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So I was wondering how to best create a list of blank lists:
[[],[],[]...] Because of how Python works with lists in memory, this doesn't work:
[[]]*n This does create [[],[],...] but each element is the same list:
d = [[]]*n d[0].append(1) #[[1],[1],...] Something like a list comprehension works:
d = [[] for x in xrange(0,n)] But this uses the Python VM for looping. Is there any way to use an implied loop (taking advantage of it being written in C)?
d = [] map(lambda n: d.append([]),xrange(0,10)) This is actually slower. :(
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To create a list of n placeholder elements, multiply the list of a single placeholder element with n . For example, use [None] * 5 to create a list [None, None, None, None, None] with five elements None . You can then overwrite some elements with index assignments.
To access the first n elements from a list, we can use the slicing syntax [ ] by passing a 0:n as an arguments to it . 0 is the start index (it is inculded). n is end index (it is excluded).
Using append() function to create a list of lists in Python. What append() function does is that it combines all the lists as elements into one list. It adds a list at the end of the list.
The probably only way which is marginally faster than
d = [[] for x in xrange(n)] is
from itertools import repeat d = [[] for i in repeat(None, n)] It does not have to create a new int object in every iteration and is about 15 % faster on my machine.
Edit: Using NumPy, you can avoid the Python loop using
d = numpy.empty((n, 0)).tolist() but this is actually 2.5 times slower than the list comprehension.
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