B `?@sdZddlZddlmZddlmZmZmZmZm Z m Z m Z m Z m Z mZddlZddlmZmZmZdddd d d d d ddddddddddddddddddd d!d"gZd#d ZdDd$dZd%dZdEd&dZdFd'dZd(dZefd)dZd*dZeZd+dZd,d Z d-d Z!dGd.dZ"d/d0Z#ydd1lm$Z%Wne&k rDe#Z$YnXd2dZ$e#je$_dHd3d Z'd4dZ(d5dZ)d6dZ*dId7d!Z+dJd8d"Z,dKd9d Z-dLd:d Z.d;d<d=dZ/dMd>dZ0d?dZ1d@dZ2dAdZ3dBdZ4dCdZ5dS)NaImported from the recipes section of the itertools documentation. All functions taken from the recipes section of the itertools library docs [1]_. Some backward-compatible usability improvements have been made. .. [1] http://docs.python.org/library/itertools.html#recipes N)deque) chain combinationscountcyclegroupbyislicerepeatstarmaptee zip_longest) randrangesamplechoice all_equalconsumeconvolve dotproduct first_trueflattengrouper iter_exceptncyclesnthnth_combinationpadnonepad_nonepairwise partitionpowersetprependquantify#random_combination_with_replacementrandom_combinationrandom_permutationrandom_product repeatfunc roundrobintabulatetailtakeunique_everseenunique_justseencCstt||S)zReturn first *n* items of the iterable as a list. >>> take(3, range(10)) [0, 1, 2] If there are fewer than *n* items in the iterable, all of them are returned. >>> take(10, range(3)) [0, 1, 2] )listr)niterabler0Q/tmp/pip-install-r497f_s2/setuptools/setuptools/_vendor/more_itertools/recipes.pyr*<s cCst|t|S)aReturn an iterator over the results of ``func(start)``, ``func(start + 1)``, ``func(start + 2)``... *func* should be a function that accepts one integer argument. If *start* is not specified it defaults to 0. It will be incremented each time the iterator is advanced. >>> square = lambda x: x ** 2 >>> iterator = tabulate(square, -3) >>> take(4, iterator) [9, 4, 1, 0] )mapr)functionstartr0r0r1r(LscCstt||dS)zReturn an iterator over the last *n* items of *iterable*. >>> t = tail(3, 'ABCDEFG') >>> list(t) ['E', 'F', 'G'] )maxlen)iterr)r.r/r0r0r1r)^scCs,|dkrt|ddntt|||ddS)aXAdvance *iterable* by *n* steps. If *n* is ``None``, consume it entirely. Efficiently exhausts an iterator without returning values. Defaults to consuming the whole iterator, but an optional second argument may be provided to limit consumption. >>> i = (x for x in range(10)) >>> next(i) 0 >>> consume(i, 3) >>> next(i) 4 >>> consume(i) >>> next(i) Traceback (most recent call last): File "", line 1, in StopIteration If the iterator has fewer items remaining than the provided limit, the whole iterator will be consumed. >>> i = (x for x in range(3)) >>> consume(i, 5) >>> next(i) Traceback (most recent call last): File "", line 1, in StopIteration Nr)r5)rnextr)iteratorr.r0r0r1ris cCstt||d|S)zReturns the nth item or a default value. >>> l = range(10) >>> nth(l, 3) 3 >>> nth(l, 20, "zebra") 'zebra' N)r7r)r/r.defaultr0r0r1rs cCst|}t|dot|d S)z Returns ``True`` if all the elements are equal to each other. >>> all_equal('aaaa') True >>> all_equal('aaab') False TF)rr7)r/gr0r0r1rs cCstt||S)zcReturn the how many times the predicate is true. >>> quantify([True, False, True]) 2 )sumr2)r/predr0r0r1r!scCst|tdS)aReturns the sequence of elements and then returns ``None`` indefinitely. >>> take(5, pad_none(range(3))) [0, 1, 2, None, None] Useful for emulating the behavior of the built-in :func:`map` function. See also :func:`padded`. N)rr )r/r0r0r1rs cCsttt||S)zvReturns the sequence elements *n* times >>> list(ncycles(["a", "b"], 3)) ['a', 'b', 'a', 'b', 'a', 'b'] )r from_iterabler tuple)r/r.r0r0r1rscCstttj||S)zcReturns the dot product of the two iterables. >>> dotproduct([10, 10], [20, 20]) 400 )r;r2operatormul)Zvec1Zvec2r0r0r1rscCs t|S)zReturn an iterator flattening one level of nesting in a list of lists. >>> list(flatten([[0, 1], [2, 3]])) [0, 1, 2, 3] See also :func:`collapse`, which can flatten multiple levels of nesting. )rr=)Z listOfListsr0r0r1rs cGs&|dkrt|t|St|t||S)aGCall *func* with *args* repeatedly, returning an iterable over the results. If *times* is specified, the iterable will terminate after that many repetitions: >>> from operator import add >>> times = 4 >>> args = 3, 5 >>> list(repeatfunc(add, times, *args)) [8, 8, 8, 8] If *times* is ``None`` the iterable will not terminate: >>> from random import randrange >>> times = None >>> args = 1, 11 >>> take(6, repeatfunc(randrange, times, *args)) # doctest:+SKIP [2, 4, 8, 1, 8, 4] N)r r )functimesargsr0r0r1r&sccs*t|\}}t|dt||EdHdS)zReturns an iterator of paired items, overlapping, from the original >>> take(4, pairwise(count())) [(0, 1), (1, 2), (2, 3), (3, 4)] On Python 3.10 and above, this is an alias for :func:`itertools.pairwise`. N)r r7zip)r/abr0r0r1 _pairwises  rG)rccst|EdHdS)N)itertools_pairwise)r/r0r0r1rscCs<t|tr tdt||}}t|g|}t|d|iS)zCollect data into fixed-length chunks or blocks. >>> list(grouper('ABCDEFG', 3, 'x')) [('A', 'B', 'C'), ('D', 'E', 'F'), ('G', 'x', 'x')] z+grouper expects iterable as first parameter fillvalue) isinstanceintwarningswarnDeprecationWarningr6r )r/r.rIrCr0r0r1rs   cgsnt|}tdd|D}xN|rhyx|D] }|Vq(WWqtk rd|d8}tt||}YqXqWdS)aJYields an item from each iterable, alternating between them. >>> list(roundrobin('ABC', 'D', 'EF')) ['A', 'D', 'E', 'B', 'F', 'C'] This function produces the same output as :func:`interleave_longest`, but may perform better for some inputs (in particular when the number of iterables is small). css|]}t|jVqdS)N)r6__next__).0itr0r0r1 9szroundrobin..N)lenr StopIterationr) iterablespendingZnextsr7r0r0r1r',s  csFdkr tfdd|D}t|\}}dd|Ddd|DfS)a Returns a 2-tuple of iterables derived from the input iterable. The first yields the items that have ``pred(item) == False``. The second yields the items that have ``pred(item) == True``. >>> is_odd = lambda x: x % 2 != 0 >>> iterable = range(10) >>> even_items, odd_items = partition(is_odd, iterable) >>> list(even_items), list(odd_items) ([0, 2, 4, 6, 8], [1, 3, 5, 7, 9]) If *pred* is None, :func:`bool` is used. >>> iterable = [0, 1, False, True, '', ' '] >>> false_items, true_items = partition(None, iterable) >>> list(false_items), list(true_items) ([0, False, ''], [1, True, ' ']) Nc3s|]}||fVqdS)Nr0)rPx)r<r0r1rRZszpartition..css|]\}}|s|VqdS)Nr0)rPcondrXr0r0r1rR]scss|]\}}|r|VqdS)Nr0)rPrYrXr0r0r1rR^s)boolr )r<r/Z evaluationst1t2r0)r<r1rCs   cs,t|tfddttdDS)aYields all possible subsets of the iterable. >>> list(powerset([1, 2, 3])) [(), (1,), (2,), (3,), (1, 2), (1, 3), (2, 3), (1, 2, 3)] :func:`powerset` will operate on iterables that aren't :class:`set` instances, so repeated elements in the input will produce repeated elements in the output. Use :func:`unique_everseen` on the input to avoid generating duplicates: >>> seq = [1, 1, 0] >>> list(powerset(seq)) [(), (1,), (1,), (0,), (1, 1), (1, 0), (1, 0), (1, 1, 0)] >>> from more_itertools import unique_everseen >>> list(powerset(unique_everseen(seq))) [(), (1,), (0,), (1, 0)] c3s|]}t|VqdS)N)r)rPr)sr0r1rRvszpowerset..rS)r-rr=rangerT)r/r0)r^r1rbsc cst}|j}g}|j}|dk }xb|D]Z}|r4||n|}y||krP|||VWq$tk r|||krx|||VYq$Xq$WdS)a Yield unique elements, preserving order. >>> list(unique_everseen('AAAABBBCCDAABBB')) ['A', 'B', 'C', 'D'] >>> list(unique_everseen('ABBCcAD', str.lower)) ['A', 'B', 'C', 'D'] Sequences with a mix of hashable and unhashable items can be used. The function will be slower (i.e., `O(n^2)`) for unhashable items. Remember that ``list`` objects are unhashable - you can use the *key* parameter to transform the list to a tuple (which is hashable) to avoid a slowdown. >>> iterable = ([1, 2], [2, 3], [1, 2]) >>> list(unique_everseen(iterable)) # Slow [[1, 2], [2, 3]] >>> list(unique_everseen(iterable, key=tuple)) # Faster [[1, 2], [2, 3]] Similary, you may want to convert unhashable ``set`` objects with ``key=frozenset``. For ``dict`` objects, ``key=lambda x: frozenset(x.items())`` can be used. N)setaddappend TypeError) r/keyZseensetZ seenset_addZseenlistZ seenlist_addZuse_keyelementkr0r0r1r+ys  cCsttttdt||S)zYields elements in order, ignoring serial duplicates >>> list(unique_justseen('AAAABBBCCDAABBB')) ['A', 'B', 'C', 'D', 'A', 'B'] >>> list(unique_justseen('ABBCcAD', str.lower)) ['A', 'B', 'C', 'A', 'D'] rS)r2r7r? itemgetterr)r/rdr0r0r1r,s ccs<y"|dk r|Vx |VqWWn|k r6YnXdS)aXYields results from a function repeatedly until an exception is raised. Converts a call-until-exception interface to an iterator interface. Like ``iter(func, sentinel)``, but uses an exception instead of a sentinel to end the loop. >>> l = [0, 1, 2] >>> list(iter_except(l.pop, IndexError)) [2, 1, 0] Nr0)rA exceptionfirstr0r0r1rs cCstt|||S)a Returns the first true value in the iterable. If no true value is found, returns *default* If *pred* is not None, returns the first item for which ``pred(item) == True`` . >>> first_true(range(10)) 1 >>> first_true(range(10), pred=lambda x: x > 5) 6 >>> first_true(range(10), default='missing', pred=lambda x: x > 9) 'missing' )r7filter)r/r9r<r0r0r1rsrS)r cGs$dd|D|}tdd|DS)aDraw an item at random from each of the input iterables. >>> random_product('abc', range(4), 'XYZ') # doctest:+SKIP ('c', 3, 'Z') If *repeat* is provided as a keyword argument, that many items will be drawn from each iterable. >>> random_product('abcd', range(4), repeat=2) # doctest:+SKIP ('a', 2, 'd', 3) This equivalent to taking a random selection from ``itertools.product(*args, **kwarg)``. cSsg|] }t|qSr0)r>)rPpoolr0r0r1 sz"random_product..css|]}t|VqdS)N)r)rPrkr0r0r1rRsz!random_product..)r>)r rCpoolsr0r0r1r%scCs*t|}|dkrt|n|}tt||S)abReturn a random *r* length permutation of the elements in *iterable*. If *r* is not specified or is ``None``, then *r* defaults to the length of *iterable*. >>> random_permutation(range(5)) # doctest:+SKIP (3, 4, 0, 1, 2) This equivalent to taking a random selection from ``itertools.permutations(iterable, r)``. N)r>rTr)r/r]rkr0r0r1r$s cs8t|t}ttt||}tfdd|DS)zReturn a random *r* length subsequence of the elements in *iterable*. >>> random_combination(range(5), 3) # doctest:+SKIP (2, 3, 4) This equivalent to taking a random selection from ``itertools.combinations(iterable, r)``. c3s|]}|VqdS)Nr0)rPi)rkr0r1rRsz%random_combination..)r>rTsortedrr_)r/r]r.indicesr0)rkr1r#s cs@t|ttfddt|D}tfdd|DS)aSReturn a random *r* length subsequence of elements in *iterable*, allowing individual elements to be repeated. >>> random_combination_with_replacement(range(3), 5) # doctest:+SKIP (0, 0, 1, 2, 2) This equivalent to taking a random selection from ``itertools.combinations_with_replacement(iterable, r)``. c3s|]}tVqdS)N)r )rPrn)r.r0r1rRsz6random_combination_with_replacement..c3s|]}|VqdS)Nr0)rPrn)rkr0r1rRs)r>rTror_)r/r]rpr0)r.rkr1r"s c Cst|}t|}|dks ||kr$td}t|||}x*td|dD]}|||||}qFW|dkrr||7}|dks||krtg}xj|r||||d|d}}}x.||kr||8}|||||d}}qW||d|qWt|S)aEquivalent to ``list(combinations(iterable, r))[index]``. The subsequences of *iterable* that are of length *r* can be ordered lexicographically. :func:`nth_combination` computes the subsequence at sort position *index* directly, without computing the previous subsequences. >>> nth_combination(range(5), 3, 5) (0, 3, 4) ``ValueError`` will be raised If *r* is negative or greater than the length of *iterable*. ``IndexError`` will be raised if the given *index* is invalid. rrS)r>rT ValueErrorminr_ IndexErrorrb) r/r]indexrkr.crfrnresultr0r0r1r"s(  cCs t|g|S)aYield *value*, followed by the elements in *iterator*. >>> value = '0' >>> iterator = ['1', '2', '3'] >>> list(prepend(value, iterator)) ['0', '1', '2', '3'] To prepend multiple values, see :func:`itertools.chain` or :func:`value_chain`. )r)valuer8r0r0r1r Ls ccslt|ddd}t|}tdg|d|}x:t|td|dD]"}||tttj ||VqBWdS)aBConvolve the iterable *signal* with the iterable *kernel*. >>> signal = (1, 2, 3, 4, 5) >>> kernel = [3, 2, 1] >>> list(convolve(signal, kernel)) [3, 8, 14, 20, 26, 14, 5] Note: the input arguments are not interchangeable, as the *kernel* is immediately consumed and stored. Nrqr)r5rS) r>rTrrr rbr;r2r?r@)signalkernelr.ZwindowrXr0r0r1r[s  )r)N)N)N)N)N)N)N)NN)N)6__doc__rL collectionsr itertoolsrrrrrrr r r r r?randomr rr__all__r*r(r)rrrrZr!rrrrrr&rGrrH ImportErrorrr'rrr+r,rrr%r$r#r"rr rr0r0r0r1 s 0   (        -   *