[Numpy-discussion] contiguous true
Charles R Harris
charlesr.harris@gmail....
Sat Mar 1 00:21:20 CST 2008
On Fri, Feb 29, 2008 at 11:12 PM, Charles R Harris <
charlesr.harris@gmail.com> wrote:
>
>
> On Fri, Feb 29, 2008 at 10:53 AM, John Hunter <jdh2358@gmail.com> wrote:
>
> > [apologies if this is a resend, my mail just flaked out]
> >
> > I have a boolean array and would like to find the lowest index "ind"
> > where N contiguous elements are all True. Eg, if x is
> >
> > In [101]: x = np.random.rand(20)>.4
> >
> > In [102]: x
> > Out[102]:
> > array([False, True, True, False, False, True, True, False, False,
> > True, False, True, False, True, True, True, False, True,
> > False, True], dtype=bool)
> >
> > I would like to find ind=1 for N=2 and ind=13 for N=2. I assume with
> > the right cumsum, diff and maybe repeat magic, this can be vectorized,
> > but the proper incantation is escaping me.
> >
> > for N==3, I thought of
> >
> > In [110]: x = x.astype(int)
> > In [112]: y = x[:-2] + x[1:-1] + x[2:]
> >
> > In [125]: ind = (y==3).nonzero()[0]
> >
> > In [126]: if len(ind): ind = ind[0]
> >
> > In [128]: ind
> > Out[128]: 13
> >
>
>
> This may be more involved than you want, but
>
> In [37]: prng = random.RandomState(1234567890)
>
> In [38]: x = prng.random_sample(50) < 0.5
>
> In [39]: y1 = concatenate(([False], x[:-1]))
>
> In [40]: y2 = concatenate((x[1:], [False]))
>
> In [41]: beg = ind[x & ~y1]
>
> In [42]: end = ind[x & ~y2]
>
> In [43]: cnt = end - beg + 1
>
> In [44]: i = beg[cnt == 4]
>
> In [45]: i
> Out[45]: array([28])
>
> In [46]: x
> Out[46]:
> array([False, False, False, False, True, False, True, False, False,
> False, True, False, True, False, True, True, True, True,
> True, False, False, False, True, False, True, False, False,
> False, True, True, True, True, False, False, True, False,
> False, False, False, False, False, False, False, True, False,
> False, True, False, True, False], dtype=bool)
>
> produces a list of the indices where sequences of length 4 begin.
>
> Chuck
>
Oops, ind = arange(len(x)). I suppose nonzero would work as well.
Chuck
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