Tim Hochberg wrote:
> Keith Goodman wrote:
>
>> I have a very long list that contains many repeated elements. The
>> elements of the list can be either all numbers, or all strings, or all
>> dates [datetime.date].
>>
>> I want to convert the list into a matrix where each unique element of
>> the list is assigned a consecutive integer starting from zero.
>>
>>
> If what you want is that the first unique element get's zero, the second
> one, I don't think the code below will work in general since the dict
> does not preserve order. You might want to look at the results for the
> character case to see what I mean. If you're looking for something else,
> you'll need to elaborate a bit. Since list2index doesn't return
> anything, it's not entirely clear what the answer consists of. Just idx?
> Idx plus uL?
>
>
>> I've done it by brute force below. Any tips for making it faster? (5x
>> would make it useful; 10x would be a dream.)
>>
>>
> Assuming I understand what you're trying to do, this might help:
>
> def list2index2(L):
> idx = ones([len(L)])
> map = {}
> for i, x in enumerate(L):
> index = map.get(x)
> if index is None:
> map[x] = index = len(map)
> idx[i] = index
> return idx
>
>
> It's almost 10x faster for numbers and about 40x faster for characters
> and dates. However it produces different results from list2index in the
> second two cases. That may or may not be a good thing depending on what
> you're really trying to do.
>
Ugh! I fell victim to premature optimization disease. The following is
both clearer and faster: Sigh.
def list2index3(L):
idx = ones([len(L)])
map = {}
for i, x in enumerate(L):
if x not in map:
map[x] = len(map)
idx[i] = map[x]
return idx
> -tim
>
>
>>
>>
>>>> list2index.test()
>>>>
>>>>
>> Numbers: 5.84955787659 seconds
>> Characters: 24.3192870617 seconds
>> Dates: 39.288228035 seconds
>>
>>
>> import datetime, time
>> from numpy import nan, asmatrix, ones
>>
>> def list2index(L):
>>
>> # Find unique elements in list
>> uL = dict.fromkeys(L).keys()
>>
>> # Convert list to matrix
>> L = asmatrix(L).T
>>
>> # Initialize return matrix
>> idx = nan * ones((L.size, 1))
>>
>> # Assign numbers to unique L values
>> for i, uLi in enumerate(uL):
>> idx[L == uLi,:] = i
>>
>> def test():
>>
>> L = 5000*range(255)
>> t1 = time.time()
>> idx = list2index(L)
>> t2 = time.time()
>> print 'Numbers:', t2-t1, 'seconds'
>>
>> L = 5000*[chr(z) for z in range(255)]
>> t1 = time.time()
>> idx = list2index(L)
>> t2 = time.time()
>> print 'Characters:', t2-t1, 'seconds'
>>
>> d = datetime.date
>> step = datetime.timedelta
>> L = 5000*[d(2006,1,1)+step(z) for z in range(255)]
>> t1 = time.time()
>> idx = list2index(L)
>> t2 = time.time()
>> print 'Dates:', t2-t1, 'seconds'
>>
>> -------------------------------------------------------------------------
>> Using Tomcat but need to do more? Need to support web services, security?
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>>
>>
>>
>>
>
>
>
> -------------------------------------------------------------------------
> Using Tomcat but need to do more? Need to support web services, security?
> Get stuff done quickly with pre-integrated technology to make your job easier
> Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo
> http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642
> _______________________________________________
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>
>
>
-------------------------------------------------------------------------
Using Tomcat but need to do more? Need to support web services, security?
Get stuff done quickly with pre-integrated technology to make your job easier
Download IBM WebSphere Application Server v.1.0.1 based on Apache Geronimo
http://sel.as-us.falkag.net/sel?cmd=lnk&kid=120709&bid=263057&dat=121642
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