Convert large numpy arrays of BCD to decimal - python

I have binary data files in the multiple GB range that I am memory mapping with numpy. The start of each data packet contains a BCD timestamp. Where each hex number is coded into the time format of 0DDD:HH:MM:SS.ssss I need this timestamp turned into total seconds of the current year.
Example:
The the first time stamp 0x0261 1511 2604 6002 Would be: 261:15:11:26.046002 or
261*86400 + 15*3600 + 11*60 + 26.046002 = 22551986.046002
Currently I am doing this to compute the timestamps:
import numpy as np
rawData = np.memmap('dataFile.bin',dtype='u1',mode='r')
#findFrameStart returns the index to the start of each data packet [0,384,768,...]
fidx = findFrameStart(rawData)
# Do lots of bit shifting and multiplying and type casting....
day1 = ((rawData[fidx ]>>4)*10 + (rawData[fidx ]&0x0F)).astype('f8')
day2 = ((rawData[fidx+1]>>4)*10 + (rawData[fidx+1]&0x0F)).astype('f8')
hour = ((rawData[fidx+2]>>4)*10 + (rawData[fidx+2]&0x0F)).astype('f8')
mins = ((rawData[fidx+3]>>4)*10 + (rawData[fidx+3]&0x0F)).astype('f8')
sec1 = ((rawData[fidx+4]>>4)*10 + (rawData[fidx+4]&0x0F)).astype('f8')
sec2 = ((rawData[fidx+5]>>4)*10 + (rawData[fidx+5]&0x0F)).astype('f8')
sec3 = ((rawData[fidx+6]>>4)*10 + (rawData[fidx+6]&0x0F)).astype('f8')
sec4 = ((rawData[fidx+7]>>4)*10 + (rawData[fidx+7]&0x0F)).astype('f8')
time = (day1*100+day2)*86400 + hour*3600 + mins*60 + sec1 + sec2/100 + sec3/10000 + sec4/1000000
Note I had to cast each of the intermediate vars (day1, day2, etc.) to double to get the time to compute correctly.
Given that there are lots of frames, fidx can get kind of large (~10e6 elements or more). This results in lots of math operations, bit shifts, casting, etc. in my current method. So far it is working OK on a smaller test file (~180ms on a 150MB data file). However, I am worried about when I hit some larger data(4-5GB) there might be memory issues with all of the intermediate arrays.
So if possible I was looking for a different method that might shortcut some of the overhead. The BCD to decimal operations are similar for each byte so it seems I should maybe be able to iterate over something and maybe convert an array in place ... at least reducing the memory footprint.
Any help would be appreciated. FYI, I am using Python 3.7

I made the following adjustments to my code. This modifies the time array in place & removed the need for all of the intermediate arrays. I haven't timed the result but it should require less memory.
time = np.zeros(fidx.shape,dtype='f8')
scale = np.array([8640000, 86400, 3600, 60, 1, .01, .0001, .000001],dtype='f8')
for ii,sf in enumerate(scale):
time = time + ((rawData[fidx+ii]>>4)*10 + (rawData[fidx+ii]&0x0F))*sf

Related

Astropy time conversion very slow

I have a script that reads some data from a binary stream of "packets" containing "parameters". The parameters read are stored in a dictionary for each packet, which is appended to an array representing the packet stream.
At the end this array of dict is written to an output CSV file.
Among the read data is a CUC7 datetime, stored as coarse/fine integer parts of a GPS time, which I also want to convert to a UTC ISO string.
from astropy.time import Time
def cuc2gps_time(coarse, fine):
return Time(coarse + fine / (2**24), format='gps')
def gps2utc_time(gps):
return Time(gps, format='isot', scale='utc')
The issue is that I realized that these two time conversions make up 90% of the total run time of my script, most of the work of my script being done in the 10% remaining (read binary file, decode 15 other parameters, write to CSV).
I somehow improved the situation by making the conversions in batches on Numpy arrays, instead of on each packet. This reduces the total runtime by about half.
import numpy as np
while end_not_reached:
# Read 1 packet
# (...)
nb_packets += 1
end_not_reached = ... # boolean
# Process in batches for better performance
if not nb_packets%1000 or not end_not_reached:
# Convert CUC7 time to GPS and UTC times
all_coarse = np.array([packet['lobt_coarse'] for packet in packets])
all_fine = np.array([packet['lobt_fine'] for packet in packets])
all_gps = cuc2gps_time(all_coarse, all_fine)
all_utc = gps2utc_time(all_gps)
# Add times to each packet
for packet, gps_time, utc_time in zip(packets, all_gps, all_utc):
packet.update({'gps_time': gps_time, 'utc_time': utc_time})
But my script is still absurdly slow. Reading 60000 packets from a 1.2GB file and writing it as a CSV takes 12s, against only 2.5s if I remove the time conversion.
So:
Is it expected that Astropy time conversions are so slow? Am I using it wrong? Is there a better library?
Is there a way to improve my current implementation? I suspect that the remaining "for" loop in there is very costly, but could not find a good way to replace it.
I think the problem is that you're doing multiple loops over your sequence of packets. In Python I would recommend having arrays representing each parameter, instead of having a list of objects, each with a bunch of scalar parameters.
If you can read all the packets in at once, I would recommend something like:
num_bytes = ...
num_bytes_per_packet = ...
num_packets = num_bytes / num_bytes_per_packet
param1 = np.empty(num_packets)
param2 = np.empty(num_packets)
...
time_coarse = np.empty(num_packets)
time_fine = np.empty(num_packets)
...
param_N = np.empty(num_packets)
for i in range(num_packets):
param_1[i], param_2[i], ..., time_coarse[i], time_fine[i], ... param_N[i] = decode_packet(...)
time_gps = Time(time_coarse + time_fine / (2**24), format='gps')
time_utc = time_gps.utc.isot

Dramatic drop in numpy fromfile performance when switching from python 2 to python 3

Background
I am analyzing large (between 0.5 and 20 GB) binary files, which contain information about particle collisions from a simulation. The number of collisions, number of incoming and outgoing particles can vary, so the files consist of variable length records. For analysis I use python and numpy. After switching from python 2 to python 3 I have noticed a dramatic decrease in performance of my scripts and traced it down to numpy.fromfile function.
Simplified code to reproduce the problem
This code, iotest.py
Generates a file of a similar structure to what I have in my studies
Reads it using numpy.fromfile
Reads it using numpy.frombuffer
Compares timing of both
import numpy as np
import os
def generate_binary_file(filename, nrecords):
n_records = np.random.poisson(lam = nrecords)
record_lengths = np.random.poisson(lam = 10, size = n_records).astype(dtype = 'i4')
x = np.random.normal(size = record_lengths.sum()).astype(dtype = 'd')
with open(filename, 'wb') as f:
s = 0
for i in range(n_records):
f.write(record_lengths[i].tobytes())
f.write(x[s:s+record_lengths[i]].tobytes())
s += record_lengths[i]
# Trick for testing: make sum of records equal to 0
f.write(np.array([1], dtype = 'i4').tobytes())
f.write(np.array([-x.sum()], dtype = 'd').tobytes())
return os.path.getsize(filename)
def read_binary_npfromfile(filename):
checksum = 0.0
with open(filename, 'rb') as f:
while True:
try:
record_length = np.fromfile(f, 'i4', 1)[0]
x = np.fromfile(f, 'd', record_length)
checksum += x.sum()
except:
break
assert(np.abs(checksum) < 1e-6)
def read_binary_npfrombuffer(filename):
checksum = 0.0
with open(filename, 'rb') as f:
while True:
try:
record_length = np.frombuffer(f.read(np.dtype('i4').itemsize), dtype = 'i4', count = 1)[0]
x = np.frombuffer(f.read(np.dtype('d').itemsize * record_length), dtype = 'd', count = record_length)
checksum += x.sum()
except:
break
assert(np.abs(checksum) < 1e-6)
if __name__ == '__main__':
from timeit import Timer
from functools import partial
fname = 'testfile.tmp'
print("# File size[MB], Timings and errors [s]: fromfile, frombuffer")
for i in [10**3, 3*10**3, 10**4, 3*10**4, 10**5, 3*10**5, 10**6, 3*10**6]:
fsize = generate_binary_file(fname, i)
t1 = Timer(partial(read_binary_npfromfile, fname))
t2 = Timer(partial(read_binary_npfrombuffer, fname))
a1 = np.array(t1.repeat(5, 1))
a2 = np.array(t2.repeat(5, 1))
print('%8.3f %12.6f %12.6f %12.6f %12.6f' % (1.0 * fsize / (2**20), a1.mean(), a1.std(), a2.mean(), a2.std()))
Results
Conclusions
In Python 2 numpy.fromfile was probably the fastest way to deal with binary files of variable structure. It was approximately 3 times faster than numpy.frombuffer. Performance of both scaled linearly with file size.
In Python 3 numpy.frombuffer became around 10% slower, while numpy.fromfile became around 9.3 times slower compared to Python 2! Performance of both still scales linearly with file size.
In the documentation of numpy.fromfile it is described as "A highly efficient way of reading binary data with a known data-type". It is not correct in Python 3 anymore. This was in fact noticed earlier by other people already.
Questions
In Python 3 how to obtain a comparable (or better) performance to Python 2, when reading binary files of variable structure?
What happened in Python 3 so that numpy.fromfile became an order of magnitude slower?
TL;DR: np.fromfile and np.frombuffer are not optimized to read many small buffers. You can load the whole file in a big buffer and then decode it very efficiently using Numba.
Analysis
The main issue is that the benchmark measure overheads. Indeed, it perform a lot of system/C calls that are very inefficient. For example, on the 24 MiB file, the while loops calls 601_214 times np.fromfile and np.frombuffer. The timing on my machine are 10.5s for read_binary_npfromfile and 1.2s for read_binary_npfrombuffer. This means respectively 17.4 us and 2.0 us per call for the two function. Such timing per call are relatively reasonable considering Numpy is not designed to efficiently operate on very small arrays (it needs to perform many checks, call some functions, wrap/unwrap CPython types, allocate some objects, etc.). The overhead of these functions can change from one version to another and unless it becomes huge, this is not a bug. The addition of new features to Numpy and CPython often impact overheads and this appear to be the case here (eg. buffering interface). The point is that it is not really a problem because there is a way to use a different approach that is much much faster (as it does not pay huge overheads).
Faster Numpy code
The main solution to write a fast implementation is to read the whole file once in a big byte buffer and then decode it using np.view. That being said, this is a bit tricky because of data alignment and the fact that nearly all Numpy function needs to be prohibited in the while loop due to their overhead. Here is an example:
def read_binary_faster_numpy(filename):
buff = np.fromfile(filename, dtype=np.uint8)
buff_int32 = buff.view(np.int32)
buff_double_1 = buff[0:len(buff)//8*8].view(np.float64)
buff_double_2 = buff[4:4+(len(buff)-4)//8*8].view(np.float64)
nblocks = buff.size // 4 # Number of 4-byte blocks
pos = 0 # Displacement by block of 4 bytes
lst = []
while pos < nblocks:
record_length = buff_int32[pos]
pos += 1
if pos + record_length * 2 > nblocks:
break
offset = pos // 2
if pos % 2 == 0: # Aligned with buff_double_1
x = buff_double_1[offset:offset+record_length]
else: # Aligned with buff_double_2
x = buff_double_2[offset:offset+record_length]
lst.append(x) # np.sum is too expensive here
pos += record_length * 2
checksum = np.sum(np.concatenate(lst))
assert(np.abs(checksum) < 1e-6)
The above implementation should be faster but it is a bit tricky to understand and it is still bounded by the latency of Numpy operations. Indeed, the loop is still calling Numpy functions due to operations like buff_int32[pos] or buff_double_1[offset:offset+record_length]. Even though the overheads of indexing is much smaller than the one of previous functions, it is still quite big for such a critical loop (with ~300_000 iterations)...
Better performance with... a basic pure-Python code
It turns out that the following pure-python implementation is faster, safer and simpler:
from struct import unpack_from
def read_binary_python_struct(filename):
checksum = 0.0
with open(filename, 'rb') as f:
data = f.read()
offset = 0
while offset < len(data):
record_length = unpack_from('#i', data, offset)[0]
checksum += sum(unpack_from(f'{record_length}d', data, offset + 4))
offset += 4 + record_length * 8
assert(np.abs(checksum) < 1e-6)
This is because the overhead of unpack_from is far lower than the one of Numpy functions but it is still not great.
In fact, now the main issue is actually the CPython interpreter. It is clearly not designed with high-performance in mind. The above code push it to the limit. Allocating millions of temporary reference-counted dynamic objects like variable-sized integers and strings is very expensive. This is not reasonable to let CPython do such an operation.
Writing a high-performance code with Numba
We can drastically speed it up using Numba which can compile Numpy-based Python codes to native ones using a just-in-time compiler! Here is an example:
#nb.njit('float64(uint8[::1])')
def decode_buffer(buff):
checksum = 0.0
offset = 0
while offset + 4 < buff.size:
record_length = buff[offset:offset+4].view(np.int32)[0]
start = offset + 4
end = start + record_length * 8
if end > buff.size:
break
x = buff[start:end].view(np.float64)
checksum += x.sum()
offset = end
return checksum
def read_binary_numba(filename):
buff = np.fromfile(filename, dtype=np.uint8)
checksum = decode_buffer(buff)
assert(np.abs(checksum) < 1e-6)
Numba removes nearly all Numpy overheads thanks to a native compiled code. That being said note that Numba does not implement all Numpy functions yet. This include np.fromfile which need to be called outside a Numba-compiled function.
Benchmark
Here are the performance results on my machine (i5-9600KF with a high-performance Nvme SSD) with Python 3.8.1, Numpy 1.20.3 and Numba 0.54.1.
read_binary_npfromfile: 10616 ms ( x1)
read_binary_npfrombuffer: 1132 ms ( x9)
read_binary_faster_numpy: 509 ms ( x21)
read_binary_python_struct: 222 ms ( x48)
read_binary_numba: 12 ms ( x885)
Optimal time: 7 ms (x1517)
One can see that the Numba implementation is extremely fast compared to the initial Python implementation and even to the fastest alternative Python implementation. This is especially true considering that 8 ms is spent in np.fromfile and only 4 ms in decode_buffer!

Numba Python - how to exploit parallelism effectively?

I have been trying to exploit Numba to speed up large array calculations. I have been measuring the calculation speed in GFLOPS, and it consistently falls far short of my expectations for my CPU.
My processor is i9-9900k, which according to float32 benchmarks should be capable of over 200 GFLOPS. In my tests I have never exceeded about 50 GFLOPS. This is running on all 8 cores.
On a single core I achieve about 17 GFLOPS, which (I believe) is 50% of the theoretical performance. I'm not sure if this is improvable, but the fact that it doesn't extend well to multi-core is a problem.
I am trying to learn this because I am planning to write some image processing code that desperately needs every speed boost possible. I also feel I should understand this first, before I dip my toes into GPU computing.
Here is some example code with a few of my attempts at writing fast functions. The operation I am testing, is multiplying an array by a float32 then summing the whole array, i.e. a MAC operation.
How can I get better results?
import os
# os.environ["NUMBA_ENABLE_AVX"] = "1"
import numpy as np
import timeit
from timeit import default_timer as timer
import numba
# numba.config.NUMBA_ENABLE_AVX = 1
# numba.config.LOOP_VECTORIZE = 1
# numba.config.DUMP_ASSEMBLY = 1
from numba import float32, float64
from numba import jit, njit, prange
from numba import vectorize
from numba import cuda
lengthY = 16 # 2D array Y axis
lengthX = 2**16 # X axis
totalops = lengthY * lengthX * 2 # MAC operation has 2 operations
iters = 100
doParallel = True
#njit(fastmath=True, parallel=doParallel)
def MAC_numpy(testarray):
output = (float)(0.0)
multconst = (float)(.99)
output = np.sum(np.multiply(testarray, multconst))
return output
#njit(fastmath=True, parallel=doParallel)
def MAC_01(testarray):
lengthX = testarray.shape[1]
lengthY = testarray.shape[0]
output = (float)(0.0)
multconst = (float)(.99)
for y in prange(lengthY):
for x in prange(lengthX):
output += multconst*testarray[y,x]
return output
#njit(fastmath=True, parallel=doParallel)
def MAC_04(testarray):
lengthX = testarray.shape[1]
lengthY = testarray.shape[0]
output = (float)(0.0)
multconst = (float)(.99)
for y in prange(lengthY):
for x in prange(int(lengthX/4)):
xn = x*4
output += multconst*testarray[y,xn] + multconst*testarray[y,xn+1] + multconst*testarray[y,xn+2] + multconst*testarray[y,xn+3]
return output
# ======================================= TESTS =======================================
testarray = np.random.rand(lengthY, lengthX)
# ==== MAC_numpy ====
time = 1000
for n in range(iters):
start = timer()
output = MAC_numpy(testarray)
end = timer()
if((end-start) < time): #get shortest time
time = end-start
print("\nMAC_numpy")
print("output = %f" % (output))
print(type(output))
print("fastest time = %16.10f us" % (time*10**6))
print("Compute Rate = %f GFLOPS" % ((totalops/time)/10**9))
# ==== MAC_01 ====
time = 1000
lengthX = testarray.shape[1]
lengthY = testarray.shape[0]
for n in range(iters):
start = timer()
output = MAC_01(testarray)
end = timer()
if((end-start) < time): #get shortest time
time = end-start
print("\nMAC_01")
print("output = %f" % (output))
print(type(output))
print("fastest time = %16.10f us" % (time*10**6))
print("Compute Rate = %f GFLOPS" % ((totalops/time)/10**9))
# ==== MAC_04 ====
time = 1000
for n in range(iters):
start = timer()
output = MAC_04(testarray)
end = timer()
if((end-start) < time): #get shortest time
time = end-start
print("\nMAC_04")
print("output = %f" % (output))
print(type(output))
print("fastest time = %16.10f us" % (time*10**6))
print("Compute Rate = %f GFLOPS" % ((totalops/time)/10**9))
Q : How can I get better results?
1st : Learn how to avoid doing useless work - you can straight eliminate HALF of the FLOP-s not speaking about also the half of all the RAM-I/O-s avoided, each one being at a cost of +100~350 [ns] per writeback
Due to the distributive nature of MUL and ADD ( a.C + b.C ) == ( a + b ).C, better first np.sum( A ) and only after that then MUL the sum by the (float) constant.
#utput = np.sum(np.multiply(testarray, multconst)) # AWFULLY INEFFICIENT
output = np.sum( testarray)*multconst #######################
2nd : Learn how to best align data along the order of processing ( cache-line reuses get you ~100x faster re-use of pre-fetched data. Not aligning vectorised-code along these already pre-fetched data side-effects just let your code pay many times the RAM-access latencies, instead of smart re-using the already paid for data-blocks. Designing work-units aligned according to this principle means a few SLOCs more, but the rewards are worth that - who gets ~100x faster CPUs+RAMs for free and right now or about a ~100x speedup for free, just from not writing a badly or naively designed looping iterators?
3rd : Learn how to efficiently harness vectorised (block-directed) operations inside numpy or numba code-blocks and avoid pressing numba to spend time on auto-analysing the call-signatures ( you pay an extra time for this auto-analyses per call, while you have designed the code and knew exactly what data-types are going to go there, so why to pay an extra time for auto-analysis each time a numba-block gets called???)
4th : Learn where the extended Amdahl's Law, having all the relevant add-on costs and processing atomicity put into the game, supports your wish to get speedups, not to ever pay way more than you will get back (to at least justify the add-on costs... ) - paying extra costs for not getting any reward is possible, yet has no beneficial impact on your code's performance ( rather the opposite )
5th : Learn when and how the manually created inline(s) may save your code, once the steps 1-4 are well learnt and routinely excersised with proper craftmanship ( Using popular COTS frameworks is fine, yet these may deliver results after a few days of work, while a hand-crafted single purpose smart designed assembly code was able to get the same results in about 12 minutes(!), not several days without any GPU/CPU tricks etc - yes, that faster - just by not doing a single step more than what was needed for the numerical processing of the large matrix data )
Did I mention float32 may surprise at being processed slower on small scales than float64, while on larger data-scales ~ n [GB] the RAM I/O-times grow slower for more efficient float32 pre-fetches? This never happens here, as float64 array gets processed here. Sure, unless one explicitly instructs the constructor(s) to downconvert the default data type, like this: np.random.rand( lengthY, lengthX ).astype( dtype = np.float32 )>>> np.random.rand( 10, 2 ).dtypedtype('float64')Avoiding extensive memory allocations is another performance trick, supported in numpy call-signatures. Using this option for large arrays will save you a lot of extra time wasted on mem-allocs for large interim arrays. Reusing already pre-allocated memory-zones and wisely controlled gc-policing are another signs of a professional, focused on low-latency & design-for-performance

How to generate a time-ordered uid in Python?

Is this possible? I've heard Cassandra has something similar : https://datastax.github.io/python-driver/api/cassandra/util.html
I have been using a ISO timestamp concatenated with a uuid4, but that ended up way too large (58 characters) and probably overkill.
Keeping a sequential number doesn't work in my context (DynamoDB NoSQL)
Worth noticing that for my application it doesn't matter if items created in batch/same second are in a random order, as long as the uid don't collapse.
I have no specific restriction on maximum length, ideally I would like to see the different collision chance for different lengths, but it needs to be smaller than 58 (my original attempt)
This is to use with DynamoDB(NoSQL Database) as Sort-key
Why uuid.uuid1 is not sequential
uuid.uuid1(node=None, clock_seq=None) is effectively:
60 bits of timestamp (representing number of 100-ns intervals after 1582-10-15 00:00:00)
14 bits of "clock sequence"
48 bits of "Node info" (generated from network card's mac-address or from hostname or from RNG).
If you don't provide any arguments, then System function is called to generate uuid. In that case:
It's unclear if "clock sequence" is sequential or random.
It's unclear if it's safe to be used in multiple processes (can clock_seq be repeated in different processes or not?). In Python 3.7 this info is now available.
If you provide clock_seq or node, then "pure python implementation is used". IN this case even with "fixed value" for clock_seq:
timestamp part is guaranteed to be sequential for all the calls in current process even in threaded execution.
clock_seq part is randomly generated. But that is not critical annymore because timestamp is sequential and unique.
It's NOT safe for multiple processes (processes that call uuid1 with the same clock_seq, node might return conflicting values if called during the "same 100-ns time interval")
Solution that reuses uuid.uuid1
It's easy to see, that you can make uuid1 sequential by providing clock_seq or node arguments (to use python implementation).
import time
from uuid import uuid1, getnode
_my_clock_seq = getrandbits(14)
_my_node = getnode()
def sequential_uuid(node=None):
return uuid1(node=node, clock_seq=_my_clock_seq)
# .hex attribute of this value is 32-characters long string
def alt_sequential_uuid(clock_seq=None):
return uuid1(node=_my_node, clock_seq=clock_seq)
if __name__ == '__main__':
from itertools import count
old_n = uuid1() # "Native"
old_s = sequential_uuid() # Sequential
native_conflict_index = None
t_0 = time.time()
for x in count():
new_n = uuid1()
new_s = sequential_uuid()
if old_n > new_n and not native_conflict_index:
native_conflict_index = x
if old_s >= new_s:
print("OOops: non-sequential results for `sequential_uuid()`")
break
if (x >= 10*0x3fff and time.time() - t_0 > 30) or (native_conflict_index and x > 2*native_conflict_index):
print('No issues for `sequential_uuid()`')
break
old_n = new_n
old_s = new_s
print(f'Conflicts for `uuid.uuid1()`: {bool(native_conflict_index)}')
Multiple processes issues
BUT if you are running some parallel processes on the same machine, then:
node which defaults to uuid.get_node() will be the same for all the processes;
clock_seq has small chance to be the same for some processes (chance of 1/16384)
That might lead to conflicts! That is general concern for using
uuid.uuid1 in parallel processes on the same machine unless you have access to SafeUUID from Python3.7.
If you make sure to also set node to unique value for each parallel process that runs this code, then conflicts should not happen.
Even if you are using SafeUUID, and set unique node, it's still possible to have non-sequential (but unique) ids if they are generated in different processes.
If some lock-related overhead is acceptable, then you can store clock_seq in some external atomic storage (for example in "locked" file) and increment it with each call: this allows to have same value for node on all parallel processes and also will make id-s sequential. For cases when all parallel processes are subprocesses created using multiprocessing: clock_seq can be "shared" using multiprocessing.Value
As a result you always have to remember:
If you are running multiple processes on the same machine, then you must:
Ensure uniqueness of node. The problem for this solution: you can't be sure to have sequential ids from different processes generated during the same 100-ns interval. But this is very "light" operation executed once on process startup and achieved by: "adding" something to default node, e.g. int(time.time()*1e9) - 0x118494406d1cc000, or by adding some counter from machine-level atomic db.
Ensure "machine-level atomic clock_seq" and the same node for all processes on one machine. That way you'll have some overhead for "locking" clock_seq, but id-s are guaranteed to be sequential even if generated in different processes during the same 100-ns interval (unless you are calling uuid from several threads in the same process).
For processes on different machines:
either you have to use some "global counter service";
or it's not possible to have sequential ids generated on different machines during the same 100-ns interval.
Reducing size of the id
General approach to generate UUIDs is quite simple, so it's easy to implement something similar from scratch, and for example use less bits for node_info part:
import time
from random import getrandbits
_my_clock_seq = getrandbits(14)
_last_timestamp_part = 0
_used_clock_seq = 0
timestamp_multiplier = 1e7 # I'd recommend to use this value
# Next values are enough up to year 2116:
if timestamp_multiplier == 1e9:
time_bits = 62 # Up to year 2116, also reduces chances for non-sequential id-s generated in different processes
elif timestamp_multiplier == 1e8:
time_bits = 60 # up to year 2335
elif timestamp_multiplier == 1e7:
time_bits = 56 # Up to year 2198.
else:
raise ValueError('Please calculate and set time_bits')
time_mask = 2**time_bits - 1
seq_bits = 16
seq_mask = 2**seq_bits - 1
node_bits = 12
node_mask = 2**node_bits - 1
max_hex_len = len(hex(2**(node_bits+seq_bits+time_bits) - 1)) - 2 # 21
_default_node_number = getrandbits(node_bits) # or `uuid.getnode() & node_mask`
def sequential_uuid(node_number=None):
"""Return 21-characters long hex string that is sequential and unique for each call in current process.
Results from different processes may "overlap" but are guaranteed to
be unique if `node_number` is different in each process.
"""
global _my_clock_seq
global _last_timestamp_part
global _used_clock_seq
if node_number is None:
node_number = _default_node_number
if not 0 <= node_number <= node_mask:
raise ValueError("Node number out of range")
timestamp_part = int(time.time() * timestamp_multiplier) & time_mask
_my_clock_seq = (_my_clock_seq + 1) & seq_mask
if _last_timestamp_part >= timestamp_part:
timestamp_part = _last_timestamp_part
if _used_clock_seq == _my_clock_seq:
timestamp_part = (timestamp_part + 1) & time_mask
else:
_used_clock_seq = _my_clock_seq
_last_timestamp_part = timestamp_part
return hex(
(timestamp_part << (node_bits+seq_bits))
|
(_my_clock_seq << (node_bits))
|
node_number
)[2:]
Notes:
Maybe it's better to simply store integer value (not hex-string) in the database
If you are storing it as text/char, then its better to convert integer to base64-string instead of converting it to hex-string. That way it will be shorter (21 chars hex-string → 16 chars b64-encoded string):
from base64 import b64encode
total_bits = time_bits+seq_bits+node_bits
total_bytes = total_bits // 8 + 1 * bool(total_bits % 8)
def int_to_b64(int_value):
return b64encode(int_value.to_bytes(total_bytes, 'big'))
Collision chances
Single process: collisions not possible
Multiple processes with manually set unique clock_seq or unique node in each process: collisions not possible
Multiple processes with randomly set node (48-bits, "fixed" in time):
Chance to have the node collision in several processes:
in 2 processes out of 10000: ~0.000018%
in 2 processes out of 100000: 0.0018%
Chance to have single collision of the id per second in 2 processes with the "colliding" node:
for "timestamp" interval of 100-ns (default for uuid.uuid1 , and in my code when timestamp_multiplier == 1e7): proportional to 3.72e-19 * avg_call_frequency²
for "timestamp" interval of 10-ns (timestamp_multiplier == 1e8): proportional to 3.72e-21 * avg_call_frequency²
In the article you've linked too, the cassandra.util.uuid_from_time(time_arg, node=None, clock_seq=None)[source] seems to be exactly what you're looking for.
def uuid_from_time(time_arg, node=None, clock_seq=None):
"""
Converts a datetime or timestamp to a type 1 :class:`uuid.UUID`.
:param time_arg:
The time to use for the timestamp portion of the UUID.
This can either be a :class:`datetime` object or a timestamp
in seconds (as returned from :meth:`time.time()`).
:type datetime: :class:`datetime` or timestamp
:param node:
None integer for the UUID (up to 48 bits). If not specified, this
field is randomized.
:type node: long
:param clock_seq:
Clock sequence field for the UUID (up to 14 bits). If not specified,
a random sequence is generated.
:type clock_seq: int
:rtype: :class:`uuid.UUID`
"""
if hasattr(time_arg, 'utctimetuple'):
seconds = int(calendar.timegm(time_arg.utctimetuple()))
microseconds = (seconds * 1e6) + time_arg.time().microsecond
else:
microseconds = int(time_arg * 1e6)
# 0x01b21dd213814000 is the number of 100-ns intervals between the
# UUID epoch 1582-10-15 00:00:00 and the Unix epoch 1970-01-01 00:00:00.
intervals = int(microseconds * 10) + 0x01b21dd213814000
time_low = intervals & 0xffffffff
time_mid = (intervals >> 32) & 0xffff
time_hi_version = (intervals >> 48) & 0x0fff
if clock_seq is None:
clock_seq = random.getrandbits(14)
else:
if clock_seq > 0x3fff:
raise ValueError('clock_seq is out of range (need a 14-bit value)')
clock_seq_low = clock_seq & 0xff
clock_seq_hi_variant = 0x80 | ((clock_seq >> 8) & 0x3f)
if node is None:
node = random.getrandbits(48)
return uuid.UUID(fields=(time_low, time_mid, time_hi_version,
clock_seq_hi_variant, clock_seq_low, node), version=1)
There's nothing Cassandra specific to a Type 1 UUID...
You should be able to encode a timestamp precise to the second for a time range of 135 years in 32 bits. That will only take 8 characters to represent in hex. Added to the hex representation of the uuid (32 hex characters) that will amount to only 40 hex characters.
Encoding the time stamp that way requires that you pick a base year (e.g. 2000) and compute the number of days up to the current date (time stamp). Multiply this number of days by 86400, then add the seconds since midnight. This will give you values that are less than 2^32 until you reach year 2135.
Note that you have to keep leading zeroes in the hex encoded form of the timestamp prefix in order for alphanumeric sorting to preserve the chronology.
With a few bits more in the time stamp, you could increase the time range and/or the precision. With 8 more bits (two hex characters), you could go up to 270 years with a precision to the hundredth of a second.
Note that you don't have to model the fraction of seconds in a base 10 range. You will get optimal bit usage by breaking it down in 128ths instead of 100ths for the same number of characters. With the doubling of the year range, this still fits within 8 bits (2 hex characters)
The collision probability, within the time precision (i.e. per second or per 100th or 128th of a second) is driven by the range of the uuid so it will be 1 in 2^128 for the chosen precision. Increasing the precision of the time stamp has the most impact on reducing the collision chances. It is also the factor that has the lowest impact on total size of the key.
More efficient character encoding: 27 to 29 character keys
You could significantly reduce the size of the key by encoding it in base 64 instead of 16 which would give you 27 to 29 characters (depending on you choice of precision)
Note that, for the timestamp part, you need to use an encoding function that takes an integer as input and that preserves the collating sequence of digit characters.
For example:
def encode64(number, size):
chars = "+-0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
result = list()
for _ in range(size):
result.append(chars[number%64])
number //= 64
return "".join(reversed(result))
a = encode64(1234567890,6) # '-7ZU9G'
b = encode64(9876543210,6) # '7Ag-Pe'
print(a < b) # True
u = encode64(int(uuid.uuid4()),22) # '1QA2LtMg30ztnugxaokVMk'
key = a+u # '-7ZU9G1QA2LtMg30ztnugxaokVMk' (28 characters)
You can save some more characters by combining the time stamp and uuid into a single number before encoding instead of concatenating the two encoded values.
The encode64() function needs one character every 6 bits.
So, for 135 years with precision to the second: (32+128)/6 = 26.7 --> 27 characters
instead of (32/6 = 5.3 --> 6) + (128/6 = 21.3 --> 22) ==> 28 characters
uid = uuid.uuid4()
timeStamp = daysSince2000 * 86400 + int(secondsSinceMidnight)
key = encode64( timeStamp<<128 | int(uid) ,27)
with a 270 year span and 128th of a second precision: (40+128)/6 = 28 characters
uid = uuid.uuid4()
timeStamp = daysSince2000 * 86400 + int(secondsSinceMidnight)
precision = 128
timeStamp = timeStamp * precision + int(factionOfSecond * precision)
key = encode64( timeStamp<<128 | int(uid) ,28)
With 29 characters you can raise precision to 1024th of a second and year range to 2160 years.
UUID masking: 17 to 19 characters keys
To be even more efficient, you could strip out the first 64 bits of the uuid (which is already a time stamp) and combine it with your own time stamp. This would give you keys with a length of 17 to 19 characters with practically no loss of collision avoidance (depending on your choice of precision).
mask = (1<<64)-1
key = encode64( timeStamp<<64 | (int(uid) & mask) ,19)
Integer/Numeric keys ?
As a final note, if your database supports very large integers or numeric fields (140 bits or more) as keys, you don't have to convert the combined number to a string. Just use it directly as the key. The numerical sequence of timeStamp<<128 | int(uid) will respect the chronology.
The uuid6 module (pip install uuid6) solves the problem. It aims at implementing the corresponding draft for a new uuid variant standard, see here.
Example code:
import uuid6
for i in range(0, 30):
u = uuid6.uuid7()
print(u)
time.sleep(0.1)
The package suggests to use uuid6.uuid7():
Implementations SHOULD utilize UUID version 7 over UUID version 1 and
6 if possible.
UUID version 7 features a time-ordered value field derived from the
widely implemented and well known Unix Epoch timestamp source, the
number of milliseconds seconds since midnight 1 Jan 1970 UTC, leap
seconds excluded. As well as improved entropy characteristics over
versions 1 or 6.

What is the fastest way to sort strings in Python if locale is a non-concern?

I was trying to find a fast way to sort strings in Python and the locale is a non-concern i.e. I just want to sort the array lexically according to the underlying bytes. This is perfect for something like radix sort. Here is my MWE
import numpy as np
import timeit
# randChar is workaround for MemoryError in mtrand.RandomState.choice
# http://stackoverflow.com/questions/25627161/how-to-solve-memory-error-in-mtrand-randomstate-choice
def randChar(f, numGrp, N) :
things = [f%x for x in range(numGrp)]
return [things[x] for x in np.random.choice(numGrp, N)]
N=int(1e7)
K=100
id3 = randChar("id%010d", N//K, N) # small groups (char)
timeit.Timer("id3.sort()" ,"from __main__ import id3").timeit(1) # 6.8 seconds
As you can see it took 6.8 seconds which is almost 10x slower than R's radix sort below.
N = 1e7
K = 100
id3 = sample(sprintf("id%010d",1:(N/K)), N, TRUE)
system.time(sort(id3,method="radix"))
I understand that Python's .sort() doesn't use radix sort, is there an implementation somewhere that allows me to sort strings as performantly as R?
AFAIK both R and Python "intern" strings so any optimisations in R can also be done in Python.
The top google result for "radix sort strings python" is this gist which produced an error when sorting on my test array.
It is true that R interns all strings, meaning it has a "global character cache" which serves as a central dictionary of all strings ever used by your program. This has its advantages: the data takes less memory, and certain algorithms (such as radix sort) can take advantage of this structure to achieve higher speed. This is particularly true for the scenarios such as in your example, where the number of unique strings is small relative to the size of the vector. On the other hand it has its drawbacks too: the global character cache prevents multi-threaded write access to character data.
In Python, afaik, only string literals are interned. For example:
>>> 'abc' is 'abc'
True
>>> x = 'ab'
>>> (x + 'c') is 'abc'
False
In practice it means that, unless you've embedded data directly into the text of the program, nothing will be interned.
Now, for your original question: "what is the fastest way to sort strings in python"? You can achieve very good speeds, comparable with R, with python datatable package. Here's the benchmark that sorts N = 10⁸ strings, randomly selected from a set of 1024:
import datatable as dt
import pandas as pd
import random
from time import time
n = 10**8
src = ["%x" % random.getrandbits(10) for _ in range(n)]
f0 = dt.Frame(src)
p0 = pd.DataFrame(src)
f0.to_csv("test1e8.csv")
t0 = time(); f1 = f0.sort(0); print("datatable: %.3fs" % (time()-t0))
t0 = time(); src.sort(); print("list.sort: %.3fs" % (time()-t0))
t0 = time(); p1 = p0.sort_values(0); print("pandas: %.3fs" % (time()-t0))
Which produces:
datatable: 1.465s / 1.462s / 1.460s (multiple runs)
list.sort: 44.352s
pandas: 395.083s
The same dataset in R (v3.4.2):
> require(data.table)
> DT = fread("test1e8.csv")
> system.time(sort(DT$C1, method="radix"))
user system elapsed
6.238 0.585 6.832
> system.time(DT[order(C1)])
user system elapsed
4.275 0.457 4.738
> system.time(setkey(DT, C1)) # sort in-place
user system elapsed
3.020 0.577 3.600
Jeremy Mets posted in the comments of this blog post that Numpy can sort string fairly by converting the array to np.araray. This indeed improve performance, however it is still slower than Julia's implementation.
import numpy as np
import timeit
# randChar is workaround for MemoryError in mtrand.RandomState.choice
# http://stackoverflow.com/questions/25627161/how-to-solve-memory-error-in-mtrand-randomstate-choice
def randChar(f, numGrp, N) :
things = [f%x for x in range(numGrp)]
return [things[x] for x in np.random.choice(numGrp, N)]
N=int(1e7)
K=100
id3 = np.array(randChar("id%010d", N//K, N)) # small groups (char)
timeit.Timer("id3.sort()" ,"from __main__ import id3").timeit(1) # 6.8 seconds

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