author | jjkang |
Fri, 11 Jun 2010 15:22:40 +0800 | |
changeset 2 | 9da1e5517a66 |
parent 0 | ae805ac0140d |
permissions | -rw-r--r-- |
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# -*- coding: Latin-1 -*- |
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"""Heap queue algorithm (a.k.a. priority queue). |
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Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for |
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all k, counting elements from 0. For the sake of comparison, |
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non-existing elements are considered to be infinite. The interesting |
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property of a heap is that a[0] is always its smallest element. |
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Usage: |
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heap = [] # creates an empty heap |
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heappush(heap, item) # pushes a new item on the heap |
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item = heappop(heap) # pops the smallest item from the heap |
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item = heap[0] # smallest item on the heap without popping it |
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heapify(x) # transforms list into a heap, in-place, in linear time |
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item = heapreplace(heap, item) # pops and returns smallest item, and adds |
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# new item; the heap size is unchanged |
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Our API differs from textbook heap algorithms as follows: |
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- We use 0-based indexing. This makes the relationship between the |
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index for a node and the indexes for its children slightly less |
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obvious, but is more suitable since Python uses 0-based indexing. |
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- Our heappop() method returns the smallest item, not the largest. |
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These two make it possible to view the heap as a regular Python list |
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without surprises: heap[0] is the smallest item, and heap.sort() |
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maintains the heap invariant! |
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""" |
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# Original code by Kevin O'Connor, augmented by Tim Peters and Raymond Hettinger |
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__about__ = """Heap queues |
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[explanation by François Pinard] |
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Heaps are arrays for which a[k] <= a[2*k+1] and a[k] <= a[2*k+2] for |
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all k, counting elements from 0. For the sake of comparison, |
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non-existing elements are considered to be infinite. The interesting |
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property of a heap is that a[0] is always its smallest element. |
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The strange invariant above is meant to be an efficient memory |
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representation for a tournament. The numbers below are `k', not a[k]: |
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0 |
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1 2 |
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3 4 5 6 |
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7 8 9 10 11 12 13 14 |
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15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 |
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In the tree above, each cell `k' is topping `2*k+1' and `2*k+2'. In |
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an usual binary tournament we see in sports, each cell is the winner |
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over the two cells it tops, and we can trace the winner down the tree |
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to see all opponents s/he had. However, in many computer applications |
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of such tournaments, we do not need to trace the history of a winner. |
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To be more memory efficient, when a winner is promoted, we try to |
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replace it by something else at a lower level, and the rule becomes |
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that a cell and the two cells it tops contain three different items, |
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but the top cell "wins" over the two topped cells. |
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If this heap invariant is protected at all time, index 0 is clearly |
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the overall winner. The simplest algorithmic way to remove it and |
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find the "next" winner is to move some loser (let's say cell 30 in the |
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diagram above) into the 0 position, and then percolate this new 0 down |
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the tree, exchanging values, until the invariant is re-established. |
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This is clearly logarithmic on the total number of items in the tree. |
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By iterating over all items, you get an O(n ln n) sort. |
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A nice feature of this sort is that you can efficiently insert new |
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items while the sort is going on, provided that the inserted items are |
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not "better" than the last 0'th element you extracted. This is |
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especially useful in simulation contexts, where the tree holds all |
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incoming events, and the "win" condition means the smallest scheduled |
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time. When an event schedule other events for execution, they are |
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scheduled into the future, so they can easily go into the heap. So, a |
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heap is a good structure for implementing schedulers (this is what I |
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used for my MIDI sequencer :-). |
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Various structures for implementing schedulers have been extensively |
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studied, and heaps are good for this, as they are reasonably speedy, |
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the speed is almost constant, and the worst case is not much different |
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than the average case. However, there are other representations which |
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are more efficient overall, yet the worst cases might be terrible. |
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Heaps are also very useful in big disk sorts. You most probably all |
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know that a big sort implies producing "runs" (which are pre-sorted |
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sequences, which size is usually related to the amount of CPU memory), |
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followed by a merging passes for these runs, which merging is often |
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very cleverly organised[1]. It is very important that the initial |
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sort produces the longest runs possible. Tournaments are a good way |
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to that. If, using all the memory available to hold a tournament, you |
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replace and percolate items that happen to fit the current run, you'll |
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produce runs which are twice the size of the memory for random input, |
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and much better for input fuzzily ordered. |
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|
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Moreover, if you output the 0'th item on disk and get an input which |
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may not fit in the current tournament (because the value "wins" over |
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the last output value), it cannot fit in the heap, so the size of the |
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heap decreases. The freed memory could be cleverly reused immediately |
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for progressively building a second heap, which grows at exactly the |
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same rate the first heap is melting. When the first heap completely |
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vanishes, you switch heaps and start a new run. Clever and quite |
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effective! |
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In a word, heaps are useful memory structures to know. I use them in |
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a few applications, and I think it is good to keep a `heap' module |
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around. :-) |
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-------------------- |
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[1] The disk balancing algorithms which are current, nowadays, are |
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more annoying than clever, and this is a consequence of the seeking |
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119 |
capabilities of the disks. On devices which cannot seek, like big |
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120 |
tape drives, the story was quite different, and one had to be very |
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121 |
clever to ensure (far in advance) that each tape movement will be the |
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122 |
most effective possible (that is, will best participate at |
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123 |
"progressing" the merge). Some tapes were even able to read |
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124 |
backwards, and this was also used to avoid the rewinding time. |
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125 |
Believe me, real good tape sorts were quite spectacular to watch! |
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126 |
From all times, sorting has always been a Great Art! :-) |
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127 |
""" |
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128 |
|
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129 |
__all__ = ['heappush', 'heappop', 'heapify', 'heapreplace', 'nlargest', |
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130 |
'nsmallest'] |
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131 |
|
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132 |
from itertools import islice, repeat, count, imap, izip, tee |
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133 |
from operator import itemgetter, neg |
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134 |
import bisect |
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135 |
|
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136 |
def heappush(heap, item): |
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137 |
"""Push item onto heap, maintaining the heap invariant.""" |
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138 |
heap.append(item) |
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139 |
_siftdown(heap, 0, len(heap)-1) |
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140 |
|
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141 |
def heappop(heap): |
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142 |
"""Pop the smallest item off the heap, maintaining the heap invariant.""" |
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143 |
lastelt = heap.pop() # raises appropriate IndexError if heap is empty |
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144 |
if heap: |
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145 |
returnitem = heap[0] |
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146 |
heap[0] = lastelt |
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147 |
_siftup(heap, 0) |
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148 |
else: |
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149 |
returnitem = lastelt |
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150 |
return returnitem |
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151 |
|
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152 |
def heapreplace(heap, item): |
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153 |
"""Pop and return the current smallest value, and add the new item. |
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154 |
|
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155 |
This is more efficient than heappop() followed by heappush(), and can be |
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156 |
more appropriate when using a fixed-size heap. Note that the value |
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157 |
returned may be larger than item! That constrains reasonable uses of |
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158 |
this routine unless written as part of a conditional replacement: |
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159 |
|
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160 |
if item > heap[0]: |
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161 |
item = heapreplace(heap, item) |
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162 |
""" |
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163 |
returnitem = heap[0] # raises appropriate IndexError if heap is empty |
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164 |
heap[0] = item |
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165 |
_siftup(heap, 0) |
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166 |
return returnitem |
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167 |
|
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168 |
def heapify(x): |
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169 |
"""Transform list into a heap, in-place, in O(len(heap)) time.""" |
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170 |
n = len(x) |
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171 |
# Transform bottom-up. The largest index there's any point to looking at |
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172 |
# is the largest with a child index in-range, so must have 2*i + 1 < n, |
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173 |
# or i < (n-1)/2. If n is even = 2*j, this is (2*j-1)/2 = j-1/2 so |
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174 |
# j-1 is the largest, which is n//2 - 1. If n is odd = 2*j+1, this is |
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175 |
# (2*j+1-1)/2 = j so j-1 is the largest, and that's again n//2-1. |
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176 |
for i in reversed(xrange(n//2)): |
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177 |
_siftup(x, i) |
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178 |
|
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179 |
def nlargest(n, iterable): |
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180 |
"""Find the n largest elements in a dataset. |
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181 |
|
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182 |
Equivalent to: sorted(iterable, reverse=True)[:n] |
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183 |
""" |
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184 |
it = iter(iterable) |
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185 |
result = list(islice(it, n)) |
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186 |
if not result: |
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187 |
return result |
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188 |
heapify(result) |
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189 |
_heapreplace = heapreplace |
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190 |
sol = result[0] # sol --> smallest of the nlargest |
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191 |
for elem in it: |
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192 |
if elem <= sol: |
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193 |
continue |
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194 |
_heapreplace(result, elem) |
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195 |
sol = result[0] |
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196 |
result.sort(reverse=True) |
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197 |
return result |
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198 |
|
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199 |
def nsmallest(n, iterable): |
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200 |
"""Find the n smallest elements in a dataset. |
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201 |
|
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202 |
Equivalent to: sorted(iterable)[:n] |
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203 |
""" |
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204 |
if hasattr(iterable, '__len__') and n * 10 <= len(iterable): |
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205 |
# For smaller values of n, the bisect method is faster than a minheap. |
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206 |
# It is also memory efficient, consuming only n elements of space. |
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207 |
it = iter(iterable) |
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208 |
result = sorted(islice(it, 0, n)) |
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209 |
if not result: |
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210 |
return result |
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211 |
insort = bisect.insort |
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212 |
pop = result.pop |
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213 |
los = result[-1] # los --> Largest of the nsmallest |
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214 |
for elem in it: |
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215 |
if los <= elem: |
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216 |
continue |
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217 |
insort(result, elem) |
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218 |
pop() |
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219 |
los = result[-1] |
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220 |
return result |
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221 |
# An alternative approach manifests the whole iterable in memory but |
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222 |
# saves comparisons by heapifying all at once. Also, saves time |
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223 |
# over bisect.insort() which has O(n) data movement time for every |
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224 |
# insertion. Finding the n smallest of an m length iterable requires |
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225 |
# O(m) + O(n log m) comparisons. |
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226 |
h = list(iterable) |
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227 |
heapify(h) |
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228 |
return map(heappop, repeat(h, min(n, len(h)))) |
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229 |
|
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230 |
# 'heap' is a heap at all indices >= startpos, except possibly for pos. pos |
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231 |
# is the index of a leaf with a possibly out-of-order value. Restore the |
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232 |
# heap invariant. |
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233 |
def _siftdown(heap, startpos, pos): |
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234 |
newitem = heap[pos] |
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235 |
# Follow the path to the root, moving parents down until finding a place |
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236 |
# newitem fits. |
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237 |
while pos > startpos: |
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238 |
parentpos = (pos - 1) >> 1 |
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239 |
parent = heap[parentpos] |
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240 |
if parent <= newitem: |
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241 |
break |
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242 |
heap[pos] = parent |
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243 |
pos = parentpos |
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244 |
heap[pos] = newitem |
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245 |
|
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246 |
# The child indices of heap index pos are already heaps, and we want to make |
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247 |
# a heap at index pos too. We do this by bubbling the smaller child of |
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248 |
# pos up (and so on with that child's children, etc) until hitting a leaf, |
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249 |
# then using _siftdown to move the oddball originally at index pos into place. |
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250 |
# |
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251 |
# We *could* break out of the loop as soon as we find a pos where newitem <= |
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252 |
# both its children, but turns out that's not a good idea, and despite that |
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253 |
# many books write the algorithm that way. During a heap pop, the last array |
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254 |
# element is sifted in, and that tends to be large, so that comparing it |
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255 |
# against values starting from the root usually doesn't pay (= usually doesn't |
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256 |
# get us out of the loop early). See Knuth, Volume 3, where this is |
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257 |
# explained and quantified in an exercise. |
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258 |
# |
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259 |
# Cutting the # of comparisons is important, since these routines have no |
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260 |
# way to extract "the priority" from an array element, so that intelligence |
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261 |
# is likely to be hiding in custom __cmp__ methods, or in array elements |
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262 |
# storing (priority, record) tuples. Comparisons are thus potentially |
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263 |
# expensive. |
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264 |
# |
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265 |
# On random arrays of length 1000, making this change cut the number of |
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266 |
# comparisons made by heapify() a little, and those made by exhaustive |
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267 |
# heappop() a lot, in accord with theory. Here are typical results from 3 |
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268 |
# runs (3 just to demonstrate how small the variance is): |
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269 |
# |
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270 |
# Compares needed by heapify Compares needed by 1000 heappops |
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271 |
# -------------------------- -------------------------------- |
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272 |
# 1837 cut to 1663 14996 cut to 8680 |
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273 |
# 1855 cut to 1659 14966 cut to 8678 |
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274 |
# 1847 cut to 1660 15024 cut to 8703 |
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275 |
# |
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276 |
# Building the heap by using heappush() 1000 times instead required |
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277 |
# 2198, 2148, and 2219 compares: heapify() is more efficient, when |
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278 |
# you can use it. |
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279 |
# |
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280 |
# The total compares needed by list.sort() on the same lists were 8627, |
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281 |
# 8627, and 8632 (this should be compared to the sum of heapify() and |
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282 |
# heappop() compares): list.sort() is (unsurprisingly!) more efficient |
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283 |
# for sorting. |
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284 |
|
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285 |
def _siftup(heap, pos): |
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endpos = len(heap) |
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287 |
startpos = pos |
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288 |
newitem = heap[pos] |
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289 |
# Bubble up the smaller child until hitting a leaf. |
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290 |
childpos = 2*pos + 1 # leftmost child position |
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291 |
while childpos < endpos: |
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292 |
# Set childpos to index of smaller child. |
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293 |
rightpos = childpos + 1 |
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294 |
if rightpos < endpos and heap[rightpos] <= heap[childpos]: |
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295 |
childpos = rightpos |
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296 |
# Move the smaller child up. |
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297 |
heap[pos] = heap[childpos] |
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298 |
pos = childpos |
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299 |
childpos = 2*pos + 1 |
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300 |
# The leaf at pos is empty now. Put newitem there, and bubble it up |
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301 |
# to its final resting place (by sifting its parents down). |
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302 |
heap[pos] = newitem |
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303 |
_siftdown(heap, startpos, pos) |
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304 |
|
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305 |
# If available, use C implementation |
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306 |
try: |
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307 |
from _heapq import heappush, heappop, heapify, heapreplace, nlargest, nsmallest |
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308 |
except ImportError: |
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309 |
pass |
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310 |
|
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311 |
# Extend the implementations of nsmallest and nlargest to use a key= argument |
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312 |
_nsmallest = nsmallest |
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313 |
def nsmallest(n, iterable, key=None): |
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314 |
"""Find the n smallest elements in a dataset. |
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315 |
|
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316 |
Equivalent to: sorted(iterable, key=key)[:n] |
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317 |
""" |
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318 |
in1, in2 = tee(iterable) |
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319 |
it = izip(imap(key, in1), count(), in2) # decorate |
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320 |
result = _nsmallest(n, it) |
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321 |
return map(itemgetter(2), result) # undecorate |
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322 |
|
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323 |
_nlargest = nlargest |
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324 |
def nlargest(n, iterable, key=None): |
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325 |
"""Find the n largest elements in a dataset. |
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326 |
|
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327 |
Equivalent to: sorted(iterable, key=key, reverse=True)[:n] |
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328 |
""" |
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329 |
in1, in2 = tee(iterable) |
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330 |
it = izip(imap(key, in1), imap(neg, count()), in2) # decorate |
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331 |
result = _nlargest(n, it) |
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332 |
return map(itemgetter(2), result) # undecorate |
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333 |
|
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334 |
if __name__ == "__main__": |
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335 |
# Simple sanity test |
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336 |
heap = [] |
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337 |
data = [1, 3, 5, 7, 9, 2, 4, 6, 8, 0] |
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338 |
for item in data: |
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339 |
heappush(heap, item) |
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340 |
sort = [] |
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341 |
while heap: |
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342 |
sort.append(heappop(heap)) |
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343 |
print sort |