Selecting items from an array

Question

I am trying to take an array of integers such as:

[1,2,3,4,9,10,11,12,18,19,20,21] and get the values at which there is a "jump," so, the output of my program would be [1,9,18]. I wrote the following code in Python which seems to be taking forever to run:

min_indices = np.where(data[:,1] == data[:,1].min())
start_indices = np.array([0])
i = 1
while (i < len(min_indices[0])):
    if (min_indices[0][i] != (min_indices[0][i-1] + 1)):
        start_indices.append(min_indices[i])
print start_indices

You never increment i in your loop, this is causing running forever. — xbb
– xbb, Commented Nov 17, 2014 at 16:43

fredtantini · Accepted Answer · 2014-11-17 16:45:31Z

3

Using list comprehension:

>>> a=[1,2,3,4,9,10,11,12,18,19,20,21]
>>> [a[i] for i,_ in enumerate(a[1:]) if (a[i]!=a[i-1]+1)]
[1, 9, 18]

and using zip:

>>> [a[0]]+[i for (i,j) in zip(a[1:],(a[:-1])) if (i!=j+1)]
[1, 9, 18]

answered Nov 17, 2014 at 16:45

fredtantini

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shooper · Accepted Answer · 2014-11-17 16:43:34Z

3

You are not incrementing "i" that I can tell.

answered Nov 17, 2014 at 16:43

shooper

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1 Comment

okarin Over a year ago

Thanks. Forgot about that.

Ashwini Chaudhary · Accepted Answer · 2014-11-17 16:49:49Z

3

You can use numpy.diff with numpy.where here:

>>> a = np.array([1,2,3,4,9,10,11,12,18,19,20,21])
>>> indices = np.where(np.diff(a) > 1)[0] + 1
>>> np.concatenate(([a[0]], a[indices]))
array([ 1,  9, 18])

answered Nov 17, 2014 at 16:49

Ashwini Chaudhary

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