Converting integer to digit list

Question

What is the quickest and cleanest way to convert an integer into a list    For example  change 132 into  1 3 2  and 23 into  2 3   I have a variable which is an int  and I want to be able to compare the individual digits so I thought making it into a list would be best  since I can just do int number 0    int number 1   to easily convert the list element back into int for digit operations

User · Answer

Use list on a number converted to string   In  1    int x  for x in list str 123    Out 2    1  2  3

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you can use   First convert the value in a string to iterate it  Them each value can be convert to a Integer value   12345  l     int item  for item in str value

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n   int raw input  n       def int to list n       l          while n    0          l    n   10    l         n   n    10     return l  print int to list n

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There are already great methods already mentioned on this page  however it does seem a little obscure as to which to use  So I have added some mesurements so you can more easily decide for yourself    A large number has been used  for overhead  1111111111111122222222222222222333333333333333333333 Using map int  str num    import timeit  def method        num   1111111111111122222222222222222333333333333333333333     return map int  str num    print timeit timeit  quot method   quot   setup  quot from   main   import method quot   number 10000   Output  0 018631496999999997 Using list comprehension  import timeit def method        num   1111111111111122222222222222222333333333333333333333     return  int x  for x in str num    print timeit timeit  quot method   quot   setup  quot from   main   import method quot   number 10000    Output  0 28403817900000006 Code taken from this answer The results show that the first method involving inbuilt methods is much faster than list comprehension  The  quot mathematical way quot   import timeit  def method        q   1111111111111122222222222222222333333333333333333333     ret          while q    0          q  r   divmod q  10    Divide by 10  see the remainder         ret insert 0  r    The remainder is the first to the right digit     return ret  print timeit timeit  quot method   quot   setup  quot from   main   import method quot   number 10000    Output  0 38133582499999996 Code taken from this answer The list str 123   method  does not provide the right output   import timeit  def method        return list str 1111111111111122222222222222222333333333333333333333        print timeit timeit  quot method   quot   setup  quot from   main   import method quot   number 10000    Output  0 028560138000000013 Code taken from this answer The answer by Duberly Gonz  lez Molinari  import timeit  def method        n   1111111111111122222222222222222333333333333333333333     l          while n    0          l    n   10    l         n   n    10     return l  print timeit timeit  quot method   quot   setup  quot from   main   import method quot   number 10000    Output  0 37039988200000007 Code taken from this answer Remarks  In all cases the map int  str num   is the fastest method  and is therefore probably the best method to use    List comprehension is the second fastest  but the method using map int  str num   is probably the most desirable of the two  Those that reinvent the wheel are interesting but are probably not so desirable in real use

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If you have a string like this   123456  and you want a list of integers like this   1 2 3 4 5 6   use this    gt  gt  gt s    123456       gt  gt  gt list1    int i  for i in list s    gt  gt  gt print list1    1 2 3 4 5 6    or if you want a list of strings like this    1   2   3   4   5   6    use this    gt  gt  gt s    123456       gt  gt  gt list1   list s   gt  gt  gt print list1     1   2   3   4   5   6

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The shortest and best way is already answered  but the first thing I thought of was the mathematical way  so here it is   def intlist n       q   n     ret          while q    0          q  r   divmod q  10    Divide by 10  see the remainder         ret insert 0  r    The remainder is the first to the right digit     return ret  print intlist 3  print  -  print intlist 10  print  --  print intlist 137    It s just another interesting approach  you definitely don t have to use such a thing in practical use cases

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gt  gt  gt list map int  str number      number is a given integer   It returns a list of all digits of number

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By looping it can be done the following way     num1  int input  Enter the number    sum1   num1  making a alt int to store the value of the orginal so it wont be affected y       making a list  while True      if sum1  0   checking if the number is not zero so it can break if it is         break     d   sum1 10  last number of your integer is saved in d     sum1   int sum1 10   integer is now with out the last number ie 4320 10 become 432     y append d    appending the last number in the first place  y reverse   as last is in first   reversing the number to orginal form print y    Answer becomes   Enter the number2342  2  3  4  2

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Convert the integer to string first  and then use map to apply int on it    gt  gt  gt  num   132  gt  gt  gt  map int  str num       note  This will return a map object in python 3   1  3  2    or using a list comprehension    gt  gt  gt   int x  for x in str num    1  3  2

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num   list str 100   index   len num  while index  gt  0      index -  1     num index    int num index   print num    It prints  1  0  0  object

[python] Converting integer to digit list

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