ring buffer with numpy/ctypes

南楼画角 提交于 2019-12-04 06:05:12

If you need a window of N bytes, make your buffer 2*N bytes and write all input to two locations: i % N and i % N + N, where i is a byte counter. That way you always have N consecutive bytes in the buffer.

data = 'Data to buffer'
N = 4
buf = 2*N*['\00']

for i,c in enumerate(data):
    j = i % N
    buf[j] = c
    buf[j+N] = c
    if i >= N-1:
        print ''.join(buf[j+1:j+N+1]) 

prints

Data
ata 
ta t
a to
 to 
to b
o bu
 buf
buff
uffe
ffer

One possible way is to start overwriting the (already old) bytes from the begining when the write pointer reaches the end of the buffer array.

That's the only option in a fixed-size ring buffer.

I've read it's impossible to create such circular slices.

Which is why I wouldn't do this with a Numpy view. You can create a class wrapper around an ndarray instead, holding the buffer/array, the capacity and a pointer (index) to the insertion point. If you want to get the contents as a Numpy array, you'll have to make a copy like so:

buf = np.array([1,2,3,4])
indices = [3,0,1,2]
contents = buf[indices]    # copy

You can still set elements' values in-place if you implement __setitem__ and __setslice__.

I think you need to take a step back from C-style thinking here. Updating a ringbuffer for every single insertion is never going to be efficient. A ring-buffer is fundamentally different from the contiguous memory block interface that numpy arrays demand; including the fft you mention you want to do.

A natural solution is to sacrifice a little bit of memory for the sake of performance. For instance, if the number of elements you need to hold in your buffer is N, allocate an array of N+1024 (or some sensible number). Then you only need to move N elements around every 1024 insertions, and you always have a contiguous view of N elements to act upon directly available.

EDIT: here is a code snippet that implements the above, and should give good performance. Note though, that you would be well advised to append in chunks, rather than per element. Otherwise, the performance advantages of using numpy are quickly nullified, regardless of how you implement your ringbuffer.

import numpy as np

class RingBuffer(object):
    def __init__(self, size, padding=None):
        self.size = size
        self.padding = size if padding is None else padding
        self.buffer = np.zeros(self.size+self.padding)
        self.counter = 0

    def append(self, data):
        """this is an O(n) operation"""
        data = data[-self.padding:]
        n = len(data)
        if self.remaining < n: self.compact()
        self.buffer[self.counter+self.size:][:n] = data
        self.counter += n

    @property
    def remaining(self):
        return self.padding-self.counter
    @property
    def view(self):
        """this is always an O(1) operation"""
        return self.buffer[self.counter:][:self.size]
    def compact(self):
        """
        note: only when this function is called, is an O(size) performance hit incurred,
        and this cost is amortized over the whole padding space
        """
        print 'compacting'
        self.buffer[:self.size] = self.view
        self.counter = 0

rb = RingBuffer(10)
for i in range(4):
    rb.append([1,2,3])
    print rb.view

rb.append(np.arange(15))
print rb.view  #test overflow

A variant of @Janne Karila's answer, for C but not numpy:
If the ring buffer is very wide, like N x 1G, then instead of doubling the whole thing, double up an array of 2*N pointers to its rows. E.g. for N=3, initialize

bufp = { buf[0], buf[1], buf[2], buf[0], buf[1], buf[2] };

Then you write data only once, and anyfunc( bufp[j:j+3] ) sees the rows in buf in time order.

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