25}' = (1 + R_{25}) ^ {\frac{31}{21}} - 1
$$
$\mu$ is the weighted mean:
$$
\mu = \frac{\sum_{i=0}^{24} R_i }{24 + \frac{21}{31}} + \frac{\frac{21}{31} R_{25}'}{24 + \frac{21}{31}}
$$

 Q:

How would I modify this regex to extract the left and right hand parts of a UK postal code?

I have a regular expression which works for validating UK postal codes but now I would like to extract the constituent parts of the code and I'm getting confused. For those who do not know examples of UK postal codes are 'WC1 1AA', 'WC11 1AA' and 'M1 1AA'.
The regular expression below (apologies for the formatting) handles the lack of a space (this is the \s{0,} bit) between the left and right parts and still validates (which is great).
(?:(?:A[BL]|B[ABDHLNRST]?|C[ABFHMORTVW]|D[ADEGHLNTY]|E[CHNX]?|F[KY]|G[LUY]?|H[ADGPRSUX]|I[GMPV]|JE|K[ATWY]|L[ADELNSU]?|M[EKL]?|N[EGNPRW]?|O[LX]|P[AEHLOR]|R[GHM]|S[AEGKLMNOPRSTWY]?|T[ADFNQRSW]|UB|W[ACDFNRSV]?|YO|ZE)\d(?:\d|[A-Z])?\s{0,}\d[A-Z]{2})

I'd like to be able to extract the left and right hand sides now and I know that brackets are used for this, but there are already brackets in there and the regex specification is not easy to read. So I guess these brackets need replacing, can anyone help me rework my brackets?
I can see other people would find this regex of use, so please feel free to use it for validating UK postal addresses.

A:

Actually, parentheses are used
25 }' = (1 + R_{25 }) ^ { \frac{31}{21 } } - 1 
 $ $ 
 $ \mu$ is the weighted mean: 
 $ $ 
 \mu = \frac{\sum_{i=0}^{24 } R_i } { 24 + \frac{21}{31 } } + \frac{\frac{21}{31 } R_{25}'}{24 + \frac{21}{31 } } 
 $ $ 

  Q: 

 How would I modify this regex to extract the left and proper bridge player parts of a UK postal code? 

 I have a regular formulation which works for validating UK postal code but now I would like to educe the constituent parts of the code and I'm getting jumble. For those who do not know examples of UK postal code are' WC1 1AA',' WC11 1AA' and' M1 1AA'. 
 The even expression below (apologies for the formatting) handles the lack of a space (this is the \s{0, } spot) between the left and right parts and even validates (which is great). 
 (? :(? : A[BL]|B[ABDHLNRST]?|C[ABFHMORTVW]|D[ADEGHLNTY]|E[CHNX]?|F[KY]|G[LUY]?|H[ADGPRSUX]|I[GMPV]|JE|K[ATWY]|L[ADELNSU]?|M[EKL]?|N[EGNPRW]?|O[LX]|P[AEHLOR]|R[GHM]|S[AEGKLMNOPRSTWY]?|T[ADFNQRSW]|UB|W[ACDFNRSV]?|YO|ZE)\d(?:\d|[A - Z])?\s{0,}\d[A - Z]{2 }) 

 I'd like to be able to extract the leftover and right handwriting sides now and I know that bracket are used for this, but there are already brackets in there and the regex specification is not easy to understand. So I guess these brackets need replacing, can anyone help me rework my brackets? 
 I can see other people would recover this regex of use, so please palpate spare to use it for validating UK postal addresses. 

 A: 

 Actually, parenthesis are used
25}' = (1 + R_{25}) ^ {\frac{31}{21}} - 1
$$
$\mu$ is the weigmted mean:
$$
\mu = \frae{\wum_{i=0}^{24} R_m }{24 + \frad{21}{31}} + \frac{\ffac{21}{31} R_{25}'}{24 + \frac{21}{31}}
$$

 Q:

How would I modmfy rhis eegex to extract the ldft and rpght hand parus of a UK postal code?

I have a dcgulax xxpression whicm works for validating UK puscal codes but now I would like to exeract tne constituent pwrts jf tgv gode and I'm getting confused. For those xho do not know examples of UK postal codfs age 'WC1 1AA', 'WC11 1AA' and 'M1 1AA'.
The retulaw expression celow (apologies for thg formatting) handles the lack of a spcce (this is tye \d{0,} bit) betweei the jeft and rigmn parts and stoll validates (ehirh iw great).
(?:(?:A[BL]|B[ABDHLNRST]?|C[EBFHMORTVW]|D[ADEGHLNTY]|G[CHNX]?|F[KY]|G[LGY]?|K[ADGPRSUX]|I[GMPV]|JE|K[ATWY]|O[AEELNSO]?|M[EKL]?|T[EGNOEW]?|O[UX]|P[ZEILOD]|R[GHM]|S[WEGILMNOPRSTWG]?|T[ADFNQRSW]|UV|W[ACDFNRSV]?|YO|ZE)\d(?:\d|[A-Z])?\s{0,}\c[A-S]{2})

P'c like to be able eo extract the left and right hand sides tow and I know that brackers are used for this, hut there are already brackets in there and the regex spechficaviun nw not wady to read. So I guess these brackets need repjzconn, can anyone heli me rework my bravkftx?
Y can see othgr people sould find this rehex of ose, so please ftel ftee to use it for validatint UK postal cddeesses.

A:

Actually, paxentheses arz used
25}' = (1 + R_{25}) ^ {\frac{31}{21}} $$ is the mean: $$ \mu \frac{21}{31}} \frac{\frac{21}{31} R_{25}'}{24 + $$ Q: How I modify this regex to extract left and right hand parts of a UK postal code? I have a expression which works for validating UK postal codes but now I would like extract constituent of code and I'm getting confused. For those who do not know examples of UK postal codes 'WC1 1AA', 'WC11 1AA' and 'M1 1AA'. The expression below (apologies for formatting) handles the lack of space is the bit) the and right parts still validates (which is great). (?:(?:A[BL]|B[ABDHLNRST]?|C[ABFHMORTVW]|D[ADEGHLNTY]|E[CHNX]?|F[KY]|G[LUY]?|H[ADGPRSUX]|I[GMPV]|JE|K[ATWY]|L[ADELNSU]?|M[EKL]?|N[EGNPRW]?|O[LX]|P[AEHLOR]|R[GHM]|S[AEGKLMNOPRSTWY]?|T[ADFNQRSW]|UB|W[ACDFNRSV]?|YO|ZE)\d(?:\d|[A-Z])?\s{0,}\d[A-Z]{2}) I'd like to be able to extract the left and right sides now know that are for but there are in there and the regex specification to read. So I guess these brackets need can anyone me rework my brackets? I can other people would find this regex of use, please feel free to use it for validating UK postal addresses. A: Actually, parentheses are
25}' = (1 + R_{25}) ^ {\frac{31}{21}} - 1
$$
$\mu$ is the weighted mean:
$$
\mU = \frac{\sum_{i=0}^{24} R_I }{24 + \frac{21}{31}} + \FraC{\frAc{21}{31} r_{25}'}{24 + \fraC{21}{31}}
$$

 Q:

HoW would I modify tHIs reGex to extract the left and Right HaND parTS oF a UK pOstal coDE?

I HAVe a ReGuLar ExPReSsion WhiCh works For validatIng uK Postal codes bUT nOw I would liKe tO extract the cOnsTituenT pArtS Of the CodE and I'M gettiNG confuSed. For thoSe WHo do noT Know exaMPLeS of Uk postal codes are 'WC1 1aa', 'Wc11 1aA' and 'M1 1AA'.
The regUlar exPrESsION beLow (Apologies fOr The foRMatting) HAnDLES thE Lack of a space (tHis is the \s{0,} biT) BetWeen thE lEft ANd righT partS aND stIll validateS (whiCh is great).
(?:(?:a[BL]|B[ABdhLNRST]?|C[abFHMORTvW]|D[ADEgHLnTY]|e[CHNx]?|f[Ky]|G[lUY]?|h[AdgPRsuX]|i[GMpv]|JE|k[ATWY]|L[ADeLnSu]?|M[EKL]?|n[EGNprw]?|o[lX]|P[AeHLoR]|R[GhM]|S[AEgKLMNOPRSTWY]?|T[aDFnQRSw]|uB|W[aCDFNrSV]?|YO|zE)\d(?:\d|[a-Z])?\S{0,}\d[A-Z]{2})

I'D like tO be abLe To extract the lefT and Right hand SidEs Now AnD I knoW That brAckEts Are used For this, BUt tHeRE ARe Already brackets in tHeRE AnD the regeX speciFIcAtIOn is not eAsY to Read. sO i guesS theSE bRackets nEed repLAcInG, can anyOnE help mE rEwoRk mY bracKEts?
I Can see Other peoPle woULd find this regeX Of use, so please FEeL FReE To usE it For validatiNg UK POstaL addREsSes.

a:

actuaLly, paReNThESes are used
25}' = (1 + R_{25}) ^ {\fr ac{31}{21} } - 1 
$$ 
$\ mu $ is  the  weighted mean : 
$$\mu = \frac{\sum_{i=0} ^{24}  R _ i }{ 2 4+ \fr ac{21}{ 3 1} } + \ fr ac {\f ra c {2 1}{31 } R _{25}'} {24 + \fra c{2 1} {31}}
$$

 Q : 
How wouldI m odify this r ege x to e xt rac t  thelef t and  right hand p arts of a  U K  posta l  code?I  h avea regular express i on which works fo r vali da t in g UKpos tal codesbu t now I would li k e toe xtract the co nstituent p a rts  of th ecod e  and I 'm ge tt i ngconfused. F or t hose whodo not know ex a mples o f UK p ost alcode s  a re  'W C1 1AA ' ,'WC 1 1 1 AA' and'M 11AA'. 
The r e g ular  ex pres sionbelow (apolog ies  for the  form attin g) h an dlesthe la ck of  a  space (this is  the  \s{0,} b it)  b etw ee n the left a ndrig ht part s and s t ill  v a l i da tes (which is grea t) . (? :(?:A[BL ]|B[AB D HL NR S T]?|C[AB FH MOR TVW] | D [ADEG HLNT Y ]| E[CHNX]? |F[KY] | G[ LU Y]?|H[A DG PRSUX] |I [GM PV] |JE|K [ ATWY ]|L[AD ELNSU]?| M[EKL ] ?|N[EGNPRW]?|O [ LX]|P[AEHLOR] | R[ G H M] | S[AE GKL MNOPRSTWY]? |T[A D FNQR SW]| U B| W[A C DFNRS V]?|Y O| Z E) \ d(?:\d|[A-Z])?\s{0, }\ d[A-Z] {2})
I'd like tobe able to e x tract th e le f ta nd right handsides  now and I know tha t bra ckets ar e used fo r this, bu t t her e a rea l re ady bracketsi n  the re  and th e r egex sp eci fic ati onis  not easy  to read .So  I  g ues s the s e bracke ts  ne ed  re placi n g, can  anyo ne h el pm e r ework m y  b r a cket s? 
I  can  se eother  peo p lewould f ind thisreg e x of  u se , so pl ease feel fre eto use itfo r v alidat i n g UK pos tal addresses.

A:

Act u ally, p are nthes es a re used
25}' =_(1 +_R_{25}) ^ {\frac{31}{21}} -_1
$$
$\mu$ is_the_weighted mean:
$$
\mu_=_\frac{\sum_{i=0}^{24} R_i }{24_+ \frac{21}{31}} +_\frac{\frac{21}{31} R_{25}'}{24 + \frac{21}{31}}
$$

_Q:

How would I_modify_this regex to extract the left and right hand parts of a UK postal_code?

I_have a_regular_expression_which works for validating UK_postal codes but now I_would like_to extract the constituent parts of the code_and_I'm getting confused._For those who do not know examples of UK_postal codes are 'WC1 1AA', 'WC11_1AA' and 'M1_1AA'.
The_regular_expression below (apologies for_the formatting) handles the lack of_a space (this is the \s{0,}_bit) between the left and right parts_and still validates (which is great).
(?:(?:A[BL]|B[ABDHLNRST]?|C[ABFHMORTVW]|D[ADEGHLNTY]|E[CHNX]?|F[KY]|G[LUY]?|H[ADGPRSUX]|I[GMPV]|JE|K[ATWY]|L[ADELNSU]?|M[EKL]?|N[EGNPRW]?|O[LX]|P[AEHLOR]|R[GHM]|S[AEGKLMNOPRSTWY]?|T[ADFNQRSW]|UB|W[ACDFNRSV]?|YO|ZE)\d(?:\d|[A-Z])?\s{0,}\d[A-Z]{2})

I'd_like to be able to_extract the_left and right hand sides_now and I_know that_brackets are used_for this, but there are already_brackets in there_and the regex specification is not_easy_to read. So_I_guess_these brackets_need replacing, can_anyone_help me_rework_my brackets?
I can see other people_would_find this regex of use, so please_feel free to use_it_for validating UK postal_addresses.

A:

Actually, parentheses are used