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I am getting a list using a list comprehension. lat say I am getting this list using this line of code bellow:

quality, angle, distance = measurements[i]
new_data = [each_value for each_value in  measurements[i:i + 20] if angle <= each_value[1] <= angle + 30 and
                        distance - 150 <= each_value[2] <= distance + 150]

where measurements is a big data set which contains (quality, angle, distance) pair. from that, I am getting those value.

desired_list= [(1,2,3)(1,5,3),(1,8,3)(1,10,3),(1,16,3),(1,17,3)]]

Now how can I add a new condition in my list comprehension so that I will only get the value if the angle is within some offset value? let say if the difference between two respective angles is less then or equal to 5 then put them in desired_list.

with this condition my list should be like so: 

desired_list= [(1,2,3)(1,5,3),(1,8,3)(1,10,3)]

cause from 2 to 5, 5 to 8, 8 to 10 the distance is less than or equal to 5.

But the last two points are not included as they break the condition after (1,10,3) and they don't need to check.

How can I achieve this? please help me

Note: it doesn't need to be in the same list comprehension.
Kazi
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  • Try writing this as a simpler problem. There is a lot of data included here that have nothing to do with your question (e.g. the 3-tuple instead of a single integer) – Cireo Apr 12 '19 at 06:57
  • I am confused, `(1,16,3)`,`(1,17,3)` pairs are also less than 5 – recnac Apr 12 '19 at 06:57
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    @recnac He's saying the condition breaks after 10->16 so no need to check after that. – Harshith Bolar Apr 12 '19 at 06:59
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    Is there a need to do everything in the same list comprehension? Why not split it into two steps? – Harshith Bolar Apr 12 '19 at 07:00
  • no dear. its not needed. @HarshithBolar – Kazi Apr 12 '19 at 07:01
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    create [MCVE](https://stackoverflow.com/help/mcve)s to make life simpler for everyone by providing sample inputs, such that people can copy paste and run the code. – Paritosh Singh Apr 12 '19 at 07:03
  • so you means traverse from start to end, and break out when one neighor pairs break the rule? – recnac Apr 12 '19 at 07:03
  • yes!! exactly @recnac – Kazi Apr 12 '19 at 07:05
  • thanks for your suggestion. I will surely do that from my next post. Beforehand I have to learn how to do MCVE. @ParitoshSingh – Kazi Apr 12 '19 at 07:06
  • insist using list comprehension? if not use list comprehension, it is easier to solve, because here is a break out. – recnac Apr 12 '19 at 07:06
  • it does not need to do in the same list comprehension. can you show me the easy solution, please?@recnac – Kazi Apr 12 '19 at 07:07

2 Answers2

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You mention the data set is large. Depending how large you many wish to avoid creating a new list from scratch and just search for the relavant index. 

data = [(1,2,3), (1,5,3), (1,8,3), (1,10,3), (1,16,3), (1,17,3)]

MAXIMUM_ANGLE = 5

def angles_within_range(x, y):
    return abs(x[1] - y[1]) <= MAXIMUM_ANGLE

def first_angle_break_index():
    for i in range(len(data) - 1):
        if not angles_within_range(data[i], data[i+1]):
            return i+1

def valid_angles_list():
    return data[:first_angle_break_index()]

print(valid_angles_list())
Andrew Allen
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1

If you means traverse from start to end, and break out when one neighor pairs break the rule.
here is a way without list comprehension:

desired_list = [(1, 2, 3), (1, 5, 3), (1, 8, 3), (1, 10, 3), (1, 16, 3), (1, 17, 3)]
res = [desired_list[0]]
for a, b in zip(desired_list[:-1], desired_list[1:]):
    if abs(a[1] - b[1]) > 5:
        break
    res += [b]

print(res)

output:

[(1, 2, 3), (1, 5, 3), (1, 8, 3), (1, 10, 3)]

if you insist on using list comprehension with break, here is a solution of recording last pair:

res = [last.pop() and last.append(b) or b for last in [[desired_list[0]]] for a, b in 
zip([desired_list[0]] + desired_list, desired_list) if abs(a[1] - b[1]) <= 5 and a == last[0]]

another version use end condition:

res = [b for end in [[]] for a, b in zip([desired_list[0]] + desired_list, desired_list) if 
(False if end or abs(a[1] - b[1]) <= 5 else end.append(42)) or not end and abs(a[1] - b[1]) <= 5]

Note: This is a bad idea. (just for fun : ))

recnac
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