Extend numpy mask by n cells to the right for each bad value, efficiently

Question

Let\'s say I have a length 30 array with 4 bad values in it. I want to create a mask for those bad values, but since I will be using rolling window functions, I\'d also like a f

Answer 1

A few years late, but I've come up with a fully vectorized solution that requires no loops or copies (besides the mask itself). This solution is a bit (potentially) dangerous because it uses numpy.lib.stride_tricks.as_strided. It's also not as fast as @swentzel's solution.

The idea is to take the mask and create a 2D view of it, where the second dimension is just the elements that follow the current element. Then you can just set an entire column to True if the head is True. Since you are dealing with a view, setting a column will actually set the following elements in the mask.

Start with the data:

import numpy as np
a = np.array([4, 0, 8, 5, 10, 9, np.nan, 1, 4, 9, 9, np.nan, np.nan, 9,\
              9, 8, 0, 3, 7, 9, 2, 6, 7, 2, 9, 4, 1, 1, np.nan, 10])
n = 3

Now, we will make the mask a.size + n elements long, so that you don't have to process the last n elements manually:

mask = np.empty(a.size + n, dtype=np.bool)
np.isnan(a, out=mask[:a.size])
mask[a.size:] = False

Now the cool part:

view = np.lib.stride_tricks.as_strided(mask, shape=(n + 1, a.size),
                                       strides=mask.strides * 2)

That last part is crucial. mask.strides is a tuple like (1,) (since bools are usually about that many bytes across. Doubling it means that you take a 1-byte step to move one element in any dimension.

Now all you need to do is expand the mask:

view[1:, view[0]] = True

That's it. Now mask has what you want. Keep in mind that this only works because the assignment index precedes the last changed value. You could not get away with view[1:] |= view[0].

For benching purposes, it appears that the definition of n has changed from the question, so the following function takes that into account:

def madphysicist0(a, n):
    m = np.empty(a.size + n - 1, dtype=np.bool)
    np.isnan(a, out=m[:a.size])
    m[a.size:] = False

    v = np.lib.stride_tricks.as_strided(m, shape=(n, a.size), strides=m.strides * 2)
    v[1:, v[0]] = True
    return v[0]

V2

Taking a leaf out of the existing fastest answer, we only need to copy log2(n) rows, not n rows:

def madphysicist1(a, n):
    m = np.empty(a.size + n - 1, dtype=np.bool)
    np.isnan(a, out=m[:a.size])
    m[a.size:] = False

    v = np.lib.stride_tricks.as_strided(m, shape=(n, a.size), strides=m.strides * 2)

    stop = int(np.log2(n))
    for k in range(1, stop + 1):
        v[k, v[0]] = True
    if (1<<k) < n:
        v[-1, v[(1<<k) - 1]] = True
    return v[0]

Since this doubles the size of the mask at every iteration, it works a bit faster than Fibonacci: https://math.stackexchange.com/q/894743/295281

Answer 2

This could also be considered a morphological dilation problem, using here the scipy.ndimage.binary_dilation:

def dilation(a, n):
    m = np.isnan(a)
    s = np.full(n, True, bool)
    return ndimage.binary_dilation(m, structure=s, origin=-(n//2))

Note on origin: this argument ensures the structure (I would call it a kernel) starts off a bit to the left of the input (your mask m). Normally the value at out[i] would be the dilation with the center of structure (which would be structure[n//2]) at in[i], but you want the structure[0] to be at in[i].

You can also do this with a kernel that is padded on the left with Falses, which is what would be required if you used the binary_dilation from scikit-image:

def dilation_skimage(a, n):
    m = np.isnan(a)
    s = np.zeros(2*n - n%2, bool)
    s[-n:] = True
    return skimage.morphology.binary_dilation(m, selem=s)

Timing doesn't seem to change too much between the two:

dilation_scipy
small:    10 loops, best of 3: 47.9 ms per loop
large: 10000 loops, best of 3: 88.9 µs per loop

dilation_skimage
small:    10 loops, best of 3: 47.0 ms per loop
large: 10000 loops, best of 3: 91.1 µs per loop

Answer 3

OP here with the benchmark results. I have included my own ("op") which I had started out with, which loops over the bad indices and adds 1...n to them then takes the uniques to find the mask indices. You can see it in the code below with all the other responses.

Anyway here are the results. The facets are size of array along x (10 thru 10e7) and size of window along y(5, 50, 500, 5000). Then it's by coder in each facet, with a log-10 score because we're talking microseconds through minutes.

@swenzel appears to be the winner with his second answer, displacing @moarningsun's first answer (moarningsun's second answer was crashing the machine through massive memory use, but that's probably because it was not designed for large n or non-sparse a).

The chart does not do justice to the fastest of these contributions because of the (necessary) log scale. They're dozens, hundreds of times faster than even decent looping solutions. swenzel1 is 1000x faster than op in the largest case, and op is already making use of numpy.

Please note that I have used a numpy version compiled against the optimised Intel MKL libraries which make full use of the AVX instructions present since 2012. In some vector use cases this will increase an i7/Xeon speed by a factor of 5. Some of the contributions may be benefitting more than others.

Here is the full code to run all the submitted answers so far, including my own. Function allagree() makes sure that results are correct, while timeall() will give you a long-form pandas Dataframe with all the results in seconds.

You can rerun it fairly easily with new code, or change my assumptions. Please keep in mind I did not take into account other factors such as memory usage. Also, I resorted to R ggplot2 for the graphic as I don't know seaborn/matplotlib well enough to make it do what I want.

For completeness, all the results agree:

In [4]: allagree(n = 7, asize = 777)
Out[4]:
             AGML0 AGML1 askewchan0 askewchan1 askewchan2 moarningsun0  \
AGML0         True  True       True       True       True         True
AGML1         True  True       True       True       True         True
askewchan0    True  True       True       True       True         True
askewchan1    True  True       True       True       True         True
askewchan2    True  True       True       True       True         True
moarningsun0  True  True       True       True       True         True
swenzel0      True  True       True       True       True         True
swenzel1      True  True       True       True       True         True
op            True  True       True       True       True         True

             swenzel0 swenzel1    op
AGML0            True     True  True
AGML1            True     True  True
askewchan0       True     True  True
askewchan1       True     True  True
askewchan2       True     True  True
moarningsun0     True     True  True
swenzel0         True     True  True
swenzel1         True     True  True
op               True     True  True

Thank you to all who submitted!

Code for the graphic after exporting output of timeall() using pd.to_csv and read.csv in R:

ww <- read.csv("ww.csv")    
ggplot(ww, aes(x=coder, y=value, col = coder)) + geom_point(size = 3) + scale_y_continuous(trans="log10")+ facet_grid(nsize ~ asize) + theme(axis.text.x = element_text(angle = 90, hjust = 1)) + ggtitle("Fastest by coder") + ylab("time (seconds)")

Code for the test:

# test Stack Overflow 32706135 nan shift routines

import numpy as np
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
from timeit import Timer
from scipy import ndimage
from skimage import morphology
import itertools
import pdb
np.random.seed(8472)


def AGML0(a, n):                               # loop itertools
    maskleft = np.where(np.isnan(a))[0]
    maskright = maskleft + n
    mask = np.zeros(len(a),dtype=bool)
    for l,r in itertools.izip(maskleft,maskright): 
        mask[l:r] = True
    return mask


def AGML1(a, n):                               # loop n
    nn = n - 1
    maskleft = np.where(np.isnan(a))[0]
    ghost_mask = np.zeros(len(a)+nn,dtype=bool)
    for i in range(0, nn+1):
        thismask = maskleft + i
        ghost_mask[thismask] = True
    mask = ghost_mask[:len(ghost_mask)-nn]
    return mask


def askewchan0(a, n):
    m = np.isnan(a)
    i = np.arange(1, len(m)+1)
    ind = np.column_stack([i-n, i]) # may be a faster way to generate this
    ind.clip(0, len(m)-1, out=ind)
    return np.bitwise_or.reduceat(m, ind.ravel())[::2]


def askewchan1(a, n):
    m = np.isnan(a)
    s = np.full(n, True, bool)
    return ndimage.binary_dilation(m, structure=s, origin=-(n//2))


def askewchan2(a, n):
    m = np.isnan(a)
    s = np.zeros(2*n - n%2, bool)
    s[-n:] = True
    return morphology.binary_dilation(m, selem=s)


def moarningsun0(a, n):
    mask = np.isnan(a)
    cs = np.cumsum(mask)
    cs[n:] -= cs[:-n].copy()
    return cs > 0


def moarningsun1(a, n):
    mask = np.isnan(a)
    idx = np.flatnonzero(mask)
    expanded_idx = idx[:,None] + np.arange(1, n)
    np.put(mask, expanded_idx, True, 'clip')
    return mask


def swenzel0(a, n):
    m = np.isnan(a)
    k = m.copy()
    for i in range(1, n):
        k[i:] |= m[:-i]
    return k


def swenzel1(a, n=4):
    m = np.isnan(a)
    k = m.copy()

    # lenM and lenK say for each mask how many
    # subsequent Trues there are at least
    lenM, lenK = 1, 1

    # we run until a combination of both masks will give us n or more
    # subsequent Trues
    while lenM+lenK < n:
        # append what we have in k to the end of what we have in m
        m[lenM:] |= k[:-lenM]

        # swap so that m is again the small one
        m, k = k, m

        # update the lengths
        lenM, lenK = lenK, lenM+lenK

    # see how much m has to be shifted in order to append the missing Trues
    k[n-lenM:] |= m[:-n+lenM]
    return k


def op(a, n):
    m = np.isnan(a)
    for x in range(1, n):
        m = np.logical_or(m, np.r_[False, m][:-1])
    return m


# all the functions in a list. NB these are the actual functions, not their names
funcs = [AGML0, AGML1, askewchan0, askewchan1, askewchan2, moarningsun0, swenzel0, swenzel1, op]

def allagree(fns = funcs, n = 10, asize = 100):
    """ make sure result is the same from all functions """
    fnames = [f.__name__ for f in fns]
    a = np.random.rand(asize)
    a[np.random.randint(0, asize, int(asize / 10))] = np.nan
    results = dict([(f.__name__, f(a, n)) for f in fns])
    isgood = [[np.array_equal(results[f1], results[f2]) for f1 in fnames] for f2 in fnames]
    pdgood = pd.DataFrame(isgood, columns = fnames, index = fnames)
    if not all([all(x) for x in isgood]):
        print "not all results identical"
        pdb.set_trace()
    return pdgood


def timeone(f):
    """ time one of the functions across the full range of a nd n """
    print "Timing", f.__name__
    Ns = np.array([10**x for x in range(0, 4)]) * 5 # 5 to 5000 window size
    As = [np.random.rand(10 ** x) for x in range(1, 8)] # up to 10 million data data points
    for i in range(len(As)): # 10% of points are always bad
        As[i][np.random.randint(0, len(As[i]), len(As[i]) / 10)] = np.nan
    results = np.array([[Timer(lambda: f(a, n)).timeit(number = 1) if n < len(a) \
                        else np.nan for n in Ns] for a in As])
    pdresults = pd.DataFrame(results, index = [len(x) for x in As], columns = Ns)
    return pdresults


def timeall(fns = funcs):
    """ run timeone for all known funcs """
    testd = dict([(x.__name__, timeone(x)) for x in fns])
    testdf = pd.concat(testd.values(), axis = 0, keys = testd.keys())
    testdf.index.names = ["coder", "asize"]
    testdf.columns.names = ["nsize"]
    testdf.reset_index(inplace = True)
    testdf = pd.melt(testdf, id_vars = ["coder", "asize"])
    return testdf

Answer 4

You can use np.ufunc.reduceat with np.bitwise_or:

import numpy as np
a = np.array([4, 0, 8, 5, 10, 9, np.nan, 1, 4, 9, 9, np.nan, np.nan, 9,
              9, 8, 0, 3, 7, 9, 2, 6, 7, 2, 9, 4, 1, 1, np.nan, 10])
m = np.isnan(a)
n = 4
i = np.arange(1, len(m)+1)
ind = np.column_stack([i-n, i]) # may be a faster way to generate this
ind.clip(0, len(m)-1, out=ind)

np.bitwise_or.reduceat(m, ind.ravel())[::2]

On your data:

print np.column_stack([m, reduced])
[[False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [ True  True]
 [False  True]
 [False  True]
 [False  True]
 [False False]
 [ True  True]
 [ True  True]
 [False  True]
 [False  True]
 [False  True]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [False False]
 [ True  True]
 [False  True]]

Answer 5

You can use the same cumsum trick as you would for a running average filter:

def cumsum_trick(a, n):
    mask = np.isnan(a)
    cs = np.cumsum(mask)
    cs[n:] -= cs[:-n].copy()
    return cs > 0

Unfortunately the additional .copy() is needed, because of some buffering that goes on internally the order of operations. It is possible to persuade numpy to apply the subtraction in reverse, but for that to work the cs array must have a negative stride:

def cumsum_trick_nocopy(a, n):
    mask = np.isnan(a)
    cs = np.cumsum(mask, out=np.empty_like(a, int)[::-1])
    cs[n:] -= cs[:-n]
    out = cs > 0
    return out

But this seems fragile and isn't really that much faster anyway.

I wonder if there's a compiled signal processing function somewhere that does this exact operation..

---

For sparse initial masks and small n this one is also pretty fast:

def index_expansion(a, n):
    mask = np.isnan(a)
    idx = np.flatnonzero(mask)
    expanded_idx = idx[:,None] + np.arange(1, n)
    np.put(mask, expanded_idx, True, 'clip')
    return mask

Answer 6

Something like this?

maskleft = np.where(np.isnan(a))[0]
maskright = maskleft + n
mask = np.zeros(len(a),dtype=bool)
for l,r in itertools.izip(maskleft,maskright): 
   mask[l:r] = True

Or, since n is small, it might be better to loop over it instead:

maskleft = np.where(np.isnan(a))[0]
mask = np.zeros(len(a),dtype=bool)
for i in range(0,n):
  thismask = maskleft+i
  mask[thismask] = True

Except for the loop over n, the above is fully vectorized. But the loop is fully parallelizable, so you could be able to get a factor-n speedup using e.g. multiprocessing or Cython, if you're willing to go to the trouble.

Edit: per @askewchan solution 2 can potentially cause out of range errors. It also has indexing problems in the range(0,n). Possible correction:

maskleft = np.where(np.isnan(a))[0]
ghost_mask = np.zeros(len(a)+n,dtype=bool)
for i in range(0, n+1):
    thismask = maskleft + i
    ghost_mask[thismask] = True
mask = ghost_mask[:len(ghost_mask)-n]