I need to create a list of size N from already given list.
N = 3
a_list = [10,4,18,2,6,19,24,1,20]
The O/P should be:
[10,4,18] [4,18,2] [18,2,6] [2,6,19] [6,19,24] [19,24,1] [24,1,20]
It's like a window size of N=3 , which slides one step to its right.
How will I do it?
A faster way:
>>> zip(a_list,a_list[1:],a_list[2:])
[(10, 4, 18), (4, 18, 2), (18, 2, 6), (2, 6, 19), (6, 19, 24), (19, 24, 1), (24, 1, 20)]
Comparison:
In [6]: %timeit [a_list[i:i+n] for i in xrange(len(a_list)-n+1)]
100000 loops, best of 3: 9.61 us per loop
In [7]: %timeit zip(a_list,a_list[1:],a_list[2:])
100000 loops, best of 3: 5.23 us per loop
Or more general:
>>> zip(*[a_list[i:] for i in range(3)]) #3 (or 2, 4, 5, etc)is the length of step
For larger list, probably you have to use numpy to get a faster solution than @Ashwini Chaudhary's (http://www.rigtorp.se/2011/01/01/rolling-statistics-numpy.html):
import numpy as np
lista=np.array(lis)
def rolling_window(a, window):
shape = a.shape[:-1] + (a.shape[-1] - window + 1, window)
strides = a.strides + (a.strides[-1],)
return np.lib.stride_tricks.as_strided(a, shape=shape, strides=strides)
%timeit [lis[i:i+n] for i in xrange(len(lis)-n+1)]
%timeit rolling_window(lista, n)
1 loops, best of 3: 171 ms per loop
100000 loops, best of 3: 19.7 µs per loop
zip(*[a_list[i:] for i in range(3)]) be valid for n=5
3 to any number, like 5. That is essentially zip(a_list,a_list[1:],a_list[2:],.....)10 4 18 2 6 19 24 1 20 . I'm able to read that string but how should I convert this 'Line' to an List?map(int,s[:-1].split(' ')), s is your line 10 4 18 2 6 19 24 1 20\n, because when you readline() there is a \n in the end.numpy, see edit. Cheers.Use list comprehension and slicing:
>>> lis = [10,4,18,2,6,19,24,1,20]
>>> n = 3
>>> [lis[i:i+n] for i in xrange(len(lis)-n+1)]
[[10, 4, 18], [4, 18, 2], [18, 2, 6], [2, 6, 19], [6, 19, 24], [19, 24, 1], [24, 1, 20]]
>>> n = 4
>>> [lis[i:i+n] for i in xrange(len(lis)-n+1)]
[[10, 4, 18, 2], [4, 18, 2, 6], [18, 2, 6, 19], [2, 6, 19, 24], [6, 19, 24, 1], [19, 24, 1, 20]]
For a bigger list the zip based approach is actually slower:
In [27]: n = 100
In [28]: lis = [10,4,18,2,6,19,24,1,20]*10000
In [30]: %timeit zip(*[lis[i:] for i in xrange(n)])
1 loops, best of 3: 593 ms per loop
In [31]: %timeit [lis[i:i+n] for i in xrange(len(lis)-n+1)]
10 loops, best of 3: 114 ms per loop
10 4 18 2 6 19 24 1 20 . I'm able to read that string but how should I convert this 'Line' to an List?list((f1.readline()).replace(" ","")). it returns me ['1', '0', '4', '1', '8', '2', '6', '1', '9', '2', '4', '1', '2', '0', '\n']
lis = '10 4 18 2 6 19 24 1 20'.split()try this
>>> lis = [10,4,18,2,6,19,24,1,20]
>>> n=3
>>> [lis[i:i+n] for i in range(len(lis))][:-2]
the output is
[[10, 4, 18], [4, 18, 2], [18, 2, 6], [2, 6, 19], [6, 19, 24], [19, 24, 1], [24, 1, 20]]
[10,4,18],[2,6,19],[24,1,20]