Create an empty list in python with certain size

Question

I want to create an empty list  or whatever is the best way  that can hold 10 elements   After that I want to assign values in that list  for example this is supposed to display 0 to 9   s1   list    for i in range 0 9      s1 i    i  print  s1   But when I run this code  it generates an error or in another case it just displays     empty     Can someone explain why

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I m surprised nobody suggest this simple approach to creating a list of empty lists  This is an old thread  but just adding this for completeness  This will create a list of 10 empty lists  x       for i in range 10

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Not technically a list but similar to a list in terms of functionality and it s a fixed length from collections import deque my deque size 10   deque maxlen 10   If it s full  ie got 10 items then adding another item results in item  index 0 being discarded  FIFO  but you can also append in either direction  Used in say  a rolling average of stats piping a list through it aka sliding a window over a list until you get a match against another deque object   If you need a list then when full just use list deque object

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Here s my code for 2D list in python which would read no  of rows from the input    empty      row   int input     for i in range row       temp   list map int  input   split         empty append temp   for i in empty      for j in i          print j  end          print

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This was written based on the original version of the question       I want to create a empty list  or whatever is the best way  can hold 10 elements    All lists can hold as many elements as you like  subject only to the limit of available memory  The only  size  of a list that matters is the number of elements currently in it      but when I run it  the result is      print display s1 is not valid syntax  based on your description of what you re seeing  I assume you meant display s1  and then print s1  For that to run  you must have previously defined a global s1 to pass into the function   Calling display does not modify the list you pass in  as written  Your code says  s1 is a name for whatever thing was passed in to the function  ok  now the first thing we ll do is forget about that thing completely  and let s1 start referring instead to a newly created list  Now we ll modify that list   This has no effect on the value you passed in   There is no reason to pass in a value here   There is no real reason to create a function  either  but that s beside the point   You want to  create  something  so that is the output of your function  No information is required to create the thing you describe  so don t pass any information in  To get information out  return it   That would give you something like   def display        s1   list        for i in range 0  9           s1 i    i     return s1   The next problem you will note is that your list will actually have only 9 elements  because the end point is skipped by the range function   As side notes     works just as well as list    the semicolon is unnecessary  s1 is a poor name for the variable  and only one parameter is needed for range if you re starting from 0   So then you end up with  def create list        result   list       for i in range 10           result i    i     return result   However  this is still missing the mark  range is not some magical keyword that s part of the language the way for and def are  but instead it s a function  And guess what that function returns  That s right - a list of those integers  So the entire function collapses to   def create list        return range 10    and now you see why we don t need to write a function ourselves at all  range is already the function we re looking for  Although  again  there is no need or reason to  pre-size  the list

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One simple way to create a 2D matrix of size n using nested list comprehensions   m     None for   in range n   for   in range n

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The accepted answer has some gotchas  For example    gt  gt  gt  a          3  gt  gt  gt  a               gt  gt  gt  a 0   hello     5  gt  gt  gt  a    hello   5     hello   5     hello   5    gt  gt  gt     So each dictionary refers to the same object  Same holds true if you initialize with arrays or objects   You could do this instead    gt  gt  gt  b       for i in range 0  3    gt  gt  gt  b               gt  gt  gt  b 0   hello     6  gt  gt  gt  b    hello   6            gt  gt  gt

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s1      for i in range 11      s1 append i   print s1   To create a list  just use these brackets        To add something to a list  use list append

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This code generates an array that contains 10 random numbers    import random numrand    for i in range 0 10      a   random randint 1 50     numrand append a     print a i  print numrand

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You can  append element  to the list  e g   s1 append i   What you are currently trying to do is access an element  s1 i   that does not exist

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Try this instead   lst    None    10   The above will create a list of size 10  where each position is initialized to None  After that  you can add elements to it   lst    None    10 for i in range 10       lst i    i   Admittedly  that s not the Pythonic way to do things  Better do this   lst      for i in range 10       lst append i    Or even simpler  in Python 2 x you can do this to initialize a list with values from 0 to 9   lst   range 10    And in Python 3 x   lst   list range 10

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Make it more reusable as a function   def createEmptyList length fill None               return a  empty  list of a given length     Example          print createEmptyList 3 -1           gt  gt   -1  -1  -1          print createEmptyList 4           gt  gt   None  None  None  None              return  fill    length

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varunl s currently accepted answer    gt  gt  gt  l    None    10   gt  gt  gt  l   None  None  None  None  None  None  None  None  None  None    Works well for non-reference types like numbers  Unfortunately if you want to create a list-of-lists you will run into referencing errors  Example in Python 2 7 6    gt  gt  gt  a        10  gt  gt  gt  a                                           gt  gt  gt  a 0  append 0   gt  gt  gt  a   0    0    0    0    0    0    0    0    0    0    gt  gt  gt     As you can see  each element is pointing to the same list object  To get around this  you can create a method that will initialize each position to a different object reference   def init list of objects size       list of objects   list       for i in range 0 size           list of objects append  list      different object reference each time     return list of objects    gt  gt  gt  a   init list of objects 10   gt  gt  gt  a                                           gt  gt  gt  a 0  append 0   gt  gt  gt  a   0                                        gt  gt  gt     There is likely a default  built-in python way of doing this  instead of writing a function   but I m not sure what it is  Would be happy to be corrected    Edit  It s      for   in range 10    Example      gt  gt  gt     random random   for   in range 2    for   in range 5    gt  gt  gt    0 7528051908943816  0 4325669600055032    0 510983236521753  0 7789949902294716    0 09475179523690558  0 30216475640534635    0 3996890132468158  0 6374322093017013    0 3374204010027543  0 4514925173253973

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You cannot assign to a list like lst i    something  unless the list already is initialized with at least i 1 elements  You need to use append to add elements to the end of the list  lst append something     You could use the assignment notation if you were using a dictionary    Creating an empty list    gt  gt  gt  l    None    10  gt  gt  gt  l  None  None  None  None  None  None  None  None  None  None    Assigning a value to an existing element of the above list    gt  gt  gt  l 1    5  gt  gt  gt  l  None  5  None  None  None  None  None  None  None  None    Keep in mind that something like l 15    5 would still fail  as our list has only 10 elements   range x  creates a list from  0  1  2      x-1     2 X only  Use list range 10   in 3 X   gt  gt  gt  l   range 10   gt  gt  gt  l  0  1  2  3  4  5  6  7  8  9    Using a function to create a list    gt  gt  gt  def display            s1              for i in range 9     This is just to tell you how to create a list              s1 append i          return s1       gt  gt  gt  print display    0  1  2  3  4  5  6  7  8    List comprehension  Using the squares because for range you don t need to do all this  you can just return range 0 9       gt  gt  gt  def display            return  x  2 for x in range 9         gt  gt  gt  print display    0  1  4  9  16  25  36  49  64

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I m a bit surprised that the easiest way to create an initialised list is not in any of these answers  Just use a generator in the list function   list range 9

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There are two  quick  methods   x   length of your list a    None  x   or a    None for   in xrange x     It appears that  None  x is faster    gt  gt  gt  from timeit import timeit  gt  gt  gt  timeit   None  100  number 10000  0 023542165756225586  gt  gt  gt  timeit   None for   in xrange 100    number 10000  0 07616496086120605   But if you are ok with a range  e g   0 1 2 3     x-1    then range x  might be fastest    gt  gt  gt  timeit  range 100   number 10000  0 012513160705566406

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I came across this SO question while searching for a similar problem  I had to build a 2D array and then replace some elements of each list  in 2D array  with elements from a dict   I then came across this SO question which helped me  maybe this will help other beginners to get around  The key trick was to initialize the 2D array as an numpy array and then using array i j  instead of array i  j    For reference this is the piece of code where I had to use this    nd array      for i in range 30       nd array append np zeros shape    32 1    new array      for i in range len lines        new array append nd array  new array   np asarray new array  for i in range len lines        splits   lines i  split          for j in range len splits             print new array i  j           new array i j    final embeddings dictionary str splits j   -1  reshape 32 1    Now I know we can use list comprehension but for simplicity sake I am using a nested for loop  Hope this helps others who come across this post

[python] Create an empty list in python with certain size

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