我發現更短和更快的方法:
from itertools import groupby
from operator import itemgetter
from time import clock
mydict={('a', 1): 0,
('a', 2): 0,
('a', 3): 0,
('a', 4): 1,
('a', 5): 2,
('a', 6): 2,
('a', 7): 0,
('a', 8): 0,
}
A,B,C = [],[],[]
for i in xrange(1000):
t0 = clock()
data = mydict.items()
data.sort()
def groupkey(item):
return item[0][0], item[1]
result1 = {}
for v, group in groupby(data, key=groupkey):
char, value = v
nums = [item[0][1] for item in group]
result1[char, min(nums), max(nums)] = value
A.append(clock()-t0)
#----------------------------------------------------------------
t0 = clock()
data = [ [a,b,c] for ((a,b),c) in mydict.items()]
data.sort()
result2 = {}
for (char,value),group in groupby(data, key=itemgetter(0,2)):
nums = [item[1] for item in group]
result2[char,nums[0],nums[-1]] = value
B.append(clock()-t0)
#-----------------------------------------------------------------
t0 = clock()
data = [ [a,b,c] for ((a,b),c) in mydict.items()]
data.sort()
result3 = {}
for ((char,value),nums) in [ (cle,[item[1] for item in group]) for cle,group in groupby(data, key=itemgetter(0,2))]:
result3[char,nums[0],nums[-1]] = value
C.append(clock()-t0)
print 'result1==',result1
print 'result2==',result2
print 'result3==',result3
print 'result1==result2==result3==',result1==result2==result3
print id(result1)==id(result2),id(result2)==id(result3),id(result3)==id(result1)
print '{:.1%}.'.format(min(B)/min(A))
print '{:.1%}.'.format(min(C)/min(A))
結果:
RESULT1 == {( '一個',5,6):2,( '一個',4,4):1,( '一個', ('a',4,4):0,('a',1,3):0}
result2 == { ,('a',7,8):0,('a',1,3):0}
result3 == {('a',5,6}:2,('a', 4,4):1,('a',7,8):0,('a',1,3):0}
result1 == result2 == result3 == True
假假假
87.0%。
93.2%。
這些不是Python字典;它們甚至不是Python中的有效表達式。 – infrared 2010-12-15 19:41:15
'(a,2)= 0'行怎麼樣? – robert 2010-12-15 19:41:51
這看起來不像蟒蛇。什麼是? 2發生了什麼?字典的關鍵和價值是什麼? – 2010-12-15 19:42:38