'Piling' an Array, Collapse by Summation












1















Given a numpy array, I'd like to sum up uniform chunks of elements to form a new, smaller, array. It's similar to binning, but not by frequency. I'm not sure how else to describe it other than by example (below).



The question: Is there either a function for this or cleaner approach (using numpy/scipy)? I've looked into digitize and histogram, but think their implementations are lengthy. I've also thought about crafty indexing, but it's beyond me and might make a long ugly line of code.



import numpy as np  



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3

new_data = np.zeros(int(np.size(old_data) / bin_size))

for ind, val in enumerate(new_data):

    leap = ind*bin_size

    new_data[ind] = 

np.sum(old_data[leap:leap+bin_size])

print(old_data, '->', bin_size, ':', new_data)



# [0 1 2 3 4 5 6 7 8] -> 3 : [ 3. 12. 21.]





python numpy scipy sum bin







share|improve this question























  • maybe np.array([sum(old_data[bin_size*i: bin_size*(i+1)]) for i in range(old_data.size // bin_size)])

    – William Lee
    Nov 22 '18 at 2:58
















1















Given a numpy array, I'd like to sum up uniform chunks of elements to form a new, smaller, array. It's similar to binning, but not by frequency. I'm not sure how else to describe it other than by example (below).



The question: Is there either a function for this or cleaner approach (using numpy/scipy)? I've looked into digitize and histogram, but think their implementations are lengthy. I've also thought about crafty indexing, but it's beyond me and might make a long ugly line of code.



import numpy as np  



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3

new_data = np.zeros(int(np.size(old_data) / bin_size))

for ind, val in enumerate(new_data):

    leap = ind*bin_size

    new_data[ind] = 

np.sum(old_data[leap:leap+bin_size])

print(old_data, '->', bin_size, ':', new_data)



# [0 1 2 3 4 5 6 7 8] -> 3 : [ 3. 12. 21.]





python numpy scipy sum bin







share|improve this question























  • maybe np.array([sum(old_data[bin_size*i: bin_size*(i+1)]) for i in range(old_data.size // bin_size)])

    – William Lee
    Nov 22 '18 at 2:58














1












1








1








Given a numpy array, I'd like to sum up uniform chunks of elements to form a new, smaller, array. It's similar to binning, but not by frequency. I'm not sure how else to describe it other than by example (below).



The question: Is there either a function for this or cleaner approach (using numpy/scipy)? I've looked into digitize and histogram, but think their implementations are lengthy. I've also thought about crafty indexing, but it's beyond me and might make a long ugly line of code.



import numpy as np  



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3

new_data = np.zeros(int(np.size(old_data) / bin_size))

for ind, val in enumerate(new_data):

    leap = ind*bin_size

    new_data[ind] = 

np.sum(old_data[leap:leap+bin_size])

print(old_data, '->', bin_size, ':', new_data)



# [0 1 2 3 4 5 6 7 8] -> 3 : [ 3. 12. 21.]





python numpy scipy sum bin







share|improve this question














Given a numpy array, I'd like to sum up uniform chunks of elements to form a new, smaller, array. It's similar to binning, but not by frequency. I'm not sure how else to describe it other than by example (below).



The question: Is there either a function for this or cleaner approach (using numpy/scipy)? I've looked into digitize and histogram, but think their implementations are lengthy. I've also thought about crafty indexing, but it's beyond me and might make a long ugly line of code.



import numpy as np  



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3

new_data = np.zeros(int(np.size(old_data) / bin_size))

for ind, val in enumerate(new_data):

    leap = ind*bin_size

    new_data[ind] = 

np.sum(old_data[leap:leap+bin_size])

print(old_data, '->', bin_size, ':', new_data)



# [0 1 2 3 4 5 6 7 8] -> 3 : [ 3. 12. 21.]





python numpy scipy sum bin




python numpy scipy sum bin






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 22 '18 at 2:52









Captain MorganCaptain Morgan

1406




1406













  • maybe np.array([sum(old_data[bin_size*i: bin_size*(i+1)]) for i in range(old_data.size // bin_size)])

    – William Lee
    Nov 22 '18 at 2:58



















  • maybe np.array([sum(old_data[bin_size*i: bin_size*(i+1)]) for i in range(old_data.size // bin_size)])

    – William Lee
    Nov 22 '18 at 2:58

















maybe np.array([sum(old_data[bin_size*i: bin_size*(i+1)]) for i in range(old_data.size // bin_size)])

– William Lee
Nov 22 '18 at 2:58





maybe np.array([sum(old_data[bin_size*i: bin_size*(i+1)]) for i in range(old_data.size // bin_size)])

– William Lee
Nov 22 '18 at 2:58












1 Answer
1






active

oldest

votes


















1














Assuming there's an integral number of bins, you can accomplish this with a reshape:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3



new_data = old_data.reshape(-1, bin_size).sum(axis=1)


new_data will then have the desired value of:



array([ 3, 12, 21])


If bin_size doesn't divide evenly into old_data.size, you can instead use resize:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

bin_size = 3



old_data.resize(old_data.size//bin_size + 1, bin_size)

new_data = old_data.sum(axis=1)


new_data will then have a value of:



array([ 3, 12, 21, 19])


Using resize has the downside of modifying old_data in place, so if you want to keep old_data around you should probably make a copy of it before you do the resize.






share|improve this answer


























  • This is concise and does exactly as desired in a very legible manner. Thank you!

    – Captain Morgan
    Nov 22 '18 at 3:16











  • Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

    – Captain Morgan
    Nov 22 '18 at 3:22











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














Assuming there's an integral number of bins, you can accomplish this with a reshape:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3



new_data = old_data.reshape(-1, bin_size).sum(axis=1)


new_data will then have the desired value of:



array([ 3, 12, 21])


If bin_size doesn't divide evenly into old_data.size, you can instead use resize:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

bin_size = 3



old_data.resize(old_data.size//bin_size + 1, bin_size)

new_data = old_data.sum(axis=1)


new_data will then have a value of:



array([ 3, 12, 21, 19])


Using resize has the downside of modifying old_data in place, so if you want to keep old_data around you should probably make a copy of it before you do the resize.






share|improve this answer


























  • This is concise and does exactly as desired in a very legible manner. Thank you!

    – Captain Morgan
    Nov 22 '18 at 3:16











  • Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

    – Captain Morgan
    Nov 22 '18 at 3:22
















1














Assuming there's an integral number of bins, you can accomplish this with a reshape:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3



new_data = old_data.reshape(-1, bin_size).sum(axis=1)


new_data will then have the desired value of:



array([ 3, 12, 21])


If bin_size doesn't divide evenly into old_data.size, you can instead use resize:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

bin_size = 3



old_data.resize(old_data.size//bin_size + 1, bin_size)

new_data = old_data.sum(axis=1)


new_data will then have a value of:



array([ 3, 12, 21, 19])


Using resize has the downside of modifying old_data in place, so if you want to keep old_data around you should probably make a copy of it before you do the resize.






share|improve this answer


























  • This is concise and does exactly as desired in a very legible manner. Thank you!

    – Captain Morgan
    Nov 22 '18 at 3:16











  • Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

    – Captain Morgan
    Nov 22 '18 at 3:22














1












1








1







Assuming there's an integral number of bins, you can accomplish this with a reshape:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3



new_data = old_data.reshape(-1, bin_size).sum(axis=1)


new_data will then have the desired value of:



array([ 3, 12, 21])


If bin_size doesn't divide evenly into old_data.size, you can instead use resize:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

bin_size = 3



old_data.resize(old_data.size//bin_size + 1, bin_size)

new_data = old_data.sum(axis=1)


new_data will then have a value of:



array([ 3, 12, 21, 19])


Using resize has the downside of modifying old_data in place, so if you want to keep old_data around you should probably make a copy of it before you do the resize.






share|improve this answer















Assuming there's an integral number of bins, you can accomplish this with a reshape:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8])

bin_size = 3



new_data = old_data.reshape(-1, bin_size).sum(axis=1)


new_data will then have the desired value of:



array([ 3, 12, 21])


If bin_size doesn't divide evenly into old_data.size, you can instead use resize:



old_data = np.array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10])

bin_size = 3



old_data.resize(old_data.size//bin_size + 1, bin_size)

new_data = old_data.sum(axis=1)


new_data will then have a value of:



array([ 3, 12, 21, 19])


Using resize has the downside of modifying old_data in place, so if you want to keep old_data around you should probably make a copy of it before you do the resize.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 22 '18 at 3:13

























answered Nov 22 '18 at 3:01









teltel

7,41621431




7,41621431













  • This is concise and does exactly as desired in a very legible manner. Thank you!

    – Captain Morgan
    Nov 22 '18 at 3:16











  • Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

    – Captain Morgan
    Nov 22 '18 at 3:22



















  • This is concise and does exactly as desired in a very legible manner. Thank you!

    – Captain Morgan
    Nov 22 '18 at 3:16











  • Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

    – Captain Morgan
    Nov 22 '18 at 3:22

















This is concise and does exactly as desired in a very legible manner. Thank you!

– Captain Morgan
Nov 22 '18 at 3:16





This is concise and does exactly as desired in a very legible manner. Thank you!

– Captain Morgan
Nov 22 '18 at 3:16













Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

– Captain Morgan
Nov 22 '18 at 3:22





Is there a clean way to do this without altering the original array? One could simply np.copy it into a dummy array, but surely there's a slick way around this.

– Captain Morgan
Nov 22 '18 at 3:22




















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