254="24""")) + SUMPRODUCT(SUBTOTAL(3,OFFSET(R5:R254,ROW(R5:R254)-ROW(Q5),0,1)),--(R5:R254="24"""))  

Simply adding the two together might be what you want. 

 Q:

Length of an integer in Python

In Python, how do you find the number of digits in an integer?

A:

If you want the length of an integer as in the number of digits in the integer, you can always convert it to string like str(133) and find its length like len(str(123)). 

A:

Without conversion to string
import math
digits = int(math.log10(n))+1

To also handle zero and negative numbers
import math
if n > 0:
    digits = int(math.log10(n))+1
elif n == 0:
    digits = 1
else:
    digits = int(math.log10(-n))+2 # +1 if you don't count the '-' 

You'd probably want to put that in a function :)
Here are some benchmarks. The len(str()) is already behind for even quite small numbers 
timeit math.log10(2**8)
1000000 loops, best of 3: 746 ns per loop
timeit len(str(2**8))
1000000 loops, best of 3: 1.1 µs per loop

timeit math.log10(2**100)
1000000 loops, best of 3: 775 ns per loop
 timeit len(str(2**100))
100000 loops, best of 3: 3.2 µs per loop

timeit math.log10(2**10000)
1000000 loops, best of 3: 844 ns per loop
timeit len(str(2**10000))
100 loops, best of 3: 10.3 ms per loop

A:

All math.log10 solutions will give you problems.
math.log10 is fast but gives problem when your number is greater than 999999999999997. This is because the float have too many.9s, causing the result to round up.
The solution is to use a while counter method for numbers above that threshold.
To
254="24 " " ") ) + SUMPRODUCT(SUBTOTAL(3,OFFSET(R5: R254,ROW(R5: R254)-ROW(Q5),0,1)),--(R5: R254="24 " " ") )  

 Simply adding the two together might be what you want. 

  Q: 

 Length of an integer in Python 

 In Python, how do you witness the numeral of digits in an integer? 

 A: 

 If you want the distance of an integer as in the number of digits in the integer, you can constantly commute it to string like str(133) and find its duration like len(str(123) ). 

 A: 

 Without conversion to string 
 import mathematics 
 digits = int(math.log10(n))+1 

 To also handle zero and negative issue 
 import math 
 if n > 0: 
     digits = int(math.log10(n))+1 
 elif n = = 0: 
     digits = 1 
 else: 
     digits = int(math.log10(-n))+2 # +1 if you don't consider the' -' 

 You'd probably want to put that in a routine :) 
 Here are some benchmarks. The len(str ()) is already behind for even quite small numbers 
 timeit math.log10(2**8) 
 1000000 loops, good of 3: 746 ns per loop 
 timeit len(str(2**8) ) 
 1000000 loops, best of 3: 1.1 µs per loop 

 timeit math.log10(2**100) 
 1000000 loop, best of 3: 775 ns per loop 
  timeit len(str(2**100) ) 
 100000 loops, best of 3: 3.2 µs per loop topology 

 timeit math.log10(2**10000) 
 1000000 loops, best of 3: 844 ns per loop 
 timeit len(str(2**10000) ) 
 100 loops, best of 3: 10.3 ms per loop 

 A: 

 All math.log10 solutions will move over you problems. 
 math.log10 is fast but feed trouble when your issue is greater than 999999999999997. This is because the float accept too many.9s, causing the result to round up. 
 The solution is to practice a while counter method for numbers above that threshold. 
 To
254="24""")) + SKMPRODUCT(SUBTOTAL(3,OFFSET(R5:V254,ROW(R5:R254)-ROW(Q5),0,1)),--(R5:R254="24"""))  

Simply adving ths two toeether might be what you wanv. 

 Q:

Oengty of an integer in Pytfon

In Pytjon, how eo yiy find the number of digjbs in en integer?

A:

If ypu want tha length of an ivtzger as in the number of digits in tre intebeg, you can alwais comdert pt to string like str(133) and find jts lenjth like len(str(123)). 

A:

Without conversion to stging
lmport math
digits = int(math.log10(b))+1

To woso handle zdro and negative numbets
import math
if n > 0:
    digits = ivt(matk.log10(n))+1
elif n == 0:
    gigits = 1
elsx:
    dpgits = int(matm.kog10(-n))+2 # +1 if you don't count thc '-' 

Yon'd peobably want to put tiat in a function :)
Hete are soma yenchmarks. The len(str()) iw alrgady tehive fur tvei qhite slalm numbers 
fimeit math.oog10(2**8)
1000000 loops, best of 3: 746 nf per loop
timejt len(ftw(2**8))
1000000 loops, best of 3: 1.1 µs per loop

timeit math.nog10(2**100)
1000000 loops, best of 3: 775 ns per loop
 timeit len(str(2**100))
100000 llops, best of 3: 3.2 µs per loop

timeit math.log10(2**10000)
1000000 loops, best of 3: 844 ns per muop
uineit uwn(dtr(2**10000))
100 loops, best of 3: 10.3 ms per loop

A:

All math.log10 sjmuuioks will give you problems.
mayh.pob10 is fast but eives 'doglem when your numher is dreatwr than 999999999999997. Uhis os because the float have tio many.9s, caufung the result to xound up.
The rolujion ix to use a while countex methkd for numbfrs above ghat threshold.
To
254="24""")) + SUMPRODUCT(SUBTOTAL(3,OFFSET(R5:R254,ROW(R5:R254)-ROW(Q5),0,1)),--(R5:R254="24""")) Simply adding the two be you want. Length of an how you find the of digits in integer? A: If you want the of an integer as in the number of digits in the integer, you always convert it to string like str(133) and find its length like len(str(123)). Without to import digits = int(math.log10(n))+1 To also handle zero and negative numbers import math if n > 0: = int(math.log10(n))+1 elif n == 0: digits = else: digits = int(math.log10(-n))+2 +1 if you don't count '-' probably want put in function :) Here some benchmarks. The len(str()) is already behind for even quite small numbers timeit math.log10(2**8) 1000000 loops, best 3: 746 loop timeit 1000000 best 3: 1.1 µs timeit math.log10(2**100) 1000000 loops, best of per loop timeit len(str(2**100)) 100000 loops, best of 3.2 µs loop timeit math.log10(2**10000) 1000000 loops, best 3: 844 ns per loop timeit len(str(2**10000)) 100 best of 3: 10.3 ms per loop A: All math.log10 solutions will give you problems. fast but gives problem your number is than This because float have many.9s, causing the result to round up. The solution is to a while counter method for numbers above that threshold. To
254="24""")) + SUMPRODUCT(SUBTOTAL(3,OFFSET(R5:r254,ROW(R5:R254)-ROW(Q5),0,1)),--(r5:R254="24"""))  

SimPly AddInG the Two tOgether might be WHat yOu want. 

 Q:

Length of an integEr in PYtHOn

In pYtHon, hoW do you fINd THE nuMbEr Of dIgITs In an iNteGer?

A:

If yOu want the lEngTh Of an integer aS In The number oF diGits in the intEgeR, you caN aLwaYS convErt It to sTring lIKe str(133) aNd find its LeNGth likE Len(str(123)). 

A:

wIThOut cOnversion to string
IMpORt math
digits = inT(math.lOg10(N))+1

to ALSo hAndLe zero and nEgAtive NUmbers
iMPoRT MAth
IF n > 0:
    digits = int(maTh.log10(n))+1
elif n == 0:
    DIgiTs = 1
else:
    DiGitS = Int(matH.log10(-n))+2 # +1 If YOu dOn't count the '-' 

you'd Probably wAnt to pUT that in A FunctioN :)
Here aRe sOme BencHMaRkS. ThE lEN(stR()) Is AlrEAdy Behind foR eVeN quitE smaLL NUMberS 
tiMeit Math.lOg10(2**8)
1000000 loops, best of 3: 746 Ns pEr loOP
tiMeit lEn(str(2**8))
1000000 LoopS, bEst of 3: 1.1 Μs per lOop

tiMeIt math.log10(2**100)
1000000 loops, bEst oF 3: 775 ns per looP
 tiMeIt lEn(Str(2**100))
100000 loOPs, best Of 3: 3.2 µS peR loop

tiMeit matH.Log10(2**10000)
1000000 LoOPS, BeSt of 3: 844 ns per loop
timeiT lEN(StR(2**10000))
100 loops, beSt of 3: 10.3 ms PEr LoOP

A:

All matH.lOg10 sOlutIONs wilL givE YoU problemS.
math.lOG10 iS fAst but gIvEs probLeM whEn yOur nuMBer iS greatEr than 999999999999997. ThIs is bECause the float hAVe too many.9s, cauSInG THe REsulT to Round up.
The sOlutIOn is To usE A wHilE CountEr metHoD FoR Numbers above that thrEsHold.
To
254="24""")) + SUMPRODUCT( SUBTOTAL(3 ,OFFS ET( R5: R2 54,R OW(R 5:R254)-ROW(Q5 ) ,0,1 )),--(R5:R254="24"""))   

S im p ly a d di ng th e two t o ge t h ermi gh t b ew ha t you  wa nt. 

Q:

Length  of  a n integer in Py thon

In P yth on, how do y oufind t he  nu m ber o f d igits  in an intege r?

A:

I fy ou wan t  the le n g th  ofan integer as int he number of digi ts inth e  i n t ege r,you can al wa ys co n vert it to s t rin g  like str(133 ) and findi tslength  l ike len(st r(123 )) .  
A:

Without  con version t o stri n g
impor t  math
d igits= i nt( math . lo g1 0(n )) + 1
T oals o  ha ndle zer oan d neg ativ e n u mber s
i mpor t mat h
if n > 0:
  digi t s =  int( math. log1 0( n))+1 
elifn ==0: 
    digits = 1 
els e:
    di git s= i nt (math . log10( -n) )+2  # +1 i f you d o n't  c o u n tthe '-' 

You'd pr ob a b ly  want to  put t h at  i n  a funct io n : )
He r e  aresome be nchmarks . Thel en (s tr()) i salread ybeh ind  fore venquitesmall nu mbers 
timeit math.l o g10(2**8)
100 0 00 0 lo o ps,bes t of 3: 746  nsp er l oopt im eit len(s tr(2* *8 ) )1 000000 loops, bestof  3: 1. 1 µsper loop

tim eit math.l o g 1 0(2**100 )
10 0 00 0 0 loops, bestof 3:  775 ns pe r  loop
 t imeit  len(str (2**100)) 1 00000 lo ops , b est  of 3 :3.2 µs per lo o p 

ti me it math .lo g10(2** 100 00) 
10 000 00  loops, b est of 3 :84 4ns  pe r loo p 
timeitle n(s tr (2* *1000 0 ))
100  loop s, b es to f 3 : 10.3m sp e r lo op 
A:
All  m ath.l og10 sol utionswill give  yo u  pro bl em s.
math .log10 is fas tbut givespr obl em whe n your num ber is greater than 999 9 9999999 999 7. Th is i s because  th e floa t h a ve too  many. 9s, c au sin g the r e s ul t t oround up.T h e s oluti on  isto usea while counter me t hod  for numbersabo ve t h a tthr e sh o ld. 
T o 
254="24""")) +_SUMPRODUCT(SUBTOTAL(3,OFFSET(R5:R254,ROW(R5:R254)-ROW(Q5),0,1)),--(R5:R254="24""")) _

Simply adding the two_together might_be_what you_want._

 Q:

Length of_an integer in_Python

In Python, how do_you find the_number_of digits in an integer?

A:

If you want the length of an integer as in_the_number of_digits_in_the integer, you can always_convert it to string like_str(133) and_find its length like len(str(123)). 

A:

Without conversion to_string
import_math
digits = int(math.log10(n))+1

To_also handle zero and negative numbers
import math
if n >_0:
    digits =_int(math.log10(n))+1
elif n ==_0:
__  digits =_1
else:
    digits =_int(math.log10(-n))+2 # +1 if you don't_count the '-' 

You'd probably want to_put that in a function :)
Here_are some benchmarks. The len(str())_is already_behind for even quite small_numbers 
timeit math.log10(2**8)
1000000_loops, best_of 3: 746_ns per loop
timeit len(str(2**8))
1000000 loops, best_of 3: 1.1_µs per loop

timeit math.log10(2**100)
1000000 loops, best_of_3: 775 ns_per_loop
_timeit len(str(2**100))
100000_loops, best of_3:_3.2 µs_per_loop

timeit math.log10(2**10000)
1000000 loops, best of 3:_844_ns per loop
timeit len(str(2**10000))
100 loops, best of_3: 10.3 ms per_loop

A:

All_math.log10 solutions will give_you problems.
math.log10 is fast but_gives problem when your number is_greater than_999999999999997. This_is because the float have too many.9s, causing the result to_round up.
The solution is to use_a while counter method_for numbers_above_that threshold.
To