Inspired by Py2.3's TimSort, this implementation of sets.py uses sorted lists instead of dictionaries. For clumped data patterns, the set operations can be super-efficient (for example, two sets can be determined to be disjoint with only O(n) comparisons). Also note, that the set elements are not required to be hashable; this provides a great deal more freedom than dictionary based implementations.
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Copy to clipboard1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 | """ altsets.py -- An alternate implementation of Sets.py
Implements set operations using sorted lists as the underlying data structure.
Advantages:
* Space savings -- lists are much more compact than a dictionary
based implementation.
* Flexibility -- elements do not need to be hashable, only __cmp__
is required.
* Fast operations depending on the underlying data patterns.
Non-overlapping sets get united, intersected, or differenced
with only log(N) element comparisons. Results are built using
fast-slicing.
* Algorithms are designed to minimize the number of compares
which can be expensive.
* Natural support for sets of sets. No special accomodation needs to
be made to use a set or dict as a set member, but users need to be
careful to not mutate a member of a set since that may breaks its
sort invariant.
Disadvantages:
* Set construction uses list.sort() with potentially N log(N)
comparisons.
* Membership testing and element addition use log(N) comparisons.
Element addition uses list.insert() with takes O(N) time.
ToDo:
* Make the search routine adapt to the data; falling backing to
a linear search when encountering random data.
"""
from bisect import bisect_left, insort_left
class Set(object):
def __init__(self, iterable):
data = list(iterable)
data.sort()
result = data[:1]
for elem in data[1:]:
if elem == result[-1]:
continue
result.append(elem)
self.data = result
def __repr__(self):
return 'Set(' + repr(self.data) + ')'
def __iter__(self):
return iter(self.data)
def __contains__(self, elem):
data = self.data
i = bisect_left(self.data, elem, 0)
return i<len(data) and data[i] == elem
def add(self, elem):
if elem not in self:
insort_left(self.data, elem)
def remove(self, elem):
data = self.data
i = bisect_left(self.data, elem, 0)
if i<len(data) and data[i] == elem:
del data[i]
def _getotherdata(other):
if not isinstance(other, Set):
other = Set(other)
return other.data
_getotherdata = staticmethod(_getotherdata)
def __cmp__(self, other, cmp=cmp):
return cmp(self.data, Set._getotherdata(other))
def union(self, other, find=bisect_left):
i = j = 0
x = self.data
y = Set._getotherdata(other)
result = Set([])
append = result.data.append
extend = result.data.extend
try:
while 1:
if x[i] == y[j]:
append(x[i])
i += 1
j += 1
elif x[i] > y[j]:
cut = find(y, x[i], j)
extend(y[j:cut])
j = cut
else:
cut = find(x, y[j], i)
extend(x[i:cut])
i = cut
except IndexError:
extend(x[i:])
extend(y[j:])
return result
def intersection(self, other, find=bisect_left):
i = j = 0
x = self.data
y = Set._getotherdata(other)
result = Set([])
append = result.data.append
try:
while 1:
if x[i] == y[j]:
append(x[i])
i += 1
j += 1
elif x[i] > y[j]:
j = find(y, x[i], j)
else:
i = find(x, y[j], i)
except IndexError:
pass
return result
def difference(self, other, find=bisect_left):
i = j = 0
x = self.data
y = Set._getotherdata(other)
result = Set([])
extend = result.data.extend
try:
while 1:
if x[i] == y[j]:
i += 1
j += 1
elif x[i] > y[j]:
j = find(y, x[i], j)
else:
cut = find(x, y[j], i)
extend(x[i:cut])
i = cut
except IndexError:
extend(x[i:])
return result
def symmetric_difference(self, other, find=bisect_left):
i = j = 0
x = self.data
y = Set._getotherdata(other)
result = Set([])
extend = result.data.extend
try:
while 1:
if x[i] == y[j]:
i += 1
j += 1
elif x[i] > y[j]:
cut = find(y, x[i], j)
extend(y[j:cut])
j = cut
else:
cut = find(x, y[j], i)
extend(x[i:cut])
i = cut
except IndexError:
extend(x[i:])
extend(y[j:])
return result
a = Set('abracadabra')
b = Set('alacazam')
print a < b
print a
print b
print map(a.__contains__, list('abcdr'))
print map(a.__contains__, list('0ey'))
print list(a)
print a.union(b), ' :union'
print b.union(a), ' :union'
print a.intersection(b), ' :intersection'
print a.difference(b), ' :difference'
print b.difference(a), ' :difference'
print a.symmetric_difference(b), ' :symmetric_difference'
print b.symmetric_difference(a), ' :symmetric_difference'
print a.intersection(b).union(a.symmetric_difference(b)) == a.union(b)
print a.intersection(b).intersection(a.symmetric_difference(b)) == Set([])
|
In its current form, the set operations perform abysmally with random data. Timsort style adaptive search behavior is expected to be the cure.
Set construction and membership testing take longer than dictionary based implementations. This implementation is more appropriate for applications that need to be space efficient and that need fast set operations.
Tags: algorithms
Set.add problem? Doesn't the add method need to check if the new element is already in the set? As it's inserted with insort_left, it could be equal to its immediate successor in the list. Keith
Yes. Will fix it. Thanks.
Fix for Set.add problem. I didn't see any fix from Raymond, so I suggest:
The insertion here takes O(N), which is not very efficient
The sortedcontainers module does something similar using lists of lists. It implements sorted list, sorted dict, and sorted set data types in pure-Python and is fast-as-C implementations (even faster!). Learn more about sortedcontainers, available on PyPI and github.