Cython code for adaptive binning
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3
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Here is my Cython code used for adaptive binning. The calcAdaptiveBinnedRateMap
function is called from another Python script. The script is compiled using Cython but the speed I am expecting is still not great. How can I improve the speed of execution?
import numpy as np
import scipy.ndimage.morphology as ndimmor
#define sampline rate and alpha
cdef float samplingRate = 30.0
cdef double alpha = 0.0001 #skaggs and sachin use this value while Jim uses 0.001
"""
runs the iteration for adaptive binning till the criteria mentioned in Skaggs et al 1996 is met
INPUT: spike map, occupancy map, alpha, Number of occupancy (Nocc), Euclidean distance transform (dists)
OUTPUT: Nspikes (number of spikes), Nocc (occupancy count)
"""
def mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists):
#initialize the variable
cdef int Nspikes2 = 1
cdef double rsq = 0
cdef int EnoughPoints = 0
cdef int Nspikes = 0
#find the row and column count for occupancy map
cdef int rowLen = int(np.shape(occMap)[0])
cdef int colLen = int(np.shape(occMap)[1])
#while the radius is less than 200 and enough points are not covered
while rsq<200.00 and EnoughPoints==0:
r = np.sqrt(rsq)
#need to set these Nocc and Nspikes to zero, otherwise the two for loops above add the same spikes over again.
Nspikes = 0
Nocc = 0
for i in range(rowLen):
for j in range(colLen):
#if the distance is less than radius
if dists[i,j]<=r:
#add Nspikes
Nspikes = Nspikes + spikeMap[i,j]
#add occupancy
Nocc = Nocc +occMap[i,j]
#if number spikes is greater than 0, then set nspikes2 = number of spikes
if Nspikes > 0:
Nspikes2 = Nspikes
#check for the condition from skaggs et al 1996
if (alpha*alpha*Nocc*Nocc*rsq*Nspikes2 > 1):
EnoughPoints = 1 #set the flag to end the loop
#keep increasing the radius
rsq = rsq + 1
#output occupancy = occupancy
Noccout = Nocc
return Nspikes, Noccout
"""calculate adaptive binned rate map
INPUT: spikemap, occupancy map (both of them just binned for 2cm/4cm no other operation applied on them)
NOTE: occupancy is still in terms of number of frames
OUTPUT: adaptive binned rate Map which is color adjusted
"""
def calcAdaptiveBinnedRateMap(spikeMap, occMap):
#set unoccupied occupancy and spike corresponding to unoccupied position = 0
spikeMap[0,0] = 0
occMap[0,0] = 0
#find the row, col index of minimum occupied and maximum occupied pixel
row, col = np.where(occMap)
minrow = np.min(row)
maxrow = np.max(row)
mincol = np.min(col)
maxcol = np.max(col)
#select the best fitting rectangle according to occupied area
occMap = occMap[minrow:maxrow+1,mincol:maxcol+1]
spikeMap = spikeMap[minrow:maxrow+1,mincol:maxcol+1]
#matrix of zeros same size as of occupancy map
z = np.zeros(np.shape(occMap))
#variable to hold adaptive binned rate map value
abrMap = np.copy(z)
#variale to hold adaptive binned occupancy map
abrOcc = np.copy(z)
#check to endure if number of spikes is greater than 1
if np.max(np.max(spikeMap))>0:
#iterate over the values
for x in range(int(np.shape(occMap)[1])):
for y in range(int(np.shape(occMap)[0])):
if occMap[y,x] > 0:
#pretend there's atleas 1 spike, and 1 occ.needed to avoid 0 threshold.
Nspikes2 = 1
Nocc = occMap[y,x]
d = np.copy(z)
d[y,x] = 1
#computes the Euclidean distance transform of the input matrix.
#For each pixel in BW, the distance transform assigns a number that is the
#distance between that pixel and the nearest nonzero pixel of BW.
dists = ndimmor.distance_transform_edt(d==0)
# function to keep on iterating while the condition mentioned in Skaggs et al 1996 is met
Nspikes, Nocc = mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists)
if Nocc < 12: #occupancy cutoff = 0.4seconds
#if less than 0.4 seconds set it to 0
abrMap[y,x] = 0
abrOcc[y,x] = 0
else:
#else equal to number of spikes/occupancy map
abrMap[y,x] = samplingRate*float(Nspikes)/float(Nocc)
#adaptive binned ocuupancy map = nocc
abrOcc[y,x] = Nocc
#find the maximum value of adaptive binned rate map
cmax = np.max(np.max(abrMap))
if cmax > 0:
#minimum = maximum value found above/60
cmin = -(cmax/60.0);
else:
cmin = -1;
#set adaptive binned rate map = cmin wherever adaptive binned occupancy map = 0
abrMap[abrOcc==0] = cmin
#return the adaptive binned rate map
return abrMap
python performance signal-processing cython scipy
add a comment |
up vote
3
down vote
favorite
Here is my Cython code used for adaptive binning. The calcAdaptiveBinnedRateMap
function is called from another Python script. The script is compiled using Cython but the speed I am expecting is still not great. How can I improve the speed of execution?
import numpy as np
import scipy.ndimage.morphology as ndimmor
#define sampline rate and alpha
cdef float samplingRate = 30.0
cdef double alpha = 0.0001 #skaggs and sachin use this value while Jim uses 0.001
"""
runs the iteration for adaptive binning till the criteria mentioned in Skaggs et al 1996 is met
INPUT: spike map, occupancy map, alpha, Number of occupancy (Nocc), Euclidean distance transform (dists)
OUTPUT: Nspikes (number of spikes), Nocc (occupancy count)
"""
def mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists):
#initialize the variable
cdef int Nspikes2 = 1
cdef double rsq = 0
cdef int EnoughPoints = 0
cdef int Nspikes = 0
#find the row and column count for occupancy map
cdef int rowLen = int(np.shape(occMap)[0])
cdef int colLen = int(np.shape(occMap)[1])
#while the radius is less than 200 and enough points are not covered
while rsq<200.00 and EnoughPoints==0:
r = np.sqrt(rsq)
#need to set these Nocc and Nspikes to zero, otherwise the two for loops above add the same spikes over again.
Nspikes = 0
Nocc = 0
for i in range(rowLen):
for j in range(colLen):
#if the distance is less than radius
if dists[i,j]<=r:
#add Nspikes
Nspikes = Nspikes + spikeMap[i,j]
#add occupancy
Nocc = Nocc +occMap[i,j]
#if number spikes is greater than 0, then set nspikes2 = number of spikes
if Nspikes > 0:
Nspikes2 = Nspikes
#check for the condition from skaggs et al 1996
if (alpha*alpha*Nocc*Nocc*rsq*Nspikes2 > 1):
EnoughPoints = 1 #set the flag to end the loop
#keep increasing the radius
rsq = rsq + 1
#output occupancy = occupancy
Noccout = Nocc
return Nspikes, Noccout
"""calculate adaptive binned rate map
INPUT: spikemap, occupancy map (both of them just binned for 2cm/4cm no other operation applied on them)
NOTE: occupancy is still in terms of number of frames
OUTPUT: adaptive binned rate Map which is color adjusted
"""
def calcAdaptiveBinnedRateMap(spikeMap, occMap):
#set unoccupied occupancy and spike corresponding to unoccupied position = 0
spikeMap[0,0] = 0
occMap[0,0] = 0
#find the row, col index of minimum occupied and maximum occupied pixel
row, col = np.where(occMap)
minrow = np.min(row)
maxrow = np.max(row)
mincol = np.min(col)
maxcol = np.max(col)
#select the best fitting rectangle according to occupied area
occMap = occMap[minrow:maxrow+1,mincol:maxcol+1]
spikeMap = spikeMap[minrow:maxrow+1,mincol:maxcol+1]
#matrix of zeros same size as of occupancy map
z = np.zeros(np.shape(occMap))
#variable to hold adaptive binned rate map value
abrMap = np.copy(z)
#variale to hold adaptive binned occupancy map
abrOcc = np.copy(z)
#check to endure if number of spikes is greater than 1
if np.max(np.max(spikeMap))>0:
#iterate over the values
for x in range(int(np.shape(occMap)[1])):
for y in range(int(np.shape(occMap)[0])):
if occMap[y,x] > 0:
#pretend there's atleas 1 spike, and 1 occ.needed to avoid 0 threshold.
Nspikes2 = 1
Nocc = occMap[y,x]
d = np.copy(z)
d[y,x] = 1
#computes the Euclidean distance transform of the input matrix.
#For each pixel in BW, the distance transform assigns a number that is the
#distance between that pixel and the nearest nonzero pixel of BW.
dists = ndimmor.distance_transform_edt(d==0)
# function to keep on iterating while the condition mentioned in Skaggs et al 1996 is met
Nspikes, Nocc = mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists)
if Nocc < 12: #occupancy cutoff = 0.4seconds
#if less than 0.4 seconds set it to 0
abrMap[y,x] = 0
abrOcc[y,x] = 0
else:
#else equal to number of spikes/occupancy map
abrMap[y,x] = samplingRate*float(Nspikes)/float(Nocc)
#adaptive binned ocuupancy map = nocc
abrOcc[y,x] = Nocc
#find the maximum value of adaptive binned rate map
cmax = np.max(np.max(abrMap))
if cmax > 0:
#minimum = maximum value found above/60
cmin = -(cmax/60.0);
else:
cmin = -1;
#set adaptive binned rate map = cmin wherever adaptive binned occupancy map = 0
abrMap[abrOcc==0] = cmin
#return the adaptive binned rate map
return abrMap
python performance signal-processing cython scipy
2
the speed I am expecting is still not great, How can I improve the speed of execution? - What does this exactly mean? How many days does this function need to complete and to how many days would be optimal for it to run?
– t3chb0t
Feb 19 '17 at 8:12
1
Please update the title so it describes what the code is doing.
– t3chb0t
Feb 19 '17 at 8:14
2
Crosspost of stackoverflow.com/q/42324609
– user7138814
Feb 19 '17 at 12:17
Did you profile your code?
– Mast
Feb 19 '17 at 19:50
I am still a learner, will do the profiling asap and get back
– raj
Feb 20 '17 at 7:16
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Here is my Cython code used for adaptive binning. The calcAdaptiveBinnedRateMap
function is called from another Python script. The script is compiled using Cython but the speed I am expecting is still not great. How can I improve the speed of execution?
import numpy as np
import scipy.ndimage.morphology as ndimmor
#define sampline rate and alpha
cdef float samplingRate = 30.0
cdef double alpha = 0.0001 #skaggs and sachin use this value while Jim uses 0.001
"""
runs the iteration for adaptive binning till the criteria mentioned in Skaggs et al 1996 is met
INPUT: spike map, occupancy map, alpha, Number of occupancy (Nocc), Euclidean distance transform (dists)
OUTPUT: Nspikes (number of spikes), Nocc (occupancy count)
"""
def mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists):
#initialize the variable
cdef int Nspikes2 = 1
cdef double rsq = 0
cdef int EnoughPoints = 0
cdef int Nspikes = 0
#find the row and column count for occupancy map
cdef int rowLen = int(np.shape(occMap)[0])
cdef int colLen = int(np.shape(occMap)[1])
#while the radius is less than 200 and enough points are not covered
while rsq<200.00 and EnoughPoints==0:
r = np.sqrt(rsq)
#need to set these Nocc and Nspikes to zero, otherwise the two for loops above add the same spikes over again.
Nspikes = 0
Nocc = 0
for i in range(rowLen):
for j in range(colLen):
#if the distance is less than radius
if dists[i,j]<=r:
#add Nspikes
Nspikes = Nspikes + spikeMap[i,j]
#add occupancy
Nocc = Nocc +occMap[i,j]
#if number spikes is greater than 0, then set nspikes2 = number of spikes
if Nspikes > 0:
Nspikes2 = Nspikes
#check for the condition from skaggs et al 1996
if (alpha*alpha*Nocc*Nocc*rsq*Nspikes2 > 1):
EnoughPoints = 1 #set the flag to end the loop
#keep increasing the radius
rsq = rsq + 1
#output occupancy = occupancy
Noccout = Nocc
return Nspikes, Noccout
"""calculate adaptive binned rate map
INPUT: spikemap, occupancy map (both of them just binned for 2cm/4cm no other operation applied on them)
NOTE: occupancy is still in terms of number of frames
OUTPUT: adaptive binned rate Map which is color adjusted
"""
def calcAdaptiveBinnedRateMap(spikeMap, occMap):
#set unoccupied occupancy and spike corresponding to unoccupied position = 0
spikeMap[0,0] = 0
occMap[0,0] = 0
#find the row, col index of minimum occupied and maximum occupied pixel
row, col = np.where(occMap)
minrow = np.min(row)
maxrow = np.max(row)
mincol = np.min(col)
maxcol = np.max(col)
#select the best fitting rectangle according to occupied area
occMap = occMap[minrow:maxrow+1,mincol:maxcol+1]
spikeMap = spikeMap[minrow:maxrow+1,mincol:maxcol+1]
#matrix of zeros same size as of occupancy map
z = np.zeros(np.shape(occMap))
#variable to hold adaptive binned rate map value
abrMap = np.copy(z)
#variale to hold adaptive binned occupancy map
abrOcc = np.copy(z)
#check to endure if number of spikes is greater than 1
if np.max(np.max(spikeMap))>0:
#iterate over the values
for x in range(int(np.shape(occMap)[1])):
for y in range(int(np.shape(occMap)[0])):
if occMap[y,x] > 0:
#pretend there's atleas 1 spike, and 1 occ.needed to avoid 0 threshold.
Nspikes2 = 1
Nocc = occMap[y,x]
d = np.copy(z)
d[y,x] = 1
#computes the Euclidean distance transform of the input matrix.
#For each pixel in BW, the distance transform assigns a number that is the
#distance between that pixel and the nearest nonzero pixel of BW.
dists = ndimmor.distance_transform_edt(d==0)
# function to keep on iterating while the condition mentioned in Skaggs et al 1996 is met
Nspikes, Nocc = mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists)
if Nocc < 12: #occupancy cutoff = 0.4seconds
#if less than 0.4 seconds set it to 0
abrMap[y,x] = 0
abrOcc[y,x] = 0
else:
#else equal to number of spikes/occupancy map
abrMap[y,x] = samplingRate*float(Nspikes)/float(Nocc)
#adaptive binned ocuupancy map = nocc
abrOcc[y,x] = Nocc
#find the maximum value of adaptive binned rate map
cmax = np.max(np.max(abrMap))
if cmax > 0:
#minimum = maximum value found above/60
cmin = -(cmax/60.0);
else:
cmin = -1;
#set adaptive binned rate map = cmin wherever adaptive binned occupancy map = 0
abrMap[abrOcc==0] = cmin
#return the adaptive binned rate map
return abrMap
python performance signal-processing cython scipy
Here is my Cython code used for adaptive binning. The calcAdaptiveBinnedRateMap
function is called from another Python script. The script is compiled using Cython but the speed I am expecting is still not great. How can I improve the speed of execution?
import numpy as np
import scipy.ndimage.morphology as ndimmor
#define sampline rate and alpha
cdef float samplingRate = 30.0
cdef double alpha = 0.0001 #skaggs and sachin use this value while Jim uses 0.001
"""
runs the iteration for adaptive binning till the criteria mentioned in Skaggs et al 1996 is met
INPUT: spike map, occupancy map, alpha, Number of occupancy (Nocc), Euclidean distance transform (dists)
OUTPUT: Nspikes (number of spikes), Nocc (occupancy count)
"""
def mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists):
#initialize the variable
cdef int Nspikes2 = 1
cdef double rsq = 0
cdef int EnoughPoints = 0
cdef int Nspikes = 0
#find the row and column count for occupancy map
cdef int rowLen = int(np.shape(occMap)[0])
cdef int colLen = int(np.shape(occMap)[1])
#while the radius is less than 200 and enough points are not covered
while rsq<200.00 and EnoughPoints==0:
r = np.sqrt(rsq)
#need to set these Nocc and Nspikes to zero, otherwise the two for loops above add the same spikes over again.
Nspikes = 0
Nocc = 0
for i in range(rowLen):
for j in range(colLen):
#if the distance is less than radius
if dists[i,j]<=r:
#add Nspikes
Nspikes = Nspikes + spikeMap[i,j]
#add occupancy
Nocc = Nocc +occMap[i,j]
#if number spikes is greater than 0, then set nspikes2 = number of spikes
if Nspikes > 0:
Nspikes2 = Nspikes
#check for the condition from skaggs et al 1996
if (alpha*alpha*Nocc*Nocc*rsq*Nspikes2 > 1):
EnoughPoints = 1 #set the flag to end the loop
#keep increasing the radius
rsq = rsq + 1
#output occupancy = occupancy
Noccout = Nocc
return Nspikes, Noccout
"""calculate adaptive binned rate map
INPUT: spikemap, occupancy map (both of them just binned for 2cm/4cm no other operation applied on them)
NOTE: occupancy is still in terms of number of frames
OUTPUT: adaptive binned rate Map which is color adjusted
"""
def calcAdaptiveBinnedRateMap(spikeMap, occMap):
#set unoccupied occupancy and spike corresponding to unoccupied position = 0
spikeMap[0,0] = 0
occMap[0,0] = 0
#find the row, col index of minimum occupied and maximum occupied pixel
row, col = np.where(occMap)
minrow = np.min(row)
maxrow = np.max(row)
mincol = np.min(col)
maxcol = np.max(col)
#select the best fitting rectangle according to occupied area
occMap = occMap[minrow:maxrow+1,mincol:maxcol+1]
spikeMap = spikeMap[minrow:maxrow+1,mincol:maxcol+1]
#matrix of zeros same size as of occupancy map
z = np.zeros(np.shape(occMap))
#variable to hold adaptive binned rate map value
abrMap = np.copy(z)
#variale to hold adaptive binned occupancy map
abrOcc = np.copy(z)
#check to endure if number of spikes is greater than 1
if np.max(np.max(spikeMap))>0:
#iterate over the values
for x in range(int(np.shape(occMap)[1])):
for y in range(int(np.shape(occMap)[0])):
if occMap[y,x] > 0:
#pretend there's atleas 1 spike, and 1 occ.needed to avoid 0 threshold.
Nspikes2 = 1
Nocc = occMap[y,x]
d = np.copy(z)
d[y,x] = 1
#computes the Euclidean distance transform of the input matrix.
#For each pixel in BW, the distance transform assigns a number that is the
#distance between that pixel and the nearest nonzero pixel of BW.
dists = ndimmor.distance_transform_edt(d==0)
# function to keep on iterating while the condition mentioned in Skaggs et al 1996 is met
Nspikes, Nocc = mexAdaptwhile(spikeMap, occMap, alpha, Nocc, dists)
if Nocc < 12: #occupancy cutoff = 0.4seconds
#if less than 0.4 seconds set it to 0
abrMap[y,x] = 0
abrOcc[y,x] = 0
else:
#else equal to number of spikes/occupancy map
abrMap[y,x] = samplingRate*float(Nspikes)/float(Nocc)
#adaptive binned ocuupancy map = nocc
abrOcc[y,x] = Nocc
#find the maximum value of adaptive binned rate map
cmax = np.max(np.max(abrMap))
if cmax > 0:
#minimum = maximum value found above/60
cmin = -(cmax/60.0);
else:
cmin = -1;
#set adaptive binned rate map = cmin wherever adaptive binned occupancy map = 0
abrMap[abrOcc==0] = cmin
#return the adaptive binned rate map
return abrMap
python performance signal-processing cython scipy
python performance signal-processing cython scipy
edited Feb 19 '17 at 21:30
Jamal♦
30.2k11115226
30.2k11115226
asked Feb 19 '17 at 7:54
raj
161
161
2
the speed I am expecting is still not great, How can I improve the speed of execution? - What does this exactly mean? How many days does this function need to complete and to how many days would be optimal for it to run?
– t3chb0t
Feb 19 '17 at 8:12
1
Please update the title so it describes what the code is doing.
– t3chb0t
Feb 19 '17 at 8:14
2
Crosspost of stackoverflow.com/q/42324609
– user7138814
Feb 19 '17 at 12:17
Did you profile your code?
– Mast
Feb 19 '17 at 19:50
I am still a learner, will do the profiling asap and get back
– raj
Feb 20 '17 at 7:16
add a comment |
2
the speed I am expecting is still not great, How can I improve the speed of execution? - What does this exactly mean? How many days does this function need to complete and to how many days would be optimal for it to run?
– t3chb0t
Feb 19 '17 at 8:12
1
Please update the title so it describes what the code is doing.
– t3chb0t
Feb 19 '17 at 8:14
2
Crosspost of stackoverflow.com/q/42324609
– user7138814
Feb 19 '17 at 12:17
Did you profile your code?
– Mast
Feb 19 '17 at 19:50
I am still a learner, will do the profiling asap and get back
– raj
Feb 20 '17 at 7:16
2
2
the speed I am expecting is still not great, How can I improve the speed of execution? - What does this exactly mean? How many days does this function need to complete and to how many days would be optimal for it to run?
– t3chb0t
Feb 19 '17 at 8:12
the speed I am expecting is still not great, How can I improve the speed of execution? - What does this exactly mean? How many days does this function need to complete and to how many days would be optimal for it to run?
– t3chb0t
Feb 19 '17 at 8:12
1
1
Please update the title so it describes what the code is doing.
– t3chb0t
Feb 19 '17 at 8:14
Please update the title so it describes what the code is doing.
– t3chb0t
Feb 19 '17 at 8:14
2
2
Crosspost of stackoverflow.com/q/42324609
– user7138814
Feb 19 '17 at 12:17
Crosspost of stackoverflow.com/q/42324609
– user7138814
Feb 19 '17 at 12:17
Did you profile your code?
– Mast
Feb 19 '17 at 19:50
Did you profile your code?
– Mast
Feb 19 '17 at 19:50
I am still a learner, will do the profiling asap and get back
– raj
Feb 20 '17 at 7:16
I am still a learner, will do the profiling asap and get back
– raj
Feb 20 '17 at 7:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Make sure to declare all variables with Cython (otherwise the lines with such variables basically run with Python-speed). Especially important for the "for"-loop variables (i.e. i
, j
, x
, y
).
One can also use cython -a mycode.pyx
to create an annotated HTML page that shows, which lines are running with C or Python speed, respectively.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Make sure to declare all variables with Cython (otherwise the lines with such variables basically run with Python-speed). Especially important for the "for"-loop variables (i.e. i
, j
, x
, y
).
One can also use cython -a mycode.pyx
to create an annotated HTML page that shows, which lines are running with C or Python speed, respectively.
add a comment |
up vote
1
down vote
Make sure to declare all variables with Cython (otherwise the lines with such variables basically run with Python-speed). Especially important for the "for"-loop variables (i.e. i
, j
, x
, y
).
One can also use cython -a mycode.pyx
to create an annotated HTML page that shows, which lines are running with C or Python speed, respectively.
add a comment |
up vote
1
down vote
up vote
1
down vote
Make sure to declare all variables with Cython (otherwise the lines with such variables basically run with Python-speed). Especially important for the "for"-loop variables (i.e. i
, j
, x
, y
).
One can also use cython -a mycode.pyx
to create an annotated HTML page that shows, which lines are running with C or Python speed, respectively.
Make sure to declare all variables with Cython (otherwise the lines with such variables basically run with Python-speed). Especially important for the "for"-loop variables (i.e. i
, j
, x
, y
).
One can also use cython -a mycode.pyx
to create an annotated HTML page that shows, which lines are running with C or Python speed, respectively.
edited Nov 19 at 0:45
Sᴀᴍ Onᴇᴌᴀ
7,74961748
7,74961748
answered Nov 18 at 22:29
Benjamin Winkel
112
112
add a comment |
add a comment |
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2
the speed I am expecting is still not great, How can I improve the speed of execution? - What does this exactly mean? How many days does this function need to complete and to how many days would be optimal for it to run?
– t3chb0t
Feb 19 '17 at 8:12
1
Please update the title so it describes what the code is doing.
– t3chb0t
Feb 19 '17 at 8:14
2
Crosspost of stackoverflow.com/q/42324609
– user7138814
Feb 19 '17 at 12:17
Did you profile your code?
– Mast
Feb 19 '17 at 19:50
I am still a learner, will do the profiling asap and get back
– raj
Feb 20 '17 at 7:16