Python中最快的字典排序方法，python字典排序,def sbv0(adi

Python中最快的字典排序方法，python字典排序,def sbv0(adi

def sbv0(adict,reverse=False):”’ proposed at Digital Sanitation Engineering http://blog.modp.com/2007/11/sorting-python-dict-by-value.html ”’return sorted(adict.iteritems(), key=lambda (k,v): (v,k), reverse=reverse)def sbv1(d,reverse=False):”’  explicit list expansion ”’L = [(k,v) for (k,v) in d.iteritems()]return sorted(L, key=lambda x: x[1] , reverse=reverse)def sbv2(d,reverse=False):”’  generator ”’L = ((k,v) for (k,v) in d.iteritems())return sorted(L, key=lambda x: x[1] , reverse=reverse)def sbv3(d,reverse=False):”’ using a lambda to get the key, rather than “double-assignment” ”’return sorted(d.iteritems(), key=lambda x: x[1] , reverse=reverse)def sbv4(d,reverse=False):”’ using a formal function to get the sorting key, rather than a lambda”’def sk(x): return x[1]return sorted(d.iteritems(), key=sk , reverse=reverse)def sk(x): return x[1]def sbv5(d,reverse=False):”’ using a formal function, defined in outer scope to get the sorting key, rather than a lambda ”’return sorted(d.iteritems(), key=sk , reverse=reverse)from operator import itemgetterdef sbv6(d,reverse=False):”’ proposed in PEP 265, using  the itemgetter ”’return sorted(d.iteritems(), key=itemgetter(1), reverse=True)D = dict(zip(range(100),range(100)))from profile import runrun(“for ii in xrange(10000):  sbv0(D, reverse=True)”)run(“for ii in xrange(10000):  sbv1(D, reverse=True)”)run(“for ii in xrange(10000):  sbv2(D, reverse=True)”)run(“for ii in xrange(10000):  sbv3(D, reverse=True)”)run(“for ii in xrange(10000):  sbv4(D, reverse=True)”)run(“for ii in xrange(10000):  sbv5(D, reverse=True)”)run(“for ii in xrange(10000):  sbv6(D, reverse=True)”)

                                具体的结果, 直接拷贝代码到你的本机下运行即可知道. 在我这里, sbv6的时间是0.578s, 最慢的是用generator的sbv2, 22s左右, 其他时间是差不多的, 都是11s左右.

总的来说, PEP 265 介绍的方法基本是别的方法十倍左右的速度. 看来, 还是官网的东西比较可信阿:)