# [Numpy-discussion] For loop tips

Torgil Svensson torgil.svensson at gmail.com
Tue Aug 29 13:59:55 CDT 2006

```def list2index(L):
idx=dict((y,x) for x,y in enumerate(set(L)))
return asmatrix(fromiter((idx[x] for x in L),dtype=int))

# old
\$ python test.py
Numbers: 29.4062280655 seconds
Characters: 84.6239070892 seconds
Dates: 117.560418844 seconds

# new
\$ python test.py
Numbers: 1.79700994492 seconds
Characters: 1.6025249958 seconds
Dates: 1.7974088192 seconds

16, 52 and 100 times faster

//Torgil

On 8/29/06, Keith Goodman <kwgoodman at gmail.com> 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.
>
> I've done it by brute force below. Any tips for making it faster? (5x
> would make it useful; 10x would be a dream.)
>
> >> 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'
>
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