使用itertools.groupby性能numpy的分组 [英] Numpy grouping using itertools.groupby performance
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问题描述
我有一个整数,将包含重复的很多大型(> 35000000)名单。我需要得到一个计数列表中的每个整数。下面code的作品,但似乎慢。能否任何人都更使用Python和$ P $基准pferably numpy的?
I have many large (>35,000,000) lists of integers that will contain duplicates. I need to get a count for each integer in a list. The following code works, but seems slow. Can anyone else better the benchmark using Python and preferably Numpy?
def group():
import numpy as np
from itertools import groupby
values = np.array(np.random.randint(0,1<<32,size=35000000),dtype='u4')
values.sort()
groups = ((k,len(list(g))) for k,g in groupby(values))
index = np.fromiter(groups,dtype='u4,u2')
if __name__=='__main__':
from timeit import Timer
t = Timer("group()","from __main__ import group")
print t.timeit(number=1)
返回:
$ python bench.py
111.377498865
干杯!
修改基于响应:
def group_original():
import numpy as np
from itertools import groupby
values = np.array(np.random.randint(0,1<<32,size=35000000),dtype='u4')
values.sort()
groups = ((k,len(list(g))) for k,g in groupby(values))
index = np.fromiter(groups,dtype='u4,u2')
def group_gnibbler():
import numpy as np
from itertools import groupby
values = np.array(np.random.randint(0,1<<32,size=35000000),dtype='u4')
values.sort()
groups = ((k,sum(1 for i in g)) for k,g in groupby(values))
index = np.fromiter(groups,dtype='u4,u2')
def group_christophe():
import numpy as np
values = np.array(np.random.randint(0,1<<32,size=35000000),dtype='u4')
values.sort()
counts=values.searchsorted(values, side='right') - values.searchsorted(values, side='left')
index = np.zeros(len(values),dtype='u4,u2')
index['f0']=values
index['f1']=counts
#Erroneous result!
def group_paul():
import numpy as np
values = np.array(np.random.randint(0,1<<32,size=35000000),dtype='u4')
values.sort()
diff = np.concatenate(([1],np.diff(values)))
idx = np.concatenate((np.where(diff)[0],[len(values)]))
index = np.empty(len(idx)-1,dtype='u4,u2')
index['f0']=values[idx[:-1]]
index['f1']=np.diff(idx)
if __name__=='__main__':
from timeit import Timer
timings=[
("group_original","Original"),
("group_gnibbler","Gnibbler"),
("group_christophe","Christophe"),
("group_paul","Paul"),
]
for method,title in timings:
t = Timer("%s()"%method,"from __main__ import %s"%method)
print "%s: %s secs"%(title,t.timeit(number=1))
返回:
$ python bench.py
Original: 113.385262966 secs
Gnibbler: 71.7464978695 secs
Christophe: 27.1690568924 secs
Paul: 9.06268405914 secs
尽管克里斯托夫给出不正确的结果目前
Although Christophe gives incorrect results currently
推荐答案
我得到了三倍的改善做这样的事情:
i get a 3x improvement doing something like this:
def group():
import numpy as np
values = np.array(np.random.randint(0,3298,size=35000000),dtype='u4')
values.sort()
dif = np.ones(values.shape,values.dtype)
dif[1:] = np.diff(values)
idx = np.where(dif>0)
vals = values[idx]
count = np.diff(idx)
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