मैंने सभी आज्ञाकारी समाधानों के लिए एक छोटा परीक्षण सूट बनाया। मुझे पायथन में काम करने के लिए उन्हें कार्यों को थोड़ा बदलना पड़ा। 3. दिलचस्प बात यह है कि, PyPy में सबसे तेज़ कार्य कुछ मामलों में पायथन 2/3 का सबसे धीमा कार्य है।
import itertools
import time
from collections import OrderedDict
def tokland_org(lst, n):
if not lst:
yield []
else:
for group in (((lst[0],) + xs) for xs in itertools.combinations(lst[1:], n-1)):
for groups in tokland_org([x for x in lst if x not in group], n):
yield [group] + groups
tokland = lambda x: tokland_org(x, 2)
def gatoatigrado(lst):
N = len(lst)
choice_indices = itertools.product(*[
range(k) for k in reversed(range(1, N, 2)) ])
for choice in choice_indices:
tmp = list(lst)
result = []
for index in choice:
result.append( (tmp.pop(0), tmp.pop(index)) )
yield result
def shang(X):
lst = list(X)
if len(lst) < 2:
yield lst
return
a = lst[0]
for i in range(1,len(lst)):
pair = (a,lst[i])
for rest in shang(lst[1:i]+lst[i+1:]):
yield [pair] + rest
def smichr(X):
lst = list(X)
if not lst:
yield [tuple()]
elif len(lst) == 1:
yield [tuple(lst)]
elif len(lst) == 2:
yield [tuple(lst)]
else:
if len(lst) % 2:
for i in (None, True):
if i not in lst:
lst = lst + [i]
PAD = i
break
else:
while chr(i) in lst:
i += 1
PAD = chr(i)
lst = lst + [PAD]
else:
PAD = False
a = lst[0]
for i in range(1, len(lst)):
pair = (a, lst[i])
for rest in smichr(lst[1:i] + lst[i+1:]):
rv = [pair] + rest
if PAD is not False:
for i, t in enumerate(rv):
if PAD in t:
rv[i] = (t[0],)
break
yield rv
def adeel_zafar(X):
L = list(X)
if len(L) == 2:
yield [(L[0], L[1])]
else:
first = L.pop(0)
for i, e in enumerate(L):
second = L.pop(i)
for list_of_pairs in adeel_zafar(L):
list_of_pairs.insert(0, (first, second))
yield list_of_pairs
L.insert(i, second)
L.insert(0, first)
if __name__ =="__main__":
import timeit
import pprint
candidates = dict(tokland=tokland, gatoatigrado=gatoatigrado, shang=shang, smichr=smichr, adeel_zafar=adeel_zafar)
for i in range(1,7):
results = [ frozenset([frozenset(x) for x in candidate(range(i*2))]) for candidate in candidates.values() ]
assert len(frozenset(results)) == 1
print("Times for getting all permutations of sets of unordered pairs consisting of two draws from a 6-element deck until it is empty")
times = dict([(k, timeit.timeit('list({0}(range(6)))'.format(k), setup="from __main__ import {0}".format(k), number=10000)) for k in candidates.keys()])
pprint.pprint([(k, "{0:.3g}".format(v)) for k,v in OrderedDict(sorted(times.items(), key=lambda t: t[1])).items()])
print("Times for getting the first 2000 permutations of sets of unordered pairs consisting of two draws from a 52-element deck until it is empty")
times = dict([(k, timeit.timeit('list(islice({0}(range(52)), 800))'.format(k), setup="from itertools import islice; from __main__ import {0}".format(k), number=100)) for k in candidates.keys()])
pprint.pprint([(k, "{0:.3g}".format(v)) for k,v in OrderedDict(sorted(times.items(), key=lambda t: t[1])).items()])
"""
print("The 10000th permutations of the previous series:")
gens = dict([(k,v(range(52))) for k,v in candidates.items()])
tenthousands = dict([(k, list(itertools.islice(permutations, 10000))[-1]) for k,permutations in gens.items()])
for pair in tenthousands.items():
print(pair[0])
print(pair[1])
"""
वे सभी सटीक समान आदेश उत्पन्न करते हैं, इसलिए सेट आवश्यक नहीं हैं, लेकिन इस तरह से यह भविष्य के प्रमाण हैं। मैंने पायथन 3 रूपांतरण के साथ थोड़ा प्रयोग किया, यह हमेशा स्पष्ट नहीं होता है कि सूची का निर्माण कहां किया जाए, लेकिन मैंने कुछ विकल्पों की कोशिश की और सबसे तेज चुना।
यहाँ बेंचमार्क परिणाम हैं:
% echo "pypy"; pypy all_pairs.py; echo "python2"; python all_pairs.py; echo "python3"; python3 all_pairs.py
pypy
Times for getting all permutations of sets of unordered pairs consisting of two draws from a 6-element deck until it is empty
[('gatoatigrado', '0.0626'),
('adeel_zafar', '0.125'),
('smichr', '0.149'),
('shang', '0.2'),
('tokland', '0.27')]
Times for getting the first 2000 permutations of sets of unordered pairs consisting of two draws from a 52-element deck until it is empty
[('gatoatigrado', '0.29'),
('adeel_zafar', '0.411'),
('smichr', '0.464'),
('shang', '0.493'),
('tokland', '0.553')]
python2
Times for getting all permutations of sets of unordered pairs consisting of two draws from a 6-element deck until it is empty
[('gatoatigrado', '0.344'),
('adeel_zafar', '0.374'),
('smichr', '0.396'),
('shang', '0.495'),
('tokland', '0.675')]
Times for getting the first 2000 permutations of sets of unordered pairs consisting of two draws from a 52-element deck until it is empty
[('adeel_zafar', '0.773'),
('shang', '0.823'),
('smichr', '0.841'),
('tokland', '0.948'),
('gatoatigrado', '1.38')]
python3
Times for getting all permutations of sets of unordered pairs consisting of two draws from a 6-element deck until it is empty
[('gatoatigrado', '0.385'),
('adeel_zafar', '0.419'),
('smichr', '0.433'),
('shang', '0.562'),
('tokland', '0.837')]
Times for getting the first 2000 permutations of sets of unordered pairs consisting of two draws from a 52-element deck until it is empty
[('smichr', '0.783'),
('shang', '0.81'),
('adeel_zafar', '0.835'),
('tokland', '0.969'),
('gatoatigrado', '1.3')]
% pypy --version
Python 2.7.12 (5.6.0+dfsg-0~ppa2~ubuntu16.04, Nov 11 2016, 16:31:26)
[PyPy 5.6.0 with GCC 5.4.0 20160609]
% python3 --version
Python 3.5.2
तो मैं कहता हूं, गतोतिरिग्रो के समाधान के साथ जाओ।
itertoolsयदि आप पहले से ही नहीं है, तो आप उस मानक मॉड्यूल को देखना चाहते हैं । इस कार्य में इस समस्या (संभवतःpermutations,combinationsयाproductकार्य) के साथ मदद करने में सक्षम होना चाहिए ।