_odd(series):
if not series:
return 0
else:
left, right = series[0], series[1:]
return count_odd(right) + (1 if (left & 1) else 0)
Tail recursion
def count_odd(integers):
def iter_(lst, count):
return iter_(rest(lst), count + is_odd(first(lst))) if lst else count
return iter_(integers, 0)
def is_odd(integer):
"""Whether the `integer` is odd."""
return integer % 2 != 0 # or `return integer & 1`
def first(lst):
"""Get the first element from the `lst` list.
Return `None` if there are no elements.
"""
return lst[0] if lst else None
def rest(lst):
"""Return `lst` list without the first element."""
return lst[1:]
There is no tail-call optimization in Python, so the above version is purely educational.
The call could be visualize as:
count_odd([1,2,3]) # returns
iter_([1,2,3], 0) # could be replaced by; depth=1
iter_([2,3], 0 + is_odd(1)) if [1,2,3] else 0 # `bool([1,2,3])` is True in Python
iter_([2,3], 0 + True) # `True == 1` in Python
iter_([2,3], 1) # depth=2
iter_([3], 1 + is_odd(2)) if [2,3] else 1
iter_([3], 1 + False) # `False == 0` in Python
iter_([3], 1) # depth=3
iter_([], 1 + is_odd(3)) if [3] else 1
iter_([], 2) # depth=4
iter_(rest([]), 2 + is_odd(first([])) if [] else 2 # bool([]) is False in Python
2 # the answer
Simple trampolining
To avoid 'max recursion depth exceeded' errors for large arrays all tail calls in recursive functions can be wrapped in lambda: expressions; and special trampoline() function can be used to unwrap such expressions. It effectively converts recursion into iterating over a simple loop:
import functools
def trampoline(function):
"""Resolve delayed calls."""
#functools.wraps(function)
def wrapper(*args):
f = function(*args)
while callable(f):
f = f()
return f
return wrapper
def iter_(lst, count):
#NOTE: added `lambda:` before the tail call
return (lambda:iter_(rest(lst), count+is_odd(first(lst)))) if lst else count
#trampoline
def count_odd(integers):
return iter_(integers, 0)
Example:
count_odd([1,2,3])
iter_([1,2,3], 0) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([2,3], 0+is_odd(1)) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([3], 1+is_odd(2)) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([], 1+is_odd(3))
2 # callable(2) is False
I would write it like this:
def countOddNumbers(numbers):
sum = 0
for num in numbers:
if num%2!=0:
sum += numbers.count(num)
return sum
not sure if i got your question , but as above something similar:
def countOddNumbers(numbers):
count=0
for i in numbers:
if i%2!=0:
count+=1
return count
Generator can give quick result in one line code:
sum((x%2 for x in nums))