Keep first element of array that is part of a sequence

user_4427777 0
Viewed 165

Say I have this type of array 

y 

array([299839, 667136, 665420, 665418, 665421, 667135, 299799, 665419, 667137, 299800])

as the result of a "top 10" argpartition:

y = np.argpartiton(-x, np.arange(10))[:10]

Now, I want to remove the elements that are sequential, only keeping the first (maximum) element in the series such that:

y_new
array([299839, 667136, 665420, 299799])

But while that seems like it should be simple I'm not seeing an efficient way to do it (or even a good way to start). Assume the real-world application will do the top 1000 or so and need to do it many times.

numpy python
2 Answers

Here's one approach based on sorting -

# Get the sorted indices
sidx = y.argsort()

# Get sorted array
ys = y[sidx]

# Get indices at which islands of sequential numbers start/stop
cut_idx = np.flatnonzero(np.concatenate(([True], np.diff(ys)!=1 )))

# Finally get the minimum indices for each island and then index into
# input for the desired output
y_new = y[np.minimum.reduceat(sidx, cut_idx)]

If you would like to keep the order of elements in the output, sort the indices and then index at the last step -

y[np.sort(np.minimum.reduceat(sidx, cut_idx))]

Sample input, output -

In [56]: y
Out[56]: 
array([299839, 667136, 665420, 665418, 665421, 667135, 299799, 665419,
       667137, 299800])

In [57]: y_new
Out[57]: array([299799, 299839, 665420, 667136])

In [58]: y[np.sort(np.minimum.reduceat(sidx, cut_idx))]
Out[58]: array([299839, 667136, 665420, 299799])
user_32938810

heres my implementation for that problem

from itertools import groupby
from operator import itemgetter


a = [299839, 667136, 665420, 665418, 665421, 667135, 299799, 665419, 
667137, 299800]
new = a[:]

# to keep the first number
b = a[0]
new.sort()
# to store diffrent arrays
saver = []
final_array = []

for k, g in groupby(enumerate(new), lambda (i, x): i - x):
    ac = map(itemgetter(1), g)
    saver.append(ac)

final_array.append(b)
for i in range(len(saver)):
    for j in range(len(a)):
        if a[j] in saver[i]:
            if b == a[j]:
                continue
            final_array.append(a[j])
            break


print final_array

output 

[299839, 299799, 665420, 667136]
user_44096600
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