how to do this operation in numpy (chaining of tiling operation)?

I'm trying to do fast generation of numpy array, possibly without passing through python.

I want to build an 1D index numpy array that would take this as an input:

[2,3] and this [2,4] and would return this

 [0,1,0,1,0,1,2,0,1,2,0,1,2,0,1,2]

Explanation:

I iterate from 0 to 2 (so [0,1] array) and repeat it 2 times : [0,1,0,1]

Then I iterate from 0 to 3 (so [0,1,2] array) and repeat it 4 times : [0,1,2,0,1,2,0,1,2,0,1,2]
Then I flattened everything.

Is there a way to do this fully in numpy?
For now I'm building each table separately in numpy by using np.tile() and flattening everything afterwards but I feel like there is a more efficient way that would only translate to C functions calls and no python

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

add a comment |

I'm trying to do fast generation of numpy array, possibly without passing through python.

I want to build an 1D index numpy array that would take this as an input:

[2,3] and this [2,4] and would return this

 [0,1,0,1,0,1,2,0,1,2,0,1,2,0,1,2]

Explanation:

I iterate from 0 to 2 (so [0,1] array) and repeat it 2 times : [0,1,0,1]

Then I iterate from 0 to 3 (so [0,1,2] array) and repeat it 4 times : [0,1,2,0,1,2,0,1,2,0,1,2]
Then I flattened everything.

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

add a comment |

I'm trying to do fast generation of numpy array, possibly without passing through python.

I want to build an 1D index numpy array that would take this as an input:

[2,3] and this [2,4] and would return this

 [0,1,0,1,0,1,2,0,1,2,0,1,2,0,1,2]

Explanation:

I iterate from 0 to 2 (so [0,1] array) and repeat it 2 times : [0,1,0,1]

Then I iterate from 0 to 3 (so [0,1,2] array) and repeat it 4 times : [0,1,2,0,1,2,0,1,2,0,1,2]
Then I flattened everything.

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

I'm trying to do fast generation of numpy array, possibly without passing through python.

I want to build an 1D index numpy array that would take this as an input:

[2,3] and this [2,4] and would return this

 [0,1,0,1,0,1,2,0,1,2,0,1,2,0,1,2]

Explanation:

I iterate from 0 to 2 (so [0,1] array) and repeat it 2 times : [0,1,0,1]

Then I iterate from 0 to 3 (so [0,1,2] array) and repeat it 4 times : [0,1,2,0,1,2,0,1,2,0,1,2]
Then I flattened everything.

python arrays numpy

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

edited Nov 22 '18 at 0:17

jpp

96.7k2158109

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

asked Nov 21 '18 at 23:44

lezebulon

3,25573062

add a comment |

2 Answers
2

active

oldest

votes

Here is a vectorized solution:

def cycles(spec):

    steps = np.repeat(*spec)

    ps = steps.cumsum()

    psj = np.zeros(ps[-1], int)

    psj[ps[:-1]] = steps[:-1]

    return np.arange(ps[-1]) - psj.cumsum()

Demo:

>>> cycles(((2,3),(2,4)))

array([0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2])

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

Thanks!However it's failing for some combinations: a = [4, 0, 1,0] b = [35, 72, 5, 72] cycles(a,b) I think it has to do with the 0 at the end of 'a' array

– lezebulon
Nov 22 '18 at 10:27

@lezebulon Rather pathological example and also easy to address. Just filter out the 0 positions. For example, add the line steps = steps[steps.astype(bool)] after the first line of the function body.

– Paul Panzer
Nov 22 '18 at 10:47

add a comment |

I am not entirely sure if this is what you want; here each tuple in the call to func() contains first the range and then the repeat.

import numpy





def func(tups):

    Arr = numpy.empty(numpy.sum([ele[0] * ele[1] for ele in tups]), dtype=int)

    i = 0

    for ele in tups:

        Arr[i:i + ele[0] * ele[1]] = numpy.tile(numpy.arange(ele[0]), ele[1])

        i += ele[0] * ele[1]

    return Arr





arr = func([(2, 3), (3, 4)])

print(arr)

# [0 1 0 1 0 1 0 1 2 0 1 2 0 1 2 0 1 2]

edited Nov 22 '18 at 1:38

answered Nov 22 '18 at 0:15

Patol75

6236

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Here is a vectorized solution:

def cycles(spec):

    steps = np.repeat(*spec)

    ps = steps.cumsum()

    psj = np.zeros(ps[-1], int)

    psj[ps[:-1]] = steps[:-1]

    return np.arange(ps[-1]) - psj.cumsum()

Demo:

>>> cycles(((2,3),(2,4)))

array([0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2])

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

Thanks!However it's failing for some combinations: a = [4, 0, 1,0] b = [35, 72, 5, 72] cycles(a,b) I think it has to do with the 0 at the end of 'a' array

– lezebulon
Nov 22 '18 at 10:27

@lezebulon Rather pathological example and also easy to address. Just filter out the 0 positions. For example, add the line steps = steps[steps.astype(bool)] after the first line of the function body.

– Paul Panzer
Nov 22 '18 at 10:47

add a comment |

Here is a vectorized solution:

def cycles(spec):

    steps = np.repeat(*spec)

    ps = steps.cumsum()

    psj = np.zeros(ps[-1], int)

    psj[ps[:-1]] = steps[:-1]

    return np.arange(ps[-1]) - psj.cumsum()

Demo:

>>> cycles(((2,3),(2,4)))

array([0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2])

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

Thanks!However it's failing for some combinations: a = [4, 0, 1,0] b = [35, 72, 5, 72] cycles(a,b) I think it has to do with the 0 at the end of 'a' array

– lezebulon
Nov 22 '18 at 10:27

@lezebulon Rather pathological example and also easy to address. Just filter out the 0 positions. For example, add the line steps = steps[steps.astype(bool)] after the first line of the function body.

– Paul Panzer
Nov 22 '18 at 10:47

add a comment |

Here is a vectorized solution:

def cycles(spec):

    steps = np.repeat(*spec)

    ps = steps.cumsum()

    psj = np.zeros(ps[-1], int)

    psj[ps[:-1]] = steps[:-1]

    return np.arange(ps[-1]) - psj.cumsum()

Demo:

>>> cycles(((2,3),(2,4)))

array([0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2])

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

Here is a vectorized solution:

def cycles(spec):

    steps = np.repeat(*spec)

    ps = steps.cumsum()

    psj = np.zeros(ps[-1], int)

    psj[ps[:-1]] = steps[:-1]

    return np.arange(ps[-1]) - psj.cumsum()

Demo:

>>> cycles(((2,3),(2,4)))

array([0, 1, 0, 1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2])

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

answered Nov 22 '18 at 0:32

Paul Panzer

30k21240

Thanks!However it's failing for some combinations: a = [4, 0, 1,0] b = [35, 72, 5, 72] cycles(a,b) I think it has to do with the 0 at the end of 'a' array

– lezebulon
Nov 22 '18 at 10:27

@lezebulon Rather pathological example and also easy to address. Just filter out the 0 positions. For example, add the line steps = steps[steps.astype(bool)] after the first line of the function body.

– Paul Panzer
Nov 22 '18 at 10:47

add a comment |

Thanks!However it's failing for some combinations: a = [4, 0, 1,0] b = [35, 72, 5, 72] cycles(a,b) I think it has to do with the 0 at the end of 'a' array

– lezebulon
Nov 22 '18 at 10:27

@lezebulon Rather pathological example and also easy to address. Just filter out the 0 positions. For example, add the line steps = steps[steps.astype(bool)] after the first line of the function body.

– Paul Panzer
Nov 22 '18 at 10:47

Thanks!However it's failing for some combinations: a = [4, 0, 1,0] b = [35, 72, 5, 72] cycles(a,b) I think it has to do with the 0 at the end of 'a' array

– lezebulon
Nov 22 '18 at 10:27

@lezebulon Rather pathological example and also easy to address. Just filter out the 0 positions. For example, add the line steps = steps[steps.astype(bool)] after the first line of the function body.

– Paul Panzer
Nov 22 '18 at 10:47

add a comment |

I am not entirely sure if this is what you want; here each tuple in the call to func() contains first the range and then the repeat.

import numpy





def func(tups):

    Arr = numpy.empty(numpy.sum([ele[0] * ele[1] for ele in tups]), dtype=int)

    i = 0

    for ele in tups:

        Arr[i:i + ele[0] * ele[1]] = numpy.tile(numpy.arange(ele[0]), ele[1])

        i += ele[0] * ele[1]

    return Arr





arr = func([(2, 3), (3, 4)])

print(arr)

# [0 1 0 1 0 1 0 1 2 0 1 2 0 1 2 0 1 2]

edited Nov 22 '18 at 1:38

answered Nov 22 '18 at 0:15

Patol75

6236

add a comment |

I am not entirely sure if this is what you want; here each tuple in the call to func() contains first the range and then the repeat.

import numpy





def func(tups):

    Arr = numpy.empty(numpy.sum([ele[0] * ele[1] for ele in tups]), dtype=int)

    i = 0

    for ele in tups:

        Arr[i:i + ele[0] * ele[1]] = numpy.tile(numpy.arange(ele[0]), ele[1])

        i += ele[0] * ele[1]

    return Arr





arr = func([(2, 3), (3, 4)])

print(arr)

# [0 1 0 1 0 1 0 1 2 0 1 2 0 1 2 0 1 2]

edited Nov 22 '18 at 1:38

answered Nov 22 '18 at 0:15

Patol75

6236

add a comment |

I am not entirely sure if this is what you want; here each tuple in the call to func() contains first the range and then the repeat.

import numpy





def func(tups):

    Arr = numpy.empty(numpy.sum([ele[0] * ele[1] for ele in tups]), dtype=int)

    i = 0

    for ele in tups:

        Arr[i:i + ele[0] * ele[1]] = numpy.tile(numpy.arange(ele[0]), ele[1])

        i += ele[0] * ele[1]

    return Arr





arr = func([(2, 3), (3, 4)])

print(arr)

# [0 1 0 1 0 1 0 1 2 0 1 2 0 1 2 0 1 2]

edited Nov 22 '18 at 1:38

answered Nov 22 '18 at 0:15

Patol75

6236

I am not entirely sure if this is what you want; here each tuple in the call to func() contains first the range and then the repeat.

import numpy





def func(tups):

    Arr = numpy.empty(numpy.sum([ele[0] * ele[1] for ele in tups]), dtype=int)

    i = 0

    for ele in tups:

        Arr[i:i + ele[0] * ele[1]] = numpy.tile(numpy.arange(ele[0]), ele[1])

        i += ele[0] * ele[1]

    return Arr





arr = func([(2, 3), (3, 4)])

print(arr)

# [0 1 0 1 0 1 0 1 2 0 1 2 0 1 2 0 1 2]

edited Nov 22 '18 at 1:38

answered Nov 22 '18 at 0:15

Patol75

6236

edited Nov 22 '18 at 1:38

answered Nov 22 '18 at 0:15

Patol75

6236

answered Nov 22 '18 at 0:15

Patol75

6236

answered Nov 22 '18 at 0:15

Patol75

6236

add a comment |

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