converting 1 dimensional array to 2 dimensional array
Hi All, I know how to reshape arrays, my problem is a little more complicated than that. I am looking for the most efficient way to do the following and an example will help: 1) I have a an array of bytes b = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] This bytes array represents a width of 2 and a height of 3. Each 2 bytes in that bytes array makes an unsigned 16 bit integer. 2) I want to get the following 2 dimensional array from the bytes array where each item in the two dimensional array is an unsigned 16 bit integer something like: [ [(1,2) , (3,4)], [(5,6) , (7,8)], [(9,10) , (11,12)] ] in the first row there are 2 elements each one is made of 2 bytes from the bytes array etc... meaning that (1, 2) = b[0] + b[1] << 8 (3, 4) = b[2] + b[3] << 8 ... ... What is the most efficient way to achieve that with numpy ? Thanks 0L [1] << 8 ,[2], [3, 4]
Hi Omry! You're looking for `.view()`: In [1]: import numpy as np In [2]: b = np.arange(1, 13).astype(np.uint8) In [4]: y = b.view(np.uint16).reshape((3, 2)) In [5]: y Out[5]: array([[ 513, 1027], [1541, 2055], [2569, 3083]], dtype=uint16) You can also change the endianness by replacing `np.uint16` with `'>u2'`. In [6]: z = b.view('>u2') In [7]: z Out[7]: array([ 258, 772, 1286, 1800, 2314, 2828], dtype=uint16) Hope this helps! Juan. On Mon, 15 Jul 2019, at 12:45 PM, Omry Levy wrote:
Hi Omry! You're looking for `.view()`: In [1]: import numpy as np In [2]: b = np.arange(1, 13).astype(np.uint8) In [4]: y = b.view(np.uint16).reshape((3, 2)) In [5]: y Out[5]: array([[ 513, 1027], [1541, 2055], [2569, 3083]], dtype=uint16) You can also change the endianness by replacing `np.uint16` with `'>u2'`. In [6]: z = b.view('>u2') In [7]: z Out[7]: array([ 258, 772, 1286, 1800, 2314, 2828], dtype=uint16) Hope this helps! Juan. On Mon, 15 Jul 2019, at 12:45 PM, Omry Levy wrote:
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Juan Nunez-Iglesias -
Omry Levy