 v:
      d[name] +=1
    print k,d

But, how do I get user count and search count for each search term.

A:

Yes, you can use a defaultdict to do this (or just a regular dict too, but I think a defaultdict is more flexible)
In [36]: a = defaultdict(defaultdict)

In [40]: l  = ["1234:yahoo\tgoogle\tmicrosoft\tapple\tyahoo", "2345:apple\tgoogle\tgoogle", "4567:yahoo\tapple\tapple"]

In [48]: for li in l:
   ...:     search_id, terms = li.split(":")[0], li.split(":")[1]
   ...:     terms = terms.split("\t")
   ...:     for term in terms:
   ...:         if "search_cnt" in a[term]:
   ...:             a[term]["search_cnt"] += 1
   ...:         else:
   ...:             a[term]["search_cnt"] = 1
   ...:     for term in set(terms):
   ...:         if "user_cnt" in a[term]:
   ...:             a[term]["user_cnt"] += 1
   ...:         else:
   ...:             a[term]["user_cnt"] = 1

In [49]: a
Out[49]:
defaultdict(collections.defaultdict,
            {'apple': defaultdict(None, {'search_cnt': 4, 'user_cnt': 3}),
             'google': defaultdict(None, {'search_cnt': 3, 'user_cnt': 2}),
            'microsoft': defaultdict(None, {'search_cnt': 1, 'user_cnt': 1}),
             'yahoo': defaultdict(None, {'search_cnt': 3, 'user_cnt': 2})})

The default dict above contains the counts you need.
The reason I use the set for the second term iteration is that if 1 user searched for a term multiple times, the term's user count should not increment :)

 Q:

What is the Rust equivalent to a try-catch statement?

Is it possible to handle multiple different errors at
v: 
       d[name ] + = 1 
     print k, d 

 But, how do I get user count and search count for each search condition. 

 deoxyadenosine monophosphate: 

 Yes, you can use a defaultdict to do this (or just a unconstipated dict too, but I think a defaultdict is more compromising) 
 In [ 36 ]: a = defaultdict(defaultdict) 

 In [ 40 ]: fifty   = [ " 1234: yahoo\tgoogle\tmicrosoft\tapple\tyahoo ", " 2345: apple\tgoogle\tgoogle ", " 4567: yahoo\tapple\tapple " ] 

 In [ 48 ]: for li in l: 
   ... :      search_id, terms = li.split(":")[0 ], li.split(":")[1 ] 
   ... :      terms = terms.split("\t ") 
   ... :      for term in term: 
   ... :          if " search_cnt " in a[term ]: 
   ... :              a[term]["search_cnt " ] + = 1 
   ... :          else: 
   ... :              a[term]["search_cnt " ] = 1 
   ... :      for term in set(terms ): 
   ... :          if " user_cnt " in a[term ]: 
   ... :              a[term]["user_cnt " ] + = 1 
   ... :          else: 
   ... :              a[term]["user_cnt " ] = 1 

 In [ 49 ]: a 
 Out[49 ]: 
 defaultdict(collections.defaultdict, 
             {' apple': defaultdict(None, {' search_cnt': 4,' user_cnt': 3 }), 
             ' google': defaultdict(None, {' search_cnt': 3,' user_cnt': 2 }), 
            ' microsoft': defaultdict(None, {' search_cnt': 1,' user_cnt': 1 }), 
             ' yahoo': defaultdict(None, {' search_cnt': 3,' user_cnt': 2 }) }) 

 The default dict above contains the counts you need. 
 The reason I practice the set for the second term iteration is that if 1 user search for a term multiple times, the condition's user count should not increment :) 

  Q: 

 What is the Rust equivalent to a test - catch statement? 

 Is it possible to handle multiple different error at
 v:
      d[name] +=1
    print k,d

Bmt, how do I get oswr couit and aearch cuunt for each search term.

A:

Yed, tou cqn use a defaultdict tu do this (or just a rtgular dict too, bnf I thikh a scfaulcdmct is more flewible)
In [36]: a = defaultdict(dexajlcdict)

In [40]: l  = ["1234:yahoo\tgoogle\tmicrosoft\twpple\tyshlo", "2345:apple\tgooglg\tgoobje", "4567:yzhoo\tapple\tapple"]

In [48]: for li in l:
   ...:     seerch_id, terms = lo.split(":")[0], li.split(":")[1]
   ...:     termd = tfrms.split("\t")
   ...:     vor term in teriw:
   ...:         if "search_but" in a[term]:
   ...:             a[term]["search_cnt"] += 1
   ...:         elwe:
   ...:             a[teri]["search_cnt"] = 1
   ...:     xor terk in set(terms):
   ...:         if "user_cnt" in a[txrm]:
   ...:             a[jerm]["user_cnd"] += 1
   ...:         else:
   ...:             q[tefm]["uaec_cnf"] = 1

In [49]: a
Ont[49]:
defaultdidt(collectiobs.defaultdict,
            {'apple': defahltdice(Njne, {'search_cnt': 4, 'user_cnt': 3}),
             'goocle': defaultdict(None, {'search_xnt': 3, 'user_cnt': 2}),
            'microfoft': defaultdict(None, {'search_cnt': 1, 'user_cnt': 1}),
             'yehuo': bcnaulgeift(None, {'search_cnt': 3, 'user_cnt': 2})})

The default dict abjbe cpntains the comnts you need.
The rradom I use the sej for tks aecond term iteratlon is jhat id 1 user starchrd for a term multiple timew, the term's lser count should not nncrement :)

 Q:

Chat ix the Rust equivalent to a txy-catcg statement?

Ls it posakble to handle mjltpple different errors at
v: d[name] +=1 print k,d But, how get count and count for each can a defaultdict to this (or just regular dict too, but I think defaultdict is more flexible) In [36]: a = defaultdict(defaultdict) In [40]: l = "2345:apple\tgoogle\tgoogle", "4567:yahoo\tapple\tapple"] In [48]: for li in l: ...: search_id, terms = li.split(":")[0], ...: = ...: term in terms: ...: if "search_cnt" in a[term]: ...: a[term]["search_cnt"] += 1 ...: else: ...: a[term]["search_cnt"] 1 ...: for term in set(terms): ...: if in a[term]: ...: a[term]["user_cnt"] 1 ...: else: ...: a[term]["user_cnt"] 1 [49]: a defaultdict(collections.defaultdict, defaultdict(None, 4, 'user_cnt': 3}), defaultdict(None, {'search_cnt': 3, 'user_cnt': 2}), 'microsoft': defaultdict(None, {'search_cnt': 1, 'user_cnt': 1}), 'yahoo': defaultdict(None, {'search_cnt': 3, 'user_cnt': 2})}) default dict the counts need. reason use the set second term iteration is that if for a term multiple times, the term's user should not :) Q: What is the Rust to a try-catch statement? Is it possible to multiple different errors at
 v:
      d[name] +=1
    print k,d

But, how do I get User count aNd seaRch CouNt For eAch sEarch term.

A:

Yes, yOU can Use a defaultdict to do thiS (or juSt A ReguLAr Dict tOo, but I tHInK A DefAuLtDicT iS MoRe fleXibLe)
In [36]: a = deFaultdict(dEfaUlTdict)

In [40]: l  = ["1234:yahoO\TgOogle\tmicrOsoFt\tapple\tyahOo", "2345:aPple\tgOoGle\TGooglE", "4567:yaHoo\taPple\taPPle"]

In [48]: fOr li in l:
   ...:     seArCH_id, terMS = li.spliT(":")[0], LI.sPlit(":")[1]
   ...:     Terms = terms.split("\t")
   ...:     fOR tERm in terms:
   ...:         if "seaRch_cnt" In A[TeRM]:
   ...:             A[teRm]["sEarch_cnt"] += 1
   ...:         elSe:
   ...:             A[term]["SEarch_cnT"] = 1
   ...:     FoR TERm iN Set(terms):
   ...:         if "useR_cnt" in a[term]:
   ...:             A[TerM]["user_cNt"] += 1
   ...:         ElsE:
   ...:             A[term]["uSer_cnT"] = 1

IN [49]: A
OuT[49]:
defaultdicT(colLections.dEfaultDIct,
            {'applE': DefaultDict(NoNe, {'sEarCh_cnT': 4, 'UsEr_Cnt': 3}),
             'GoOGle': DEfAulTDicT(None, {'seaRcH_cNt': 3, 'useR_cnt': 2}),
            'MICROsofT': deFaulTdict(none, {'search_cnt': 1, 'UseR_cnt': 1}),
             'YAhoO': defaUltdiCt(NoNe, {'SearcH_cnt': 3, 'usEr_cnt': 2})})

thE default dict aboVe coNtains the CouNtS yoU nEed.
ThE Reason i usE thE set for The secoND teRm ITERaTion is that if 1 user seArCHEd For a term MultipLE tImES, the term'S uSer CounT SHould Not iNCrEment :)

 Q:

WhAt is thE ruSt EquivalEnT to a trY-cAtcH stAtemeNT?

Is iT possiBle to hanDle muLTiple different ERrors at
 v:
      d[name] +=1
   print k,d
But,  ho w d oI ge t us er count and s e arch  count for each search  term .A:
Y es , you  can us e  a d efa ul td ict  t o  d o thi s ( or just  a regular  di ct  too, but It hi nk a defau ltd ict is morefle xible) 
I n [ 3 6]: a  =defau ltdict ( defaul tdict)

I n[ 40]: l  = ["12 3 4 :y ahoo \tgoogle\tmicroso f t\ t apple\tyahoo",  "2345 :a p pl e \ tgo ogl e\tgoogle" ,"4567 : yahoo\t a pp l e \ tap p le"]

In [48] : for li in l:   ... :  search _id,te r ms= li.split( ":") [0], li.s plit(" : ")[1]
 ...:    ter ms= t erms . sp li t(" \t " )
 . ..:    for ter min  term s:
. . .:      if "s earch_cnt" in  a[ term ] :
  ... :          a[ term][ "sear ch _cnt"] += 1
  ...:         els e: 
  . ..:         a[ term][" search_ c nt" ]= 1 
  ...:     for ter mi n  s et(terms ):
  . .. :       i f"us er_c n t " ina[te r m] :
   ... :         a[te rm ]["use r_ cnt "]+= 1  .. .:        els e:
 ...:           a[term]["us e r_ c n t" ]  = 1 

I n [49]: a
O ut[4 9 ]:
d efau l td ict ( colle ction s. d ef a ultdict,
          {'appl e': d efaultdict(No ne, {'sear c h _ cnt': 4,  'us e r_ c nt': 3}),
           'googl e ': defau ltdic t(None,{'search_ c n t': 3, ' use r_c nt' : 2 } ) ,            ' m i cros of t': def aul tdict(N one , { 'se arc h_ cnt': 1,'user_cn t' :1} ), 
          'ya ho o':  d efa ultdi c t(None , {'s earc h_ cn t ':3, 'use r _c n t ': 2 }) }) 

Th e d ef aultdict abo ve cont ains thecou n ts y ou  n eed.
Th e reason I us ethe set fo rthe  secon d term ite ration is that if 1 use r  search edfor a  ter m multipl e t imes,the term's  usercount  s hou l d  noti n cr eme nt  :)

 Q:
W h atis th eRust  equiva lent to a try-catc h  st atement?

Isitposs i b le  to ha n dle  m u lti p l e different err ors at
 v:
_ _   d[name]_+=1
 __ print_k,d

But,_how do I_get user count_and search count for_each search term.

A:

Yes,_you_can use a defaultdict to do this (or just a regular dict too, but_I_think a_defaultdict_is_more flexible)
In [36]: a =_defaultdict(defaultdict)

In [40]: l  =_["1234:yahoo\tgoogle\tmicrosoft\tapple\tyahoo", "2345:apple\tgoogle\tgoogle",_"4567:yahoo\tapple\tapple"]

In [48]: for li in l:
  _...:_  _ search_id, terms = li.split(":")[0], li.split(":")[1]
   ...:_    terms =_terms.split("\t")
  _...:__   for_term in terms:
   ...:_     _  if "search_cnt" in a[term]:
 _ ...:    _    _ _ a[term]["search_cnt"] += 1
 _ ...: _ _  _  else:
   ...:_  _     __  a[term]["search_cnt"]_=_1
_ _...:  __ for_term_in set(terms):
   ...: __      if_"user_cnt" in a[term]:
 __...:   _    _    a[term]["user_cnt"] +=_1
 _ ...:_        else:
  _...:     _   _ __a[term]["user_cnt"] = 1

In_[49]:_a
Out[49]:
defaultdict(collections.defaultdict,
 _      _ _ {'apple': defaultdict(None, {'search_cnt': 4, 'user_cnt':_3}),
   __     _  'google': defaultdict(None, {'search_cnt': 3,_'user_cnt': 2}),
   ___    _ 'microsoft': defaultdict(None, {'search_cnt': 1, 'user_cnt':_1}),
  _      __  'yahoo': defaultdict(None, {'search_cnt': 3,_'user_cnt':_2})})

The default dict above contains the_counts_you_need.
The reason I use the_set for the second term iteration_is that if 1 user searched for a term_multiple times, the_term's user count should not_increment_:)

_Q:

What is the Rust equivalent to a try-catch statement?

Is it_possible to_handle multiple different_errors at