Circular buffer python

Question

I want to do this with a circular buffer(that is x as deque)

i = 0
x = []

while True:
    accel_data = sensor.get_accel()
    d = datetime.utcnow().strftime('%Y-%m-%d')
    t = datetime.utcnow().strftime('%H:%M:%S.%f')
    x.append(accel_data + (d, t))
    i = i + 1

I know how to implement a simple circular buffer:

from collections import deque
import time

d = deque(maxlen=4)
bool = True
i = 1
y = 0
while bool:
    d.append(i)
    i = i + 1
    print(d)
    time.sleep(1)

But I can’t use it to reproduce the first code.

Why can’t you reproduce it?

bfontaine
– bfontaine

2017-09-19 09:26:18 +00:00
Commented Sep 19, 2017 at 9:26 — bfontaine
– bfontaine, Commented Sep 19, 2017 at 9:26

norok2 · Accepted Answer · 2017-09-19 12:39:33Z

2

Does something like this work?

from collections import deque

container = deque(maxlen=4)
while True:
    accel_data = sensor.get_accel()
    curr_date = datetime.utcnow().strftime('%Y-%m-%d')
    curr_time = datetime.utcnow().strftime('%H:%M:%S.%f')
    entry = accel_data + (curr_date, curr_time)
    container.append(entry)
    print(container)  # this is not strictly necessary

A few tips:

Use sensible names for the variables you are using.
Do not initialize / declare variables you are not going to use.
Be more specific in what you are not managing.

edited Sep 19, 2017 at 12:39

answered Sep 19, 2017 at 9:40

norok2

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9 Comments

Nick9214 Over a year ago

"Do not declare variables you are not going to use." : i forgot to add few lines before "i=i+1": for y in range(4): print x[i][y]

norok2 Over a year ago

Be careful in how you introduce that code though, as x from your code became container whose maximum length is fixed (4 in the example), so i will go out of range if you keep adding it.

Nick9214 Over a year ago

mmm I couldn't try your code because I need my raspberry, but I think there has been a misunderstanding,anyway i will explain better my problem With first code I've an array ("x"),and in evry step of while I add all those information.So after 4 steps,for example it will be: 0.48 , 0.89 , 1.2 16/12/17 20.45.34 ==> x[0] -----------0.48 , 0.90 , 1.4 16/12/17 20.45.35 ==> x[1] ---------- 0.48 , 0.91 , 1.2 16/12/17 20.45.36 ==>x[2] ------0.48 , 0.94 , 1.2 16/12/17 20.45.37==> x[3]

Nick9214 Over a year ago

So the "maxlen" is not referred to "j"( the elements in x[i],that are always 5),but to "i"

norok2 Over a year ago

This code works also if you change container = deque(maxlen=4) with container = [] as you (ideally) originally had. The list version will let container grow by 1 at each iteration, hence if you can access container with an ever growing index like container[i] (assuming you have i += 1 or similar somewhere inside the loop and you had initialized i = 0 outside the loop). On the other hand, if container is now a deque, accessing it when i is larger than its max size, will cause the container[i] to get out of range for some value of i, regardless of what is in entry.

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