How to find length of digits in an integer

Question

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

User · Answer

Let the number be n then the number of digits in n is given by   math floor math log10 n   1   Note that this will give correct answers for  ve integers  lt  10e15  Beyond that the precision limits of the return type of math log10 kicks in and the answer may be off by 1  I would simply use len str n   beyond that  this requires O log n   time which is same as iterating over powers of 10   Thanks to  SetiVolkylany for bringing my attenstion to this limitation  Its amazing how seemingly correct solutions have caveats in implementation details

User · Answer

If you have to ask an user to give input and then you have to count how many numbers are there then you can follow this   count number   input  Please enter a number t    print len count number     Note  Never take an int as user input

User · Answer

For posterity  no doubt by far the slowest solution to this problem   def num digits num  number of calls 1        Returns the number of digits of an integer num       if num    0 or num    -1          return 1 if number of calls    1 else 0     else          return 1   num digits num 10  number of calls 1

User · Answer

As mentioned the dear user  Calvintwr  the function math log10 has problem in a number outside of a range  -999999999999997  999999999999997   where we get floating point errors  I had this problem with the JavaScript  the Google V8 and the NodeJS  and the C  the GNU GCC compiler   so a  purely mathematically  solution is impossible here     Based on this gist and the answer the dear user  Calvintwr  import math   def get count digits number  int          Return number of digits in a number          if number    0          return 1      number   abs number       if number  lt   999999999999997          return math floor math log10 number     1      count   0     while number          count    1         number     10     return count     I tested it on numbers with length up to 20  inclusive  and all right  It must be enough  because the length max integer number on a 64-bit system is 19  len str sys maxsize      19    assert get count digits -99999999999999999999     20 assert get count digits -10000000000000000000     20 assert get count digits -9999999999999999999     19 assert get count digits -1000000000000000000     19 assert get count digits -999999999999999999     18 assert get count digits -100000000000000000     18 assert get count digits -99999999999999999     17 assert get count digits -10000000000000000     17 assert get count digits -9999999999999999     16 assert get count digits -1000000000000000     16 assert get count digits -999999999999999     15 assert get count digits -100000000000000     15 assert get count digits -99999999999999     14 assert get count digits -10000000000000     14 assert get count digits -9999999999999     13 assert get count digits -1000000000000     13 assert get count digits -999999999999     12 assert get count digits -100000000000     12 assert get count digits -99999999999     11 assert get count digits -10000000000     11 assert get count digits -9999999999     10 assert get count digits -1000000000     10 assert get count digits -999999999     9 assert get count digits -100000000     9 assert get count digits -99999999     8 assert get count digits -10000000     8 assert get count digits -9999999     7 assert get count digits -1000000     7 assert get count digits -999999     6 assert get count digits -100000     6 assert get count digits -99999     5 assert get count digits -10000     5 assert get count digits -9999     4 assert get count digits -1000     4 assert get count digits -999     3 assert get count digits -100     3 assert get count digits -99     2 assert get count digits -10     2 assert get count digits -9     1 assert get count digits -1     1 assert get count digits 0     1 assert get count digits 1     1 assert get count digits 9     1 assert get count digits 10     2 assert get count digits 99     2 assert get count digits 100     3 assert get count digits 999     3 assert get count digits 1000     4 assert get count digits 9999     4 assert get count digits 10000     5 assert get count digits 99999     5 assert get count digits 100000     6 assert get count digits 999999     6 assert get count digits 1000000     7 assert get count digits 9999999     7 assert get count digits 10000000     8 assert get count digits 99999999     8 assert get count digits 100000000     9 assert get count digits 999999999     9 assert get count digits 1000000000     10 assert get count digits 9999999999     10 assert get count digits 10000000000     11 assert get count digits 99999999999     11 assert get count digits 100000000000     12 assert get count digits 999999999999     12 assert get count digits 1000000000000     13 assert get count digits 9999999999999     13 assert get count digits 10000000000000     14 assert get count digits 99999999999999     14 assert get count digits 100000000000000     15 assert get count digits 999999999999999     15 assert get count digits 1000000000000000     16 assert get count digits 9999999999999999     16 assert get count digits 10000000000000000     17 assert get count digits 99999999999999999     17 assert get count digits 100000000000000000     18 assert get count digits 999999999999999999     18 assert get count digits 1000000000000000000     19 assert get count digits 9999999999999999999     19 assert get count digits 10000000000000000000     20 assert get count digits 99999999999999999999     20     All example of codes tested with the Python 3 5

User · Answer

Assuming you are asking for the largest number you can store in an integer  the value is implementation dependent  I suggest that you don t think in that way when using python  In any case  quite a large value can be stored in a python  integer   Remember  Python uses duck typing   Edit  I gave my answer before the clarification that the asker wanted the number of digits  For that  I agree with the method suggested by the accepted answer  Nothing more to add

User · Answer

def length i     return len str i

User · Answer

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 make this even faster  create 10 16  10 17 so on so forth and store as variables in a list  That way  it is like a table lookup   def getIntegerPlaces theNumber       if theNumber  lt   999999999999997          return int math log10 theNumber     1     else          counter   15         while theNumber  gt   10  counter              counter    1         return counter

User · Answer

It s been several years since this question was asked  but I have compiled a benchmark of several methods to calculate the length of an integer    def libc size i        return libc snprintf buf  100  c char p b  i    i    equivalent to  return snprintf buf  100    i   i     def str size i       return len str i     Length of  i  as a string  def math size i       return 1   math floor math log10 i     1   floor of log10 of i  def exp size i       return int     5e   format i  split  e   1     1   e g   1e10  - gt   10    1 - gt  11  def mod size i       return len   i    i    Uses string modulo instead of str i   def fmt size i       return len   0   format i     Same as above but str format    the libc function requires some setup  which I haven t included   size exp is thanks to Brian Preslopsky  size str is thanks to GeekTantra  and size math is thanks to John La Rooy  Here are the results   Time for libc size       1 2204   s Time for string size     309 41 ns Time for math size       329 54 ns Time for exp size        1 4902   s Time for mod size        249 36 ns Time for fmt size        336 63 ns In order of speed  fastest first     mod size  1 000000x    str size  1 240835x    math size  1 321577x    fmt size  1 350007x    libc size  4 894290x    exp size  5 976219x     Disclaimer  the function is run on inputs 1 to 1 000 000   Here are the results for sys maxsize - 100000 to sys maxsize   Time for libc size       1 4686   s Time for string size     395 76 ns Time for math size       485 94 ns Time for exp size        1 6826   s Time for mod size        364 25 ns Time for fmt size        453 06 ns In order of speed  fastest first     mod size  1 000000x    str size  1 086498x    fmt size  1 243817x    math size  1 334066x    libc size  4 031780x    exp size  4 619188x    As you can see  mod size  len   i    i   is the fastest  slightly faster than using str i  and significantly faster than others

User · Answer

n   3566002020360505 count   0 while n gt 0     count    1   n   n   10 print f quot The number of digits in the number are   count  quot    output  The number of digits in the number are  16

User · Answer

Without conversion to string  import math digits   int math log10 n   1   To also handle zero and negative numbers  import math if n  gt  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   s 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   s 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

User · Answer

My code for the same is as follows i have used the log10 method   from math import     def digit count number    if number gt 1 and round log10 number   gt  log10 number  and number 10  0       return round log10 number   elif  number gt 1 and round log10 number   lt log10 number  and number 10  0      return round log10 number   1 elif number 10  0 and number  0      return int log10 number  1  elif number  1 or number  0      return 1   I had to  specify in case of 1 and 0 because log10 1  0 and log10 0  ND and hence the condition mentioned isn t satisfied  However  this code works only for whole numbers

User · Answer

Count the number of digits w o convert integer to a string   x 123 x abs x  i   0 while x  gt   10  i      i   1   i is the number of digits

User · Answer

from math import log10 digits   lambda n    n  0  and 1  or int log10 abs n    1

User · Answer

gt  gt  gt  a 12345  gt  gt  gt  a   str       len     5

User · Answer

Well  without converting to string I would do something like   def lenDigits x                Assumes int x               x   abs x       if x  lt  10          return 1      return 1   lenDigits x   10    Minimalist recursion FTW

User · Answer

def count digit number     if number  gt   10      count   2   else      count   1   while number  10  gt  9      count    1     number   number  10   return count

User · Answer

It can be done for integers quickly by using   len str abs 1234567890      Which gets the length of the string of the absolute value of  1234567890   abs returns the number WITHOUT any negatives  only the magnitude of the number   str casts converts it to a string and len returns the string length of that string   If you want it to work for floats  you can use either of the following     Ignore all after decimal place len str abs 0 1234567890   split      0      Ignore just the decimal place len str abs 0 1234567890   -1   For future reference

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Here is a bulky but fast version    def nbdigit   x        if x  gt   10000000000000000     17 -         return len  str  x        if x  lt  100000000     1 - 8         if x  lt  10000     1 - 4             if x  lt  100               return  x  gt   10  1              else                     return  x  gt   1000  3         else    5 - 8                                                              if x  lt  1000000           return  x  gt   100000  5              else                     return  x  gt   10000000  7     else    9 - 16          if x  lt  1000000000000     9 - 12             if x  lt  10000000000       return  x  gt   1000000000  9              else                     return  x  gt   100000000000  11         else    13 - 16             if x  lt  100000000000000   return  x  gt   10000000000000  13              else                     return  x  gt   1000000000000000  15   Only 5 comparisons for not too big numbers  On my computer it is about 30  faster than the math log10 version and 5  faster than the len  str    one  Ok    no so attractive if you don t use it furiously   And here is the set of numbers I used to test measure my function   n     int   i 1     17 7     for i in xrange  1000000       0 10  16-1 10  16 10  16 1    NB  it does not manage negative numbers  but the adaptation is easy

User · Answer

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

User · Answer

def digits n      count   0     if n    0          return 1          if n  lt  0          n    -1      while  n  gt   10  count           count    1         n    n 10      return count  print digits 25       Should print 2 print digits 144      Should print 3 print digits 1000     Should print 4 print digits 0        Should print 1

User · Answer

Python 2   ints take either 4 or 8 bytes  32 or 64 bits   depending on your Python build   sys maxint  2  31-1 for 32-bit ints  2  63-1 for 64-bit ints  will tell you which of the two possibilities obtains   In Python 3  ints  like longs in Python 2  can take arbitrary sizes up to the amount of available memory  sys getsizeof gives you a good indication for any given value  although it does also count some fixed overhead    gt  gt  gt  import sys  gt  gt  gt  sys getsizeof 0  12  gt  gt  gt  sys getsizeof 2  99  28   If  as other answers suggests  you re thinking about some string representation of the integer value  then just take the len of that representation  be it in base 10 or otherwise

User · Answer

Top answers are saying mathlog10 faster but I got results that suggest len str n   is faster  arr      for i in range 5000000       arr append random randint 0 12345678901234567890      timeit  for n in arr      len str n     2 72 s    304 ms per loop  mean    std  dev  of 7 runs  1 loop each     timeit  for n in arr      int math log10 n   1   3 13 s    545 ms per loop  mean    std  dev  of 7 runs  1 loop each   Besides  I haven t added logic to the math way to return accurate results and I can only imagine it slows it even more  I have no idea how the previous answers proved the maths way is faster though

User · Answer

Format in scientific notation and pluck off the exponent   int     5e   format 1000000  split  e   1     1   I don t know about speed  but it s simple    Please note the number of significant digits after the decimal  the  5  in the   5e  can be an issue if it rounds up the decimal part of the scientific notation to another digit   I set it arbitrarily large  but could reflect the length of the largest number you know about

[python] How to find length of digits in an integer?

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