List of zeros in python [duplicate]

#add code here to figure out the number of 0's you need, naming the variable n.listofzeros = [0] * n

if you prefer to put it in the function, just drop in that code and add return listofzeros

Which would look like this:

def zerolistmaker(n):    listofzeros = [0] * n    return listofzeros

sample output:

>>> zerolistmaker(4)[0, 0, 0, 0]>>> zerolistmaker(5)[0, 0, 0, 0, 0]>>> zerolistmaker(15)[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]>>> 

$python 2.7.8from timeit import timeitimport numpytimeit("list(0 for i in xrange(0, 100000))", number=1000)> 8.173301935195923timeit("[0 for i in xrange(0, 100000)]", number=1000)> 4.881675958633423timeit("[0] * 100000", number=1000)> 0.6624710559844971timeit('list(itertools.repeat(0, 100000))', 'import itertools', number=1000)> 1.0820629596710205

You should use [0] * n to generate a list with n zeros.

See why [] is faster than list()

There is a gotcha though, both itertools.repeat and [0] * n will create lists whose elements refer to same id. This is not a problem with immutable objects like integers or strings but if you try to create list of mutable objects like a list of lists ([[]] * n) then all the elements will refer to the same object.

a = [[]] * 10a[0].append(1)a> [[1], [1], [1], [1], [1], [1], [1], [1], [1], [1]]

[0] * n will create the list immediately while repeat can be used to create the list lazily when it is first accessed.

If you're dealing with really large amount of data and your problem doesn't need variable length of list or multiple data types within the list it is better to use numpy arrays.

timeit('numpy.zeros(100000, numpy.int)', 'import numpy', number=1000)> 0.057849884033203125

numpy arrays will also consume less memory.

The easiest way to create a list where all values are the same is multiplying a one-element list by n.

>>> [0] * 4[0, 0, 0, 0]

So for your loop:

for i in range(10):    print [0] * i