On 05/25/2018 10:13 AM, bartc wrote:
On 25/05/2018 17:58, Rob Gaddi wrote:
So, in the spirit of explicit being better than implicit, please
assume that for actual implementation replicate would be a static
method of actual list, rather than the conveniently executable hackjob
below.
_list = list
_nodefault = object()
class list(_list):
@staticmethod
def replicate(*n, fill=_nodefault, call=list):
That seems to work, but the dimensions are created in reverse order to
what I expected. Which is to have the order of indices corresponding to
the order of dimensions. So:
x=list.replicate(2,3,4)
print (len(x))
print (len(x[0]))
print (len(x[0][0]))
Gives output of 4, 3, 2 rather than 2, 3, 4.
Which means that x[0][0][3] is a bounds error.
A fair point. Something about multidimensional arrays always makes me
semi-dyslexic. And there I was wondering why making it work right had
required I rip dimensions off the list from the end instead.
Corrected version below. list.replicate(3, 2) now means "3 of 2 of
this" as one would expect. This is one of those days when doctest is my
favorite thing about Python.
_list = list
_nodefault = object()
class list(_list):
@staticmethod
def replicate(*n, fill=_nodefault, call=list):
"""Return a list of specified dimensions.
Fill and call can be used to prime the list with initial values, the
default is to create a list of empty lists.
Parameters:
n : List of dimensions
fill: If provided, the fill object is used in all locations.
call: If fill is not provided, the result of call (a function of
no arguments) is used in all locations.
Example:
>>> a = list.replicate(3, 2)
>>> a
[[[], []], [[], []], [[], []]]
>>> a[0][0].append('a')
>>> a
[[['a'], []], [[], []], [[], []]]
>>> b = list.replicate(3, 2, fill=[])
>>> b
[[[], []], [[], []], [[], []]]
>>> b[0][0].append('a')
>>> b
[[['a'], ['a']], [['a'], ['a']], [['a'], ['a']]]
>>> c = list.replicate(3, 2, call=dict)
>>> c
[[{}, {}], [{}, {}], [{}, {}]]
>>> c[0][0]['swallow'] = 'unladen'
>>> c
[[{'swallow': 'unladen'}, {}], [{}, {}], [{}, {}]]
>>> d = list.replicate(3, 2, fill=0)
>>> d
[[0, 0], [0, 0], [0, 0]]
>>> d[0][0] = 5
>>> d
[[5, 0], [0, 0], [0, 0]]
"""
if n:
this = n[0]
future = n[1:]
return [
list.replicate(*future, fill=fill, call=call)
for _ in range(this)
]
elif fill is _nodefault:
return call()
else:
return fill
--
Rob Gaddi, Highland Technology -- www.highlandtechnology.com
Email address domain is currently out of order. See above to fix.
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