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Inefficient way to process binary data, posted on codereview.SE: http://codereview.stackexchange.com/q/117080/61165
#!/usr/bin/python3
import os
import numpy as np
# ============ CONSTANTS TO (OPTIONALLY) SET =============
WINDOW_SIZE = 100
INPUT_FILE_NAME = 'timetags5byte.bin'
# ======== CODE TO READ AND PARSE 5 BYTE BINARY FILE =========================
# process data in INPUT_FILE_NAME and print results in OUTPUT_FILE_NAME
with open(INPUT_FILE_NAME, 'rb') as input_file:
# number of bytes found in every file produced when writing the binary file
initial_offset = 40
# number of bytes occupied by each timetag
timetag_size = 5
# each timetag is stored in 5 bytes. The first 4 bytes contain the
# timestamp, with the first byte being the least significant one, and last
# byte the most significant one
file_size = os.path.getsize(INPUT_FILE_NAME)
effective_file_size = file_size - initial_offset - file_size % 5
input_file.seek(initial_offset)
timetags = np.fromfile(input_file,
dtype = np.dtype([('timestamp', np.uint32), ('channel', np.uint8)]),
count = effective_file_size
)
# output_channels will contain the data to export (in file or whatever)
output_channels = []
# the first timetag is read out of the loop, in order to avoid checking
# at each iteration if coincidences is empty.
triggering_timestamp = timetags[0][0]
channels = [timetags[0][1]]
# Loop over all the timetags, except for the first one
for timetag in timetags[1:]:
if (timetag[0] > triggering_timestamp and
timetag[0] - triggering_timestamp < WINDOW_SIZE
):
channels.append(timetag[1])
else:
if len(channels) == 4:
print("fourfold coincidence:", end='')
for channel in channels:
print('{0: >5}'.format(channel), end='')
print('')
output_channels.append(channels)
channels = [timetag[1]]
triggering_timestamp = timetag[0]
if len(channels) == 4:
print("fourfold coincidence:", end='')
for channel in channels:
print('{0: >5}'.format(channel), end='')
print('')
output_channels.append(channels)
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