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I have two long (7683200 length) 1D arrays corresponding to two clock cycles, that I wish to combine into single 1D array. The clock_ticks_fine cycles between 0-255, and corresponds to a coarse division of 7.5 msec, with the reset to 0 signified by a 1 in the clock_ticks_coarse array.

I would like this to be much much faster (I have many of these files), but I can't work out how to vectorise the code. I don't care about the clock_ticks_fine value when clock_ticks_coarse is equal to 1.

First 100 datapoints in both arrays using a basic for loop:

corse_div = 7.5 # in msec
fine_div = 7.5/255.0

clock_ticks_fine = np.array([ 36,   1,  11,  22,  39,  66,  71,  78, 107, 121, 137, 142, 163,
       190, 211, 212, 239, 246,  36,   7,  13,  47,  56,  73,  78, 110,
       128, 143, 146, 175, 192, 203, 204, 239,  36,   0,  13,  19,  43,
        58,  74,  81, 111, 130, 139, 139, 175, 189, 208, 209, 238,  36,
         2,  19,  22,  39,  66,  76,  79, 101, 130, 145, 146, 163, 184,
       201, 214, 227, 255,  36,   9,  20,  45,  64,  75,  78, 103, 122,
       143, 144, 175, 184, 205, 208, 235, 250,  36,  14,  17,  46,  62,
        71,  72, 103, 119, 143, 144, 175, 186, 200])

clock_ticks_coarse = np.array([1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0])

count = -1
for i,boolean in enumerate(clock_ticks_coarse):
    if boolean == 1:
        count +=1
        clock_ticks_coarse[i] = count
    else:
        clock_ticks_coarse[i] = boolean+count

fine_time_array = clock_ticks_fine*fine_div
coarse_time_array = clock_ticks_coarse*corse_div
time_array = fine_time_array+coarse_time_array

np.set_printoptions(precision=3, suppress = True)
print time_array

Output:

[  1.059   0.029   0.324   0.647   1.147   1.941   2.088   2.294   3.147
   3.559   4.029   4.176   4.794   5.588   6.206   6.235   7.029   7.235
   8.559   7.706   7.882   8.882   9.147   9.647   9.794  10.735  11.265
  11.706  11.794  12.647  13.147  13.471  13.5    14.529  16.059  15.
  15.382  15.559  16.265  16.706  17.176  17.382  18.265  18.824  19.088
  19.088  20.147  20.559  21.118  21.147  22.     23.559  22.559  23.059
  23.147  23.647  24.441  24.735  24.824  25.471  26.324  26.765  26.794
  27.294  27.912  28.412  28.794  29.176  30.     31.059  30.265  30.588
  31.324  31.882  32.206  32.294  33.029  33.588  34.206  34.235  35.147
  35.412  36.029  36.118  36.912  37.353  38.559  37.912  38.     38.853
  39.324  39.588  39.618  40.529  41.     41.706  41.735  42.647  42.971
  43.382]
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  • 1
    \$\begingroup\$ What is boolean doing in clock_ticks_coarse[i] = boolean+count? It should be zero, shouldn't it? \$\endgroup\$ – plonser Dec 8 '15 at 16:56
  • \$\begingroup\$ yes, you are correct - the if else is unnecessary. \$\endgroup\$ – Cornspierre Dec 9 '15 at 5:33
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Everything apart from the for loop in your code is already vectorized which leaves only the for loop to be optimized.

You define a counter which is raised by 1 every time the entry of clock_ticks_coarse is 1. There exists a function cumsum in numpy which sums all entries of the argument like this:

import numpy as np

x = np.array([1,3,0,5,7])

np.cumsum(x)
# array([ 1,  4,  4,  9, 16])

It is a vectorized function, so it is very fast. Furthermore, it allows you to replace the for loop completely which should significantly boost the speed of your program. The if else condition can entirely be droped (if boolean is always zero in the else statement).

# This line replaces your whole for-loop and is vectorized
clock_ticks_coarse = np.cumsum(clock_ticks_coarse)-1

# the following lines are already vectorized, so they are fine
fine_time_array = clock_ticks_fine*fine_div
coarse_time_array = clock_ticks_coarse*corse_div
time_array = fine_time_array+coarse_time_array

np.set_printoptions(precision=3, suppress = True)
print time_array

# output

[  1.059   0.029   0.324   0.647   1.147   1.941   2.088   2.294   3.147
   3.559   4.029   4.176   4.794   5.588   6.206   6.235   7.029   7.235
   8.559   7.706   7.882   8.882   9.147   9.647   9.794  10.735  11.265
  11.706  11.794  12.647  13.147  13.471  13.5    14.529  16.059  15.
  15.382  15.559  16.265  16.706  17.176  17.382  18.265  18.824  19.088
  19.088  20.147  20.559  21.118  21.147  22.     23.559  22.559  23.059
  23.147  23.647  24.441  24.735  24.824  25.471  26.324  26.765  26.794
  27.294  27.912  28.412  28.794  29.176  30.     31.059  30.265  30.588
  31.324  31.882  32.206  32.294  33.029  33.588  34.206  34.235  35.147
  35.412  36.029  36.118  36.912  37.353  38.559  37.912  38.     38.853
  39.324  39.588  39.618  40.529  41.     41.706  41.735  42.647  42.971
  43.382]
\$\endgroup\$

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