# [Numpy-discussion] the fast way to loop over ndarray elements?

Chao YUE chaoyuejoy at gmail.com
Sat Nov 17 08:43:04 EST 2012

```Yes, both the "base" and "target" are ascending.  Thanks!

Chao

On Sat, Nov 17, 2012 at 2:40 PM, Benjamin Root <ben.root at ou.edu> wrote:

>
>
> On Saturday, November 17, 2012, Chao YUE wrote:
>
>> Dear all,
>>
>> I need to make a linear contrast of the 2D numpy array "data" from an
>> interval to another, the approach is:
>> I have another two list: "base" & "target", then I check for each ndarray
>> element "data[i,j]",
>> if   base[m] =< data[i,j] <= base[m+1], then it will be linearly
>> converted to be in the interval of (target[m], target[m+1]),
>> using another function called "lintrans".
>>
>>
>> #The way I do is to loop each row and column of the 2D array, and finally
>> loop the intervals constituted by base list:
>>
>> for row in range(data.shape[0]):
>>     for col in range(data.shape[1]):
>>         for i in range(len(base)-1):
>>             if data[row,col]>=base[i] and data[row,col]<=base[i+1]:
>>
>> data[row,col]=lintrans(data[row,col],(base[i],base[i+1]),(target[i],target[i+1]))
>>                 break  #use break to jump out of loop as the data have to
>> be ONLY transferred ONCE.
>>
>>
>> Now the profiling result shows that most of the time has been used in
>> this loop over the array ("plot_array_transg"),
>> and less time in calling the linear transformation fuction "lintrans":
>>
>>    ncalls     tottime  percall    cumtime    percall
>> filename:lineno(function)
>>   18047    0.110    0.000      0.110        0.000
>> mathex.py:132(lintrans)
>>   1            12.495  12.495   19.061      19.061
>> mathex.py:196(plot_array_transg)
>>
>>
>> so is there anyway I can speed up this loop?  Thanks for any suggestions!!
>>
>> best,
>>
>> Chao
>>
>>
> If the values in base are ascending, you can use searchsorted() to find
> out where values from data can be placed into base while maintaining order.
>  Don't know if it is faster, but it would certainly be easier to read.
>
> Cheers!
> Ben Root
>
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>
>

--
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Chao YUE
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