Creating number masks from number range - python

I have a range like this:
1323000-1555999
It's is necessary to create "masks" covering entire range. So the "masks" list for range above should look like this:
1323***
1324***
1325***
...
14*****
...
153****
154****
1550***
1551***
And so on.
Anyone have ideas about how to solve this problem using Python?
The idea is to cover all range using minimum amount of masks. So in case of 1000-1999, algo should output 1*** but not 101*,102*... or 10**,11**....

With a bit of looping:
Code:
def wild_card_range(start, end):
while start <= end:
shift = 0
multiple = 1
done = False
over = False
still_fits = True
while not (done or over) and still_fits:
multiple *= 10
shift += 1
next_value = int(start / multiple) * multiple + multiple
done = next_value == end + 1
over = next_value > end + 1
still_fits = int(start / multiple) == \
int((start + multiple - 1) / multiple)
if over or not still_fits:
multiple = int(multiple / 10)
shift -= 1
yield str(int(start / multiple)) + '*' * shift
start += multiple
for mask in wild_card_range(1323000, 1555999):
print(mask)
Results:
1323***
1324***
1325***
1326***
1327***
1328***
1329***
133****
134****
135****
136****
137****
138****
139****
14*****
150****
151****
152****
153****
154****
1550***
1551***
1552***
1553***
1554***
1555***

Related

Creating shift patterns

I'm a little lost and would love to help build shift buildings.
The work is carried out in an 8-hour shift and employees are entitled to three 10-minute breaks during their shift, which for convenience can start every ten minutes (i.e .: breaks can start at 08: 00,09: 20, ...)
It is determined that the maximum interval between breaks and breaks or between breaks and the beginning / end of the shift does not exceed w = 160 minutes) it can be assumed that W receives a value that is double 10 minutes.
I need to build all the possible shift structures.
I say this list indicates a shift, every interval 10 minutes
print(list(range(1,49)))
Example of one shift
A1=[2,3,....,15,17,........,32,34,...48]
I.e. break 1 in the first 10 minutes of the shift, break 2 after 150 minutes and last break after 330 minutes.
Thanks
All shift structures can be obtained by performing a Exhaustive Search in three loops(a number of breaks) so that the condition is match.
work = set(range(1, 49))
w = 16
breaktime = []
check = 0
for i in range(1, w + 2):
if i > check:
break1 = i
for j in range(i, i + w + 2):
if j > break1:
break2 = j
for k in range(j, j + w + 2):
if k > break2 and k < 49:
break3 = k
if w + k > 49:
breaktime.append([break1, break2, break3])
check += 1
shift_list = []
for l in breaktime:
shift = list(work - set(l))
print(shift)
shift_list.append(shift)
I'd just be lazy and use itertools functions here since the data is quite small. At larger numbers one might want to be more clever.
The breaks (and shift) results are zero-indexed for programming convenience (as opposed to 1-indexed as in the question), but that's trivial to adjust, and makes it easier to use e.g. datetime for actual times.
import itertools
def compute_breaks(
*,
timeblock_count: int,
n_breaks: int,
min_between: int,
min_start: int = 0,
min_end: int = 0,
):
timeblocks = range(timeblock_count)
for breaks in itertools.combinations(timeblocks, n_breaks):
if any(abs(a - b) < min_between for a, b in itertools.combinations(breaks, 2)):
continue
if any(a < min_start or a > timeblock_count - min_end for a in breaks):
continue
yield set(breaks)
def render_shift(shift, timeblock_count: int):
return "".join("O" if x in shift else "." for x in range(timeblock_count))
def breaks_to_shift(breaks, timeblock_count: int):
return [i for i in range(timeblock_count) if i not in breaks]
for breaks in sorted(compute_breaks(timeblock_count=48, n_breaks=3, min_between=16)):
shift = breaks_to_shift(breaks, timeblock_count=48)
print(render_shift(shift, timeblock_count=48))
The renderings show breaks as .s and work blocks as Os, e.g. (truncated):
.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOO
.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOO.OOOOOOOOOOOOOO
.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOO.OOOOOOOOOOOOO
...
.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOOOOOOOOOOOOOO.O
.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOOOOOOOOOOOOOOO.
...
O.OOOOOOOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOO.OOOOOOO
O.OOOOOOOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOO.OOOOOO
O.OOOOOOOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOOO.OOOOO
...
OOOOO.OOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOO.OOOOOOOO
OOOOO.OOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOO.OOOOOOO
OOOOO.OOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOOOO.OOOOOO
...
OOOOOOOOOOOOO.OOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOO.
OOOOOOOOOOOOO.OOOOOOOOOOOOOOOOO.OOOOOOOOOOOOOOO.
OOOOOOOOOOOOOO.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOO.O
OOOOOOOOOOOOOO.OOOOOOOOOOOOOOO.OOOOOOOOOOOOOOOO.

Optimizing mathematical formula implementation with concatenations of summations

I'm trying to implement this following formula in Python. It's basically a long concatenation os summations, where an additional summation is added each time a new 'element' is needed. To simply explain the formula's structure, here's how this formula goes in order from 2 to 5 elements:
2 elements
3 elements
4 elements
5 elements
By the way, here's the g function shown in the formulas:
g function
Now, I foolishly tried coding this formula with my extremely barebones python programming skills. The initial goal was to try this with 15 elements, but given that it contained a lot of nested for loops and factorials, I quickly noticed that I could not really obtain a result from that.
At the end I ended up with this monstrous code, that would finish just after the heat death of the universe:
from ast import Str
import math
pNuevos = [0,2,2,2,2,1,1,1,2,2,2,1,2,2,1,1]
pTotales = [0,10,10,7,8,7,7,7,7,7,10,7,8,7,8,8]
def PTirada (personajes):
tirada = 0.05/personajes
return tirada
def Ppers1 (personajes, intentos):
p1pers = ((math.factorial(intentos-1)) / ((math.factorial(4))*(math.factorial(intentos-5)))) * (PTirada(personajes)**5) * ((1-PTirada(personajes))**(intentos-5))
return p1pers
def Ppers2 (personajes, intentos):
p2pers = 0
for i in range(10,intentos+1):
p2pers = p2pers + ( (math.factorial(intentos-1)) / ((math.factorial(4))*(math.factorial(i-5))*(math.factorial(intentos-i))) ) * (PTirada(personajes)**i) * ((1 - 2*(PTirada(personajes))) **(intentos-i))
p2pers = 2*p2pers
return p2pers
def Activate (z) :
probability1 = 0
probability2 = 0
probability3 = 0
probability4 = 0
probability5 = 0
probability6 = 0
probability7 = 0
probability8 = 0
probability9 = 0
probability10 = 0
probability11 = 0
probability12 = 0
probability13 = 0
probability14 = 0
for i in range (5*pNuevos[1], z-5*pNuevos[2]+1):
for j in range (5*pNuevos[2], z-i-5*pNuevos[3]+1):
for k in range (5*pNuevos[3], z-j-i-5*pNuevos[4]+1):
for l in range (5*pNuevos[4], z-k-j-i-5*pNuevos[5]+1):
for m in range (5*pNuevos[5], z-l-k-j-i-5*pNuevos[6]+1):
for n in range (5*pNuevos[6], z-m-l-k-j-i-5*pNuevos[7]+1):
for o in range (5*pNuevos[7], z-n-m-l-k-j-i-5*pNuevos[8]+1):
for p in range (5*pNuevos[8], z-o-n-m-l-k-j-i-5*pNuevos[9]+1):
for q in range (5*pNuevos[9], z-p-o-n-m-l-k-j-i-5*pNuevos[10]+1):
for r in range (5*pNuevos[10], z-q-p-o-n-m-l-k-j-i-5*pNuevos[11]+1):
for s in range (5*pNuevos[11], z-r-q-p-o-n-m-l-k-j-i-5*pNuevos[12]+1):
for t in range (5*pNuevos[12], z-s-r-q-p-o-n-m-l-k-j-i-5*pNuevos[13]+1):
for u in range (5*pNuevos[13], z-t-s-r-q-p-o-n-m-l-k-j-i-5*pNuevos[14]+1):
for v in range (5*pNuevos[14], z-u-t-s-r-q-p-o-n-m-l-k-j-i-5*pNuevos[15]+1):
probability14 = probability14 + eval("Ppers"+str(pNuevos[14])+"("+str(pTotales[14])+","+str(v)+")") * eval("Ppers"+str(pNuevos[15])+"("+str(pTotales[15])+","+str(z-v-u-t-s-r-q-p-o-n-m-l-k-j-i)+")")
probability13 = probability13 + eval("Ppers"+str(pNuevos[13])+"("+str(pTotales[13])+","+str(u)+")") * probability14
probability12 = probability12 + eval("Ppers"+str(pNuevos[12])+"("+str(pTotales[12])+","+str(t)+")") * probability13
probability11 = probability11 + eval("Ppers"+str(pNuevos[11])+"("+str(pTotales[11])+","+str(s)+")") * probability12
probability10 = probability10 + eval("Ppers"+str(pNuevos[10])+"("+str(pTotales[10])+","+str(r)+")") * probability11
probability9 = probability9 + eval("Ppers"+str(pNuevos[9])+"("+str(pTotales[9])+","+str(q)+")") * probability10
probability8 = probability8 + eval("Ppers"+str(pNuevos[8])+"("+str(pTotales[8])+","+str(p)+")") * probability9
probability7 = probability7 + eval("Ppers"+str(pNuevos[7])+"("+str(pTotales[7])+","+str(o)+")") * probability8
probability6 = probability6 + eval("Ppers"+str(pNuevos[6])+"("+str(pTotales[6])+","+str(n)+")") * probability7
probability5 = probability5 + eval("Ppers"+str(pNuevos[5])+"("+str(pTotales[5])+","+str(m)+")") * probability6
probability4 = probability4 + eval("Ppers"+str(pNuevos[4])+"("+str(pTotales[4])+","+str(l)+")") * probability5
probability3 += eval("Ppers"+str(pNuevos[3]) + "("+str(pTotales[3])+","+str(k)+")") * probability4
probability2 += eval("Ppers"+str(pNuevos[2]) + "("+str(pTotales[2])+","+str(j)+")") * probability3
probability1 += eval("Ppers"+str(pNuevos[1]) + "("+str(pTotales[1])+","+str(i)+")") * probability2
return probability1
print (str(Activate(700)))
Edit: Alright I think it would be helpful to explain a couple things:
-First of all, I was trying to find ways the code could run faster, as I'm aware the nested for loops are a performance hog. I was also hoping there would be a way to optimize so many factorial operations.
-Also, the P(A) function described in the g function represents the probability of an event happening, which is already in the code, in the first function from the top.
There's also the function f in the formula, which is just a simplification of the function g for specific cases.
The function f is the second function in the code, whereas g is the third function in the code.
I will try to find a way to simplify the multiple summations, and thanks for the tip of not using eval()!
I'm sorry again for not specifying the question more, and for that mess of code also.
I would expect to break it down with something like this:
def main():
A = 0.5
m = 10
result = g(A, m)
return
def sigma(k, m):
''' function to deal with the sum loop'''
for k in range(10, m+1):
# the bits in the formula
pass
return
def g(A, m):
''' function to deal with g '''
k=10
return 2 * sigma(k,m)
if __name__=='__main__':
''' This is executed when run from the command line '''
main()
Or alternatively to do similar with classes.
I expect you also need a function for p(A) and one for factorials.

How to properly add gradually increasing/decreasing space between objects?

I've trying to implement transition from an amount of space to another which is similar to acceleration and deceleration, except i failed and the only thing that i got from this was this infinite stack of mess, here is a screenshot showing this in action:
you can see a very black circle here, which are in reality something like 100 or 200 circles stacked on top of each other
and i reached this result using this piece of code:
def Place_circles(curve, circle_space, cs, draw=True, screen=None):
curve_acceleration = []
if type(curve) == tuple:
curve_acceleration = curve[1][0]
curve_intensity = curve[1][1]
curve = curve[0]
#print(curve_intensity)
#print(curve_acceleration)
Circle_list = []
idx = [0,0]
for c in reversed(range(0,len(curve))):
for p in reversed(range(0,len(curve[c]))):
user_dist = circle_space[curve_intensity[c]] + curve_acceleration[c] * p
dist = math.sqrt(math.pow(curve[c][p][0] - curve[idx[0]][idx[1]][0],2)+math.pow(curve [c][p][1] - curve[idx[0]][idx[1]][1],2))
if dist > user_dist:
idx = [c,p]
Circle_list.append(circles.circles(round(curve[c][p][0]), round(curve[c][p][1]), cs, draw, screen))
This place circles depending on the intensity (a number between 0 and 2, random) of the current curve, which equal to an amount of space (let's say between 20 and 30 here, 20 being index 0, 30 being index 2 and a number between these 2 being index 1).
This create the stack you see above and isn't what i want, i also came to the conclusion that i cannot use acceleration since the amount of time to move between 2 points depend on the amount of circles i need to click on, knowing that there are multiple circles between each points, but not being able to determine how many lead to me being unable to the the classic acceleration formula.
So I'm running out of options here and ideas on how to transition from an amount of space to another.
any idea?
PS: i scrapped the idea above and switched back to my master branch but the code for this is still available in the branch i created here https://github.com/Mrcubix/Osu-StreamGenerator/tree/acceleration .
So now I'm back with my normal code that don't possess acceleration or deceleration.
TL:DR i can't use acceleration since i don't know the amount of circles that are going to be placed between the 2 points and make the time of travel vary (i need for exemple to click circles at 180 bpm of one circle every 0.333s) so I'm looking for another way to generate gradually changing space.
First, i took my function that was generating the intensity for each curves in [0 ; 2]
Then i scrapped the acceleration formula as it's unusable.
Now i'm using a basic algorithm to determine the maximum amount of circles i can place on a curve.
Now the way my script work is the following:
i first generate a stream (multiple circles that need to be clicked at high bpm)
this way i obtain the length of each curves (or segments) of the polyline.
i generate an intensity for each curve using the following function:
def generate_intensity(Circle_list: list = None, circle_space: int = None, Args: list = None):
curve_intensity = []
if not Args or Args[0] == "NewProfile":
prompt = True
while prompt:
max_duration_intensity = input("Choose the maximum amount of curve the change in intensity will occur for: ")
if max_duration_intensity.isdigit():
max_duration_intensity = int(max_duration_intensity)
prompt = False
prompt = True
while prompt:
intensity_change_odds = input("Choose the odds of occurence for changes in intensity (1-100): ")
if intensity_change_odds.isdigit():
intensity_change_odds = int(intensity_change_odds)
if 0 < intensity_change_odds <= 100:
prompt = False
prompt = True
while prompt:
min_intensity = input("Choose the lowest amount of spacing a circle will have: ")
if min_intensity.isdigit():
min_intensity = float(min_intensity)
if min_intensity < circle_space:
prompt = False
prompt = True
while prompt:
max_intensity = input("Choose the highest amount of spacing a circle will have: ")
if max_intensity.isdigit():
max_intensity = float(max_intensity)
if max_intensity > circle_space:
prompt = False
prompt = True
if Args:
if Args[0] == "NewProfile":
return [max_duration_intensity, intensity_change_odds, min_intensity, max_intensity]
elif Args[0] == "GenMap":
max_duration_intensity = Args[1]
intensity_change_odds = Args[2]
min_intensity = Args[3]
max_intensity = Args[4]
circle_space = ([min_intensity, circle_space, max_intensity] if not Args else [Args[0][3],circle_space,Args[0][4]])
count = 0
for idx, i in enumerate(Circle_list):
if idx == len(Circle_list) - 1:
if random.randint(0,100) < intensity_change_odds:
if random.randint(0,100) > 50:
curve_intensity.append(2)
else:
curve_intensity.append(0)
else:
curve_intensity.append(1)
if random.randint(0,100) < intensity_change_odds:
if random.randint(0,100) > 50:
curve_intensity.append(2)
count += 1
else:
curve_intensity.append(0)
count += 1
else:
if curve_intensity:
if curve_intensity[-1] == 2 and not count+1 > max_duration_intensity:
curve_intensity.append(2)
count += 1
continue
elif curve_intensity[-1] == 0 and not count+1 > max_duration_intensity:
curve_intensity.append(0)
count += 1
continue
elif count+1 > 2:
curve_intensity.append(1)
count = 0
continue
else:
curve_intensity.append(1)
else:
curve_intensity.append(1)
curve_intensity.reverse()
if curve_intensity.count(curve_intensity[0]) == len(curve_intensity):
print("Intensity didn't change")
return circle_space[1]
print("\n")
return [circle_space, curve_intensity]
with this, i obtain 2 list, one with the spacing i specified, and the second one is the list of randomly generated intensity.
from there i call another function taking into argument the polyline, the previously specified spacings and the generated intensity:
def acceleration_algorithm(polyline, circle_space, curve_intensity):
new_circle_spacing = []
for idx in range(len(polyline)): #repeat 4 times
spacing = []
Length = 0
best_spacing = 0
for p_idx in range(len(polyline[idx])-1): #repeat 1000 times / p_idx in [0 ; 1000]
# Create multiple list containing spacing going from circle_space[curve_intensity[idx-1]] to circle_space[curve_intensity[idx]]
spacing.append(np.linspace(circle_space[curve_intensity[idx]],circle_space[curve_intensity[idx+1]], p_idx).tolist())
# Sum distance to find length of curve
Length += abs(math.sqrt((polyline[idx][p_idx+1][0] - polyline[idx][p_idx][0]) ** 2 + (polyline [idx][p_idx+1][1] - polyline[idx][p_idx][1]) ** 2))
for s in range(len(spacing)): # probably has 1000 list in 1 list
length_left = Length # Make sure to reset length for each iteration
for dist in spacing[s]: # substract the specified int in spacing[s]
length_left -= dist
if length_left > 0:
best_spacing = s
else: # Since length < 0, use previous working index (best_spacing), could also jsut do `s-1`
if spacing[best_spacing] == []:
new_circle_spacing.append([circle_space[1]])
continue
new_circle_spacing.append(spacing[best_spacing])
break
return new_circle_spacing
with this, i obtain a list with the space between each circles that are going to be placed,
from there, i can Call Place_circles() again, and obtain the new stream:
def Place_circles(polyline, circle_space, cs, DoDrawCircle=True, surface=None):
Circle_list = []
curve = []
next_circle_space = None
dist = 0
for c in reversed(range(0, len(polyline))):
curve = []
if type(circle_space) == list:
iter_circle_space = iter(circle_space[c])
next_circle_space = next(iter_circle_space, circle_space[c][-1])
for p in reversed(range(len(polyline[c])-1)):
dist += math.sqrt((polyline[c][p+1][0] - polyline[c][p][0]) ** 2 + (polyline [c][p+1][1] - polyline[c][p][1]) ** 2)
if dist > (circle_space if type(circle_space) == int else next_circle_space):
dist = 0
curve.append(circles.circles(round(polyline[c][p][0]), round(polyline[c][p][1]), cs, DoDrawCircle, surface))
if type(circle_space) == list:
next_circle_space = next(iter_circle_space, circle_space[c][-1])
Circle_list.append(curve)
return Circle_list
the result is a stream with varying space between circles (so accelerating or decelerating), the only issue left to be fixed is pygame not updating the screen with the new set of circle after i call Place_circles(), but that's an issue i'm either going to try to fix myself or ask in another post
the final code for this feature can be found on my repo : https://github.com/Mrcubix/Osu-StreamGenerator/tree/Acceleration_v02

How to structure python programs? Tried making it more structured, now runs 13 times slower

Im very new to programming, I wrote a simple program for a school project and wanted to make the code "prettier" by not just having the program be one giant function but instead be made up of multiple smaller functions with a singe purpose. I seemed to have messed up royally since the program now runs 13 times slower. How should I structured the program to make it run faster and just in general make programs easier to write, read and edit?
Here are the two programs:
First program (for reference values runs in ≈0:20):
import numpy as np
import matplotlib.pyplot as plt
def graf(a,b,H,p):
GM = 39.5216489684
x_0 = a + np.sqrt(a**2 - b**2)
v_0 = np.sqrt(GM*(2/x_0 - 1/a))
konstant_period = np.sqrt(a**3)*H
h = 1/H
'''starting position given by an elliptic orbit '''
stor_x_lista = [x_0]
stor_y_lista = [0]
hastighet_x = [0]
hastighet_y = [v_0]
liten_x_lista = []
liten_y_lista = []
''' a loop that approximates the points of the orbit'''
t = 0
tid_lista = []
n = 0
while n < konstant_period:
hastighet_x.append(hastighet_x[n] - h*GM* stor_x_lista[n]/(np.sqrt(stor_x_lista[n]**2 + stor_y_lista[n]**2))**3)
stor_x_lista.append(stor_x_lista[n] + h*hastighet_x[n])
hastighet_y.append(hastighet_y[n] - h*GM*stor_y_lista[n]/(np.sqrt(stor_x_lista[n]**2 + stor_y_lista[n]**2))**3)
stor_y_lista.append(stor_y_lista[n] + h*hastighet_y[n])
'''smaller list of points to run faster'''
if n % p == 0:
liten_x_lista.append(stor_x_lista[n])
liten_y_lista.append(stor_y_lista[n])
tid_lista.append(t)
n += 1
t += h
''' function that finds the angle'''
vinkel = []
siffra = 0
while siffra < len(liten_x_lista):
if liten_y_lista[siffra ] >= 0:
vinkel.append( np.arccos( liten_x_lista[siffra]/np.sqrt( liten_x_lista[siffra]**2 + liten_y_lista[siffra]**2)))
siffra += 1
elif liten_y_lista[siffra] < 0 :
vinkel.append( np.pi + np.arccos( -liten_x_lista[siffra]/np.sqrt( liten_x_lista[siffra]**2 + liten_y_lista[siffra]**2) ))
siffra += 1
'''get rid of line to find periodic function'''
mod_lista = []
modn = 0
while modn < len(vinkel):
mod_lista.append(vinkel[modn] - (2*np.pi*tid_lista[modn])/np.sqrt(a**3))
modn += 1
'''make all inputs have period 1'''
squeeze_tid = []
squeezen = 0
while squeezen < len(tid_lista):
squeeze_tid.append(tid_lista[squeezen]/np.sqrt(a**3))
squeezen += 1
del mod_lista[-1:]
del tid_lista[-1:]
del squeeze_tid[-1:]
plt.plot(squeeze_tid,mod_lista)
plt.title('p(t) där a = ' + str(a) + ' och b = ' + str(b))
plt.show
Second more split up program (for reference values runs in ≈4:20):
import numpy as np
import matplotlib.pyplot as plt
'''function that generates the points of the orbit'''
def punkt(a,b,H,p):
GM = 39.5216489684
x_0 = a + np.sqrt(a**2 - b**2)
v_0 = np.sqrt(GM*(2/x_0 - 1/a))
konstant_period = np.sqrt(a**3)*H
h = 1/H
'''starting position given by an elliptic orbit '''
stor_x_lista = [x_0]
stor_y_lista = [0]
hastighet_x = [0]
hastighet_y = [v_0]
liten_x_lista = []
liten_y_lista = []
''' a loop that approximates the points of the orbit'''
t = 0
tid_lista = []
n = 0
while n < konstant_period:
hastighet_x.append(hastighet_x[n] - h*GM* stor_x_lista[n]/(np.sqrt(stor_x_lista[n]**2 + stor_y_lista[n]**2))**3)
stor_x_lista.append(stor_x_lista[n] + h*hastighet_x[n])
hastighet_y.append(hastighet_y[n] - h*GM*stor_y_lista[n]/(np.sqrt(stor_x_lista[n]**2 + stor_y_lista[n]**2))**3)
stor_y_lista.append(stor_y_lista[n] + h*hastighet_y[n])
'''smaller list of points to run faster'''
if n % p == 0:
liten_x_lista.append(stor_x_lista[n])
liten_y_lista.append(stor_y_lista[n])
tid_lista.append(t)
n += 1
t += h
return (liten_x_lista,liten_y_lista,tid_lista)
''' function that finds the angle'''
def vinkel(a,b,H,p):
'''import lists'''
liten_x_lista = punkt(a,b,H,p)[0]
liten_y_lista = punkt(a,b,H,p)[1]
tid_lista = punkt(a,b,H,p)[2]
'''find the angle'''
vinkel_lista = []
siffra = 0
while siffra < len(liten_x_lista):
if liten_y_lista[siffra ] >= 0:
vinkel_lista.append( np.arccos( liten_x_lista[siffra]/np.sqrt( liten_x_lista[siffra]**2 + liten_y_lista[siffra]**2)))
siffra += 1
elif liten_y_lista[siffra] < 0 :
vinkel_lista.append( np.pi + np.arccos( -liten_x_lista[siffra]/np.sqrt( liten_x_lista[siffra]**2 + liten_y_lista[siffra]**2) ))
siffra += 1
return (vinkel_lista, tid_lista)
def periodisk(a,b,H,p):
'''import lists'''
tid_lista = vinkel(a,b,H,p)[1]
vinkel_lista = vinkel(a,b,H,p)[0]
'''get rid of linear line to find p(t)'''
mod_lista = []
modn = 0
while modn < len(vinkel_lista):
mod_lista.append((vinkel_lista[modn] - (2*np.pi*tid_lista[modn])/np.sqrt(a**3)))
modn += 1
'''make all inputs have period 1'''
squeeze_tid = []
squeezen = 0
while squeezen < len(tid_lista):
squeeze_tid.append(tid_lista[squeezen]/np.sqrt(a**3))
squeezen += 1
del mod_lista[-1:]
del tid_lista[-1:]
del squeeze_tid[-1:]
return (squeeze_tid,mod_lista)
'''fixa 3d-punkt av p(a,b) a är konstant b varierar??? '''
def hitta_amp(a):
x_b = []
y_b = []
n_b = 0.1
while n_b <= a:
x_b.append(n_b)
y_b.append(punkt(a,n_b,10**5,10**3))
return 0
def graf(a,b,H,p):
plt.plot(periodisk(a,b,H,p)[0],periodisk(a,b,H,p)[1])
plt.show
I would assume the thing that is going wrong is that the program is running the same, slow code multiple times instead of just running it once and then accessing the data. Is the problem that everything is done locally and nothing is stored globally or is it something else?
Just as a heads up, the only thing I know about programming is basic syntax, I have no clue how to actually write and run programs. I ran all the code in spyder if that affects anything.
plt.plot(periodisk(a,b,H,p)[0],periodisk(a,b,H,p)[1])
This code runs periodisk twice with the same arguments, thus at this point we know we run things at least 2 times slower.
You should do some_var = periodisk(a,b,H,p) and then some_var[0], some_var[1]. Or just use unpacking:
plt.plot(*periodisk(a,b,H,p))
tid_lista = vinkel(a,b,H,p)[1]
vinkel_lista = vinkel(a,b,H,p)[0]
Again doing the same thing twice (total: 4*time of (current) vinkel function). Again, smart assignment to fix this:
vinkel_lista, tid_lista = vinkel(a,b,H,p)
liten_x_lista = punkt(a,b,H,p)[0]
liten_y_lista = punkt(a,b,H,p)[1]
tid_lista = punkt(a,b,H,p)[2]
And now you repeat yourself thrice. (total: 12 * time of current punkt function)
liten_x_lista, liten_y_lista, tid_lista = punkt(a,b,H,p)
punkt function is like in original, so we arrived as total being 12 times slower - which quite matches your time estimations. :)
You are calling the functions once per returned list, you should only call them once.
When a method returns multiple variables, (e.g. punkt):
def punkt(a,b,H,p):
# Here is all your code
return (liten_x_lista,liten_y_lista,tid_lista)
You must be careful to only call the function once:
result = punkt(a,b,H,p)
liten_x_lista = result[0]
liten_y_lista = result[1]
tid_lista = result[2]
# As opposed to:
liten_x_lista = punkt(a,b,H,p)[0] # 1st call, ignoring results 2 and 3
liten_y_lista = punkt(a,b,H,p)[1] # 2nd call, ignoring results 1 and 3
tid_lista = punkt(a,b,H,p)[2] # 3rd call, ignoring results 1 and 2
Note: I would personally not return a list, but use python's unpacking:
def punkt(a,b,H,p):
# Here is all your code
return liten_x_lista, liten_y_lista, tid_lista
And you'd access it:
liten_x_lista, liten_y_lista, tid_lista = punkt(a,b,H,p)

Speeding a numpy correlation program using the fact that lists are sorted

I am currently using python and numpy for calculations of correlations between 2 lists: data_0 and data_1. Each list contains respecively sorted times t0 and t1.
I want to calculate all the events where 0 < t1 - t0 < t_max.
for time_0 in np.nditer(data_0):
delta_time = np.subtract(data_1, np.full(data_1.size, time_0))
delta_time = delta_time[delta_time >= 0]
delta_time = delta_time[delta_time < time_max]
Doing so, as the list are sorted, I am selecting a subarray of data_1 of the form data_1[index_min: index_max].
So I need in fact to find two indexes to get what I want.
And what's interesting is that when I go to the next time_0, as data_0 is also sorted, I just need to find the new index_min / index_max such as new_index_min >= index_min / new_index_max >= index_max.
Meaning that I don't need to scann again all the data_1.
(data list from scratch).
I have implemented such a solution not using the numpy methods (just with while loop) and it gives me the same results as before but not as fast than before (15 times longer!).
I think as normally it requires less calculation, there should be a way to make it faster using numpy methods but I don't know how to do it.
Does anyone have an idea?
I am not sure if I am super clear so if you have any questions, do not hestitate.
Thank you in advance,
Paul
Here is a vectorized approach using argsort. It uses a strategy similar to your avoid-full-scan idea:
import numpy as np
def find_gt(ref, data, incl=True):
out = np.empty(len(ref) + len(data) + 1, int)
total = (data, ref) if incl else (ref, data)
out[1:] = np.argsort(np.concatenate(total), kind='mergesort')
out[0] = -1
split = (out < len(data)) if incl else (out >= len(ref))
if incl:
out[~split] -= len(data)
split[0] = False
return np.maximum.accumulate(np.where(split, -1, out))[split] + 1
def find_intervals(ref, data, span, incl=(True, True)):
index_min = find_gt(ref, data, incl[0])
index_max = len(ref) - find_gt(-ref[::-1], -span-data[::-1], incl[1])[::-1]
return index_min, index_max
ref = np.sort(np.random.randint(0,20000,(10000,)))
data = np.sort(np.random.randint(0,20000,(10000,)))
span = 2
idmn, idmx = find_intervals(ref, data, span, (True, True))
print('checking')
for d,mn,mx in zip(data, idmn, idmx):
assert mn == len(ref) or ref[mn] >= d
assert mn == 0 or ref[mn-1] < d
assert mx == len(ref) or ref[mx] > d+span
assert mx == 0 or ref[mx-1] <= d+span
print('ok')
It works by
indirectly sorting both sets together
finding for each time in one set the preceding time in the other
this is done using maximum.reduce
the preceding steps are applied twice, the second time the times in
one set are shifted by span

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