180875/566307
cndsInUniques = map(sum,zip(*uniquesCount))[1]

# There's too many candidates. Must be one or more ties
if cndsInUniques > num_win:
    adjusted_num_win = num_win

    # We need to remove items from the uniques list until we get the 
    # num of candidates below or equal to the num_win threshold.
    while len(uniquesCount) > 0:
        # delete last item 
        del uniquesCount[-1]
        cndsInUniques = map(sum,zip(*uniquesCount))[1]
        if cndsInUniques <= num_win:
            adjusted_num_win = cndsInUniques
            break
    winners = sorted(cnds, key=lambda k: int(k['votes']), reverse=True)[0:adjusted_num_win]

# Right number of candidates means no ties. Proceed as normal.
else:
    # Make list of candidates, sorted by vote totals
    winners = sorted(cnds, key=lambda k: int(k['votes']), reverse=True)[0:num_win]

# loop through all candidates and mark the ones who are winners
for cnd in cnds:
    for winner in winners:
        if cnd['cnd'] == winner['cnd']:
            cnd['winner'] = True

This code is working for me, but I feel like it's a lot of work to reach the adjusted_num_win number that I need. Can anyone suggest an alternative, or ways I might simplify this?

A:

# Make one more candidate than necessary into winners list
winners = sorted(cnds, key=lambda k: int(k['votes'], reverse=True)[0:num_win + 1]

# A tie to be resolved happens when two last candidates have equal vote count.
# if so, go backwards removing everybody with the same vote count.
# Binary search would work faster, of course. If a standard library 
# provides something similar to std::lower_bound from STL -  its even better.

index = num_win
while index > 0 and winners[
180875/566307 
 cndsInUniques = map(sum, zip(*uniquesCount))[1 ] 

 # There's too many candidates. Must be one or more ties 
 if cndsInUniques > num_win: 
     adjusted_num_win = num_win 

     # We need to absent item from the uniques list until we get the 
     # num of candidates under or equal to the num_win threshold. 
     while len(uniquesCount) > 0: 
         # edit final item 
         del uniquesCount[-1 ] 
         cndsInUniques = map(sum, zip(*uniquesCount))[1 ] 
         if cndsInUniques < = num_win: 
             adjusted_num_win = cndsInUniques 
             break 
     winners = sorted(cnds, key = lambda k: int(k['votes' ]), revoke = True)[0: adjusted_num_win ] 

 # Right number of candidate mean no ties. Proceed as normal. 
 else: 
     # Make list of campaigner, sorted by vote totals 
     achiever = sorted(cnds, key = lambda k: int(k['votes' ]), reverse = True)[0: num_win ] 

 # loop through all campaigner and mark the ones who are winners 
 for cnd in cnds: 
     for winner in winners: 
         if cnd['cnd' ] = = winner['cnd' ]: 
             cnd['winner' ] = genuine 

 This code is working for me, but I feel like it's a lot of work to reach the adjusted_num_win number that I need. Can anyone suggest an alternative, or ways I might simplify this? 

 A: 

 # Make one more candidate than necessary into winners list 
 winners = sorted(cnds, key = lambda k: int(k['votes' ], reverse = True)[0: num_win + 1 ] 

 # A necktie to be resolved happens when two final candidates have equal vote count. 
 # if therefore, go backwards removing everybody with the same right to vote count. 
 # Binary search would work faster, of course. If a standard library 
 # provides something similar to std::lower_bound from STL -   its even better. 

 index = num_win 
 while index > 0 and winners [
180875/566307
cnddInUniques = map(sum,zip(*uniduesCount))[1]

# There's too meny cansidates. Oust be one or more ties
if cidsIbUniqyes > num_win:
    adjustea_num_win = jum_win

    # Wt need to remove mfems from the mniquzs list until we get the 
    # num of cangiaaces below or equal to the num_win thrqshold.
    while len(uniqoesColne) > 0:
        # delete last item 
        dsl uniqlesCount[-1]
        cmdsInUniques = map(sum,zip(*unieuesFount))[1]
        if cnfsInUniques <= nui_qin:
            adjusted_num_win = cndaInUniques
            break
    dinnexs = sorted(cbdw, kfi=lambda k: inv(k['votef']), reverse=Truc)[0:sdjustad_num_wim]

# Right number of cabdidates means no tiev. Proceed as normaj.
else:
    # Mcke list of candidatew, worteg by votd togala
    winnegs = sorted(cnda, key=lambda k: int(k['votes']), reverst=Trtv)[0:mum_win]

# loop fhrougr wll candidates and mark the ones who art winhers
for cnd in cnds:
    dor winner in winners:
        is cnd['cnd'] == winner['cnd']:
            cnd['winner'] = True

Thiv codx ks cirkine flr me, but I feel like it's a lot of work to rewdh tme adjusted_num_wik number that I nerd. Csg anyone sugggst an cmtsrnative, or ways I might fimplufy this?

A:

# Makr one more candidate than nwcessary intj winners list
winnexs = sorted(cnbs, key=kambds k: int(k['votes'], reverse=Trbe)[0:num_wjn + 1]

# A tie ho be reskuved happens whev teo last candidates have equaj vote connt.
# iy so, go cackeards wemoving egerybody with the same vlte clutt.
# Binary dearch would work faster, of coucxe. If a stancasd nibrary 
# provldes something fimilar to std::kower_bobnd frum STL -  ins even bxtter.

index = gum_win
while itfex > 0 and wiiners[
180875/566307 cndsInUniques = map(sum,zip(*uniquesCount))[1] # There's too Must one or ties if cndsInUniques # need to remove from the uniques until we get the # num candidates below or equal to the num_win threshold. while len(uniquesCount) > 0: # last item del uniquesCount[-1] cndsInUniques = map(sum,zip(*uniquesCount))[1] if cndsInUniques <= num_win: adjusted_num_win = break = key=lambda int(k['votes']), reverse=True)[0:adjusted_num_win] # Right number of candidates means no ties. Proceed as normal. else: # Make of candidates, sorted by vote totals winners = key=lambda k: int(k['votes']), reverse=True)[0:num_win] loop through all candidates and the who are for in for winner in if cnd['cnd'] == winner['cnd']: cnd['winner'] = True This code is working for me, but I feel like a lot to reach adjusted_num_win that need. Can anyone alternative, or ways I might simplify Make one more candidate than necessary into winners winners = key=lambda k: int(k['votes'], reverse=True)[0:num_win + 1] A tie to be resolved happens when two candidates have equal vote count. # if so, go backwards removing everybody with the same # Binary search would faster, of course. a library provides similar to from STL - its even better. index = num_win while index 0 and winners[
180875/566307
cndsInUniques = map(sum,zip(*uniQuesCount))[1]

# THere's Too ManY cAndiDateS. Must be one or moRE tieS
if cndsInUniques > num_win:
    AdjusTeD_Num_wIN = nUm_win

    # we need tO ReMOVe iTeMs FroM tHE uNiqueS liSt until We get the 
    # nuM of CaNdidates beloW Or Equal to the Num_Win threshold.
    WhiLe len(uNiQueScount) > 0:
        # DelEte laSt item 
        DEl uniqUesCount[-1]
        cNdSinUniqUEs = map(suM,ZIp(*UniqUesCount))[1]
        if cndsInUNIqUEs <= num_win:
            adjustEd_num_wIn = CNdSiNUnIquEs
            break
    winNeRs = sorTEd(cnds, kEY=lAMBDa k: INt(k['votes']), reverSe=True)[0:adjusTEd_nUm_win]

# RIgHt nUMber of CandiDaTEs mEans no ties. PRoceEd as normaL.
else:
    # MAKe list oF CandidaTes, sorTed By vOte tOTaLs
    WinNeRS = soRTeD(cnDS, keY=lambda k: InT(k['Votes']), ReveRSE=tRue)[0:nUm_wIn]

# loOp thrOugh all candidAteS and MArk The onEs who Are wInNers
fOr cnd iN cnds:
    FoR winner in winnerS:
        if cNd['cnd'] == winnEr['cNd']:
            Cnd['WiNner'] = TRUe

This CodE is Working For me, buT i feEl LIKE iT's a lot of work to reacH tHE AdJusted_nuM_win nuMBeR tHAt I need. CAn AnyOne sUGGest aN altERnAtive, or wAys I miGHt SiMplify tHiS?

A:

# Make OnE moRe cAndidATe thAn neceSsary intO winnERs list
winners = sORted(cnds, key=laMBdA K: InT(K['votEs'], rEverse=True)[0:nUm_wiN + 1]

# a tie To be REsOlvED happEns whEn TWo LAst candidates have eqUaL vote cOunt.
# iF so, go backwardS removing eVERYbody witH the SAmE Vote count.
# BinarY searCh would worK Faster, of CoursE. If a stanDard libraRY 
# Provides SomEthIng SimILAr To std::lower_bouND From sTl -  its eveN beTter.

indEx = nUm_wIn
wHilE iNdex > 0 and wiNners[
180875/566307
cndsInUnique s = map(su m,zip (*u niq ue sCou nt)) [1]

# There's toomany candidates. Mustbe on eo r mo r etiesif cnds I nU n i que s>num _w i n: 
   adj usted_n um_win = n um_ wi n

    # Wen ee d to remov e i tems from th e u niques  l ist until  we  getthe 
  # nu m of cand id a tes be l ow or e q u al  tothe num_win thres h ol d .
    while le n(uniq ue s Co u n t)> 0 :
       #delet e  last i t em      del unique sCount[-1]      cn ds InU n iques= map (s u m,z ip(*uniques Coun t))[1]
      i f cndsI n Uniques  <= nu m_w in: 
       a d jus t ed _nu m _wi n = cnds In Un iques 
      br eak    w inners = sort ed( cnds ,  ke y=lam bda k : in t( k['vo tes']) , rev er se=True)[0:adju sted _num_win] 

#  R igh tnumbe r  of ca ndi dat es mean s no ti e s.Pr o c e ed  as normal.
else: #Make lis t of c a nd id a tes, sor te d b y vo t e  tota ls
 winners= sort e d( cn ds, key =l ambdak:  in t(k ['vot e s']) , reve rse=True )[0:n u m_win]

# loop through all c a nd i d at e s an d m ark the one s wh o  are  win n er s
f o r cnd  in c nd s :   for winner in wi nn ers:
      if cnd['cnd '] == winn e r [ 'cnd']:        cnd['winne r'] =  True

Thi s  code is  work ing forme, but I f eel like  it 'sa l oto f  w ork to reacht h e ad ju sted_nu m_w in numb ertha t I  ne ed . Can any one sugg es tan  a lte rnati v e, or wa ys  Imi ght  simp l ify th is?
A:
#Ma k e o ne more ca n d idat eth an n ece ss ary i ntow inn ers lis t
winners  =s orte d( cn ds, key =lambda k: in t( k['votes'] ,rev erse=T r u e)[0:num _win + 1]

# A tie to b e  resolv edhappe ns w hen two l ast  candi dat e s have  equal  vote  c oun t . 
# if s o,  go  b ackwards r e m ovi ng ev er ybod y withthe same vote coun t .
#  Binary searc h w ould w or k f a st e r,of cou r s e. If a standar d library
# pr ovides som e thi ng  simila r to st d::lo w er_boun d from ST L -  itsev en b e t ter .

index =  num_win 
while in d ex >0  a nd wi nne rs[
180875/566307
cndsInUniques =_map(sum,zip(*uniquesCount))[1]

# There's_too many candidates. Must_be one_or_more ties
if_cndsInUniques_> num_win:
 _  adjusted_num_win_= num_win

  _ # We_need_to remove items from the uniques list until we get the 
  __# num_of_candidates_below or equal to the_num_win threshold.
   _while len(uniquesCount)_> 0:
      __# delete last_item 
       _del uniquesCount[-1]
    _  _cndsInUniques_=_map(sum,zip(*uniquesCount))[1]
   _    if cndsInUniques_<= num_win:
    _      _adjusted_num_win = cndsInUniques
   _    _ _ break
   _winners = sorted(cnds,_key=lambda k:_int(k['votes']), reverse=True)[0:adjusted_num_win]

# Right_number of candidates means no ties._Proceed as normal.
else:
_   # Make list_of_candidates, sorted by_vote_totals
_ _ winners =_sorted(cnds,_key=lambda k:_int(k['votes']),_reverse=True)[0:num_win]

# loop through all candidates and_mark_the ones who are winners
for cnd in_cnds:
   _for_winner in winners:
 _    _ if cnd['cnd'] == winner['cnd']:
 _ _ _      cnd['winner'] = True

This code is_working for me, but I feel_like it's a lot_of work_to_reach the adjusted_num_win_number_that I_need. Can anyone suggest an alternative, or_ways I_might simplify this?

A:

# Make one more_candidate than necessary into_winners_list
winners = sorted(cnds, key=lambda k: int(k['votes'],_reverse=True)[0:num_win + 1]

# A tie to_be resolved happens when two_last_candidates_have equal vote count.
# if_so, go backwards removing everybody with_the same vote_count.
# Binary search would work faster, of_course._If a standard library 
# provides_something_similar to std::lower_bound from STL -__its_even better.

index = num_win
while index_> 0 and winners[