from itertools import groupby
from operator import itemgetter
mylist = [0, 0, 0, 1, 2, 2, 0, 0]
def interval(v):
head = tail = next(v)
for tail in v:
pass
return head[0] + 1, tail[0] + 1
print({interval(v): k for k, v in groupby(enumerate(mylist), key=itemgetter(1))})
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))
结果是:
^{pr2}$如果将此数据存储在列表中,则会更容易:
给予
^{pr2}$我找到了一种更短更快的方法:
结果是:
结果1=={('a',5,6):2,('a',4,4):1,('a',7,8):0,('a',1,3):0}
结果2=={('a',5,6):2,('a',4,4):1,('a',7,8):0,('a',1,3):0}
结果3=={('a',5,6):2,('a',4,4):1,('a',7,8):0,('a',1,3):0}
结果1==结果2==结果3==真
假假假
87.0%。在
93.2%。在
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