How to delete rows with specific value and modify column values in a 2-D array?

Question

Putting a very simple example of what I need to do, I have a 2-D array:

array([[0, 4],[2, 3],[4, 1],[0, 8],[3, 2],[4, 5],[6, 1],[0, 3],[1, 7],[2, 9]])

The first column is time and the other is distance. Being specifically what I have is [time, distance].

For my final array, I need to delete each time=0 and to make continuation of the time. Here is what my final array should be:

array([[0, 4],[2, 3],[4, 1],[7, 2],[8, 5],[10, 1],[11, 7],[12, 9]])

In this particular case, only [0, 8] and [0, 3] are deleted because they have time=0.

Also, the previous [3, 2] becomes [7, 2] given that 4+3=7 (because previous time was 4). Same goes for previous [4, 5] becomes [8, 5] given that 4+4=8.Same goes for previous [6, 1] becomes [10, 1] given that 4+6=10.

Same goes for previous [1, 7] becomes [11, 7] given that 10+1=11 (because previous time was 10). Same goes for previous [2, 9] becomes [12, 9] given that 10+2=12.

I am just putting a very small piece of array, thus I need a Python code to do what I just explained.

Because I need the time to start from zero and then on have continuation — ananvodo
– ananvodo, Commented May 30, 2018 at 1:34
Is your time format military? because if not, some times are gonna exceed 12 and that's problem — Mansour Zayer
– Mansour Zayer, Commented May 30, 2018 at 2:37

Mansour Zayer · Accepted Answer · 2018-05-30 02:27:38Z

1

I came up with this short code without the use of any external libraries:

array = [[0, 4],[2, 3],[4, 1],[0, 8],[3, 2],[4, 5],[6, 1],[0, 3],[1, 7],[2, 9]]
temp = 0
out = [array[0]]
for i in range(1,len(array)):
    if array[i][0] != 0:
        array[i][0] += temp
        out.append(array[i])
    else:
        temp = array[i-1][0]
print(out)

which yields:

[out]: [[0, 4], [2, 3], [4, 1], [7, 2], [8, 5], [10, 1], [11, 7], [12, 9]]

answered May 30, 2018 at 2:27

Mansour Zayer

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1 Comment

ananvodo Over a year ago

Thanks. It worked perfect and it is very efficient too. Hope to code like you one day. Best

AChampion · Accepted Answer · 2018-05-30 18:30:29Z

A slightly faster but pretty much identical approach to @MansourZayer. It is generally preferred to iterate over the items in a list vs. using range() and indexing into the list, e.g.

In []:
x = [[0, 4],[2, 3],[4, 1],[0, 8],[3, 2],[4, 5],[6, 1],[0, 3],[1, 7],[2, 9]]
t0 = tminus = 0
r = [x[0]]
for t, d in x:
    if t == 0:
        t0 += tminus
        continue
    tminus = t
    r.append([t+t0, d])
r

Out[]:
[[0, 4], [2, 3], [4, 1], [7, 2], [8, 5], [10, 1], [11, 7], [12, 9]]

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