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What is the best way/method to digitize the data of a 3D surface into a grid of pixels with smaller resolution in Python?

I want to digitize (= average out over cells) photon count data into pixels given by a grid that tells how they are aligned. The photon count data is stored in a 2D array. I want to split that data into cells, each of which would correspond to a pixel. The idea is basically the same as changing an HD image to a smaller resolution. I'd like to achieve this in Python.
The digitizing function I've written:
import numpy as np
def digitize(function_data, grid_shape):
"""
function_data = 2D array of function values of some 3D shape,
eg.: exp(-(x^2 + y^2 -> want to digitize this
grid_shape: an array of length 2 which contains the dimensions of the smaller resolution
"""
l = len(function_data)
pixel_len_x = int(l/grid_shape[0])
pixel_len_y = int(l/grid_shape[1])
digitized_data = np.empty((grid_shape[0], grid_shape[1]))
for i in range(grid_shape[0]): #row-index of pixel in smaller-resolution grid
for j in range(grid_shape[1]): #column-index of pixel in smaller-resolution grid
hd_pixel = []
for k in range(pixel_len_y):
hd_pixel.append(z_data[k][j:j*pixel_len_x])
hd_pixel = np.ravel(hd_pixel) #turns 2D array into 1D to be able to compute average
pixel_avg = np.average(hd_pixel)
digitized_data[i][j] = pixel_avg
return digitized_data
In theory, this function should do what I want to achieve, but when tested it doesn't yield the expected results. Either a completed version of my function or any other method that achieves my goal would be extremely helpful.

You could also use a interpolation function, if you can use SciPy. Here we use one of the gridded data interpolating functions, RectBivariateSpline to upsample your function, but you can find numerous examples on this and other sites.
import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import RectBivariateSpline as rbs
# Sampling coordinates
x = np.linspace(-2,2,20)
y = np.linspace(-2,2,30)
# Your function
f = np.exp(-(x[:,None]**2 + y**2))
# Interpolator
interp = rbs(x, y, f)
# Higher resolution coordinates
x_hd = np.linspace(x.min(), x.max(), x.size * 5)
y_hd = np.linspace(y.min(), y.max(), y.size * 5)
# New higher res function
f_hd = interp(x_hd, y_hd, grid = True)
# Some plots
fig, ax = plt.subplots(ncols = 2)
ax[0].imshow(f)
ax[1].imshow(f_hd)

How to interpolate in n dimensions with different length axes

So I was looking at how to use scipy's interpn function, and the example they have on the documentation isn't quite working with what I need it to do.
My implementation is a bit different. I have a precomputed value array with shape [200,40,40,40] that I get from a different script.
So when I do something like:
t = np.linspace(0,1, 200)
x = np.linspace(0,1, 40)
y = np.linspace(0,1, 40)
z = np.linspace(0,1, 40)
points = (t,x,y,z)
interpn(points,values,point)
I get an error: "ValueError: There are 40 points and 200 values in dimension 0"
It seems as though the dimensions of my points tuple and value array are not lining up, but I thought since my "t" axis is first in the tuple, it should be match. Any advice?

So this works for me:
import numpy as np
from scipy.interpolate import interpn
def f(x,y,z,t):
'''Simple 3D + time dimensional function.'''
return (np.sin(x)+y+np.sqrt(z))*t
t = np.linspace(0,1,200)
x = np.linspace(0,1,40)
y = np.linspace(0,1,40)
z = np.linspace(0,1,40)
points = (x,y,z,t)
values = f(*np.meshgrid(*points))
# example point in domain
point = [0,0.5,0.75,1/3.]
print(interpn(points, values, point))
array([0.44846267])
You defined x,y,z as np.linspace(0,40,1), this means you have a single point on the interval [0,40]. The same for t. That's probably your error. Example taken from the official scipy documentation.

Counterclockwise sorting of x, y data

I have a set of points in a text file: random_shape.dat.
The initial order of points in the file is random. I would like to sort these points in a counter-clockwise order as follows (the red dots are the xy data):
I tried to achieve that by using the polar coordinates: I calculate the polar angle of each point (x,y) then sort by the ascending angles, as follows:
"""
Script: format_file.py
Description: This script will format the xy data file accordingly to be used with a program expecting CCW order of data points, By soting the points in Counterclockwise order
Example: python format_file.py random_shape.dat
"""
import sys
import numpy as np
# Read the file name
filename = sys.argv[1]
# Get the header name from the first line of the file (without the newline character)
with open(filename, 'r') as f:
header = f.readline().rstrip('\n')
angles = []
# Read the data from the file
x, y = np.loadtxt(filename, skiprows=1, unpack=True)
for xi, yi in zip(x, y):
angle = np.arctan2(yi, xi)
if angle < 0:
angle += 2*np.pi # map the angle to 0,2pi interval
angles.append(angle)
# create a numpy array
angles = np.array(angles)
# Get the arguments of sorted 'angles' array
angles_argsort = np.argsort(angles)
# Sort x and y
new_x = x[angles_argsort]
new_y = y[angles_argsort]
print("Length of new x:", len(new_x))
print("Length of new y:", len(new_y))
with open(filename.split('.')[0] + '_formatted.dat', 'w') as f:
print(header, file=f)
for xi, yi in zip(new_x, new_y):
print(xi, yi, file=f)
print("Done!")
By running the script:
python format_file.py random_shape.dat
Unfortunately I don't get the expected results in random_shape_formated.dat! The points are not sorted in the desired order.
Any help is appreciated.
EDIT: The expected resutls:
Create a new file named: filename_formatted.dat that contains the sorted data according to the image above (The first line contains the starting point, the next lines contain the points as shown by the blue arrows in counterclockwise direction in the image).
EDIT 2: The xy data added here instead of using github gist:
random_shape
0.4919261070361315 0.0861956168831175
0.4860816807027076 -0.06601587301587264
0.5023029456281289 -0.18238249845392662
0.5194784026079869 0.24347943722943777
0.5395164357511545 -0.3140611471861465
0.5570497147514262 0.36010146103896146
0.6074231036252226 -0.4142604617604615
0.6397066014669927 0.48590810704447085
0.7048302091822873 -0.5173701298701294
0.7499157837544145 0.5698170011806378
0.8000108666123336 -0.6199254449254443
0.8601249660418364 0.6500974025974031
0.9002010323281716 -0.7196585989767801
0.9703341483292582 0.7299242424242429
1.0104102146155935 -0.7931355765446666
1.0805433306166803 0.8102046438410078
1.1206193969030154 -0.865251869342778
1.1907525129041021 0.8909386068476981
1.2308285791904374 -0.9360074773711129
1.300961695191524 0.971219008264463
1.3410377614778592 -1.0076702085792988
1.4111708774789458 1.051499409681228
1.451246943765281 -1.0788793781975592
1.5213800597663678 1.1317798110979933
1.561456126052703 -1.1509956709956706
1.6315892420537896 1.2120602125147582
1.671665308340125 -1.221751279024005
1.7417984243412115 1.2923406139315234
1.7818744906275468 -1.2943211334120424
1.8520076066286335 1.3726210153482883
1.8920836729149686 -1.3596340023612745
1.9622167889160553 1.4533549783549786
2.0022928552023904 -1.4086186540731989
2.072425971203477 1.5331818181818184
2.1125020374898122 -1.451707005116095
2.182635153490899 1.6134622195985833
2.2227112197772345 -1.4884454939000387
2.292844335778321 1.6937426210153486
2.3329204020646563 -1.5192876820149541
2.403053518065743 1.774476584022039
2.443129584352078 -1.5433264462809912
2.513262700353165 1.8547569854388037
2.5533387666395 -1.561015348288075
2.6234718826405867 1.9345838252656438
2.663547948926922 -1.5719008264462806
2.7336810649280086 1.9858362849271942
2.7737571312143436 -1.5750757575757568
2.8438902472154304 2.009421487603306
2.883966313501766 -1.5687258953168035
2.954099429502852 2.023481896890988
2.9941754957891877 -1.5564797323888229
3.0643086117902745 2.0243890200708385
3.1043846780766096 -1.536523022432113
3.1745177940776963 2.0085143644234558
3.2145938603640314 -1.5088557654466737
3.284726976365118 1.9749508067689887
3.324803042651453 -1.472570838252656
3.39493615865254 1.919162731208186
3.435012224938875 -1.4285753640299088
3.5051453409399618 1.8343467138921687
3.545221407226297 -1.3786835891381335
3.6053355066557997 1.7260966810966811
3.655430589513719 -1.3197205824478546
3.6854876392284703 1.6130086580086582
3.765639771801141 -1.2544077134986225
3.750611246943765 1.5024152236652237
3.805715838087476 1.3785173160173163
3.850244800627849 1.2787337662337666
3.875848954088563 -1.1827449822904361
3.919007794704616 1.1336638361638363
3.9860581363759846 -1.1074537583628485
3.9860581363759846 1.0004485329485333
4.058012891753723 0.876878197560016
4.096267318663407 -1.0303482880755608
4.15638141809291 0.7443374218374221
4.206476500950829 -0.9514285714285711
4.256571583808748 0.6491902794175526
4.3166856832382505 -0.8738695395513574
4.36678076609617 0.593855765446675
4.426894865525672 -0.7981247540338443
4.476989948383592 0.5802489177489183
4.537104047813094 -0.72918339236521
4.587199130671014 0.5902272727272733
4.647313230100516 -0.667045454545454
4.697408312958435 0.6246979535615904
4.757522412387939 -0.6148858717040526
4.807617495245857 0.6754968516332154
4.8677315946753605 -0.5754260133805582
4.917826677533279 0.7163173947264858
4.977940776962782 -0.5500265643447455
5.028035859820701 0.7448917748917752
5.088149959250204 -0.5373268398268394
5.138245042108123 0.7702912239275879
5.198359141537626 -0.5445838252656432
5.2484542243955445 0.7897943722943728
5.308568323825048 -0.5618191656828015
5.358663406682967 0.8052154663518301
5.41877750611247 -0.5844972451790631
5.468872588970389 0.8156473829201105
5.5289866883998915 -0.6067217630853987
5.579081771257811 0.8197294372294377
5.639195870687313 -0.6248642266824076
5.689290953545233 0.8197294372294377
5.749405052974735 -0.6398317591499403
5.799500135832655 0.8142866981503349
5.859614235262157 -0.6493565525383702
5.909709318120076 0.8006798504525783
5.969823417549579 -0.6570670995670991
6.019918500407498 0.7811767020857934
6.080032599837001 -0.6570670995670991
6.13012768269492 0.7562308146399057
6.190241782124423 -0.653438606847697
6.240336864982342 0.7217601338055886
6.300450964411845 -0.6420995670995664
6.350546047269764 0.6777646595828419
6.410660146699267 -0.6225964187327819
6.4607552295571855 0.6242443919716649
6.520869328986689 -0.5922077922077915
6.570964411844607 0.5548494687131056
6.631078511274111 -0.5495730027548205
6.681173594132029 0.4686727666273125
6.7412876935615325 -0.4860743801652889
6.781363759847868 0.3679316979316982
6.84147785927737 -0.39541245791245716
6.861515892420538 0.25880333951762546
6.926639500135833 -0.28237987012986965
6.917336127605076 0.14262677798392165
6.946677533279001 0.05098957832291173
6.967431210462995 -0.13605442176870675
6.965045730326905 -0.03674603174603108

I find that an easy way to sort points with x,y-coordinates like that is to sort them dependent on the angle between the line from the points and the center of mass of the whole polygon and the horizontal line which is called alpha in the example. The coordinates of the center of mass (x0 and y0) can easily be calculated by averaging the x,y coordinates of all points. Then you calculate the angle using numpy.arccos for instance. When y-y0 is larger than 0 you take the angle directly, otherwise you subtract the angle from 360° (2𝜋). I have used numpy.where for the calculation of the angle and then numpy.argsort to produce a mask for indexing the initial x,y-values. The following function sort_xy sorts all x and y coordinates with respect to this angle. If you want to start from any other point you could add an offset angle for that. In your case that would be zero though.
def sort_xy(x, y):
x0 = np.mean(x)
y0 = np.mean(y)
r = np.sqrt((x-x0)**2 + (y-y0)**2)
angles = np.where((y-y0) > 0, np.arccos((x-x0)/r), 2*np.pi-np.arccos((x-x0)/r))
mask = np.argsort(angles)
x_sorted = x[mask]
y_sorted = y[mask]
return x_sorted, y_sorted
Plotting x, y before sorting using matplotlib.pyplot.plot (points are obvisously not sorted):
Plotting x, y using matplotlib.pyplot.plot after sorting with this method:

If it is certain that the curve does not cross the same X coordinate (i.e. any vertical line) more than twice, then you could visit the points in X-sorted order and append a point to one of two tracks you follow: to the one whose last end point is the closest to the new one. One of these tracks will represent the "upper" part of the curve, and the other, the "lower" one.
The logic would be as follows:
dist2 = lambda a,b: (a[0]-b[0])*(a[0]-b[0]) + (a[1]-b[1])*(a[1]-b[1])
z = list(zip(x, y)) # get the list of coordinate pairs
z.sort() # sort by x coordinate
cw = z[0:1] # first point in clockwise direction
ccw = z[1:2] # first point in counter clockwise direction
# reverse the above assignment depending on how first 2 points relate
if z[1][1] > z[0][1]:
cw = z[1:2]
ccw = z[0:1]
for p in z[2:]:
# append to the list to which the next point is closest
if dist2(cw[-1], p) < dist2(ccw[-1], p):
cw.append(p)
else:
ccw.append(p)
cw.reverse()
result = cw + ccw
This would also work for a curve with steep fluctuations in the Y-coordinate, for which an angle-look-around from some central point would fail, like here:
No assumption is made about the range of the X nor of the Y coordinate: like for instance, the curve does not necessarily have to cross the X axis (Y = 0) for this to work.

Counter-clock-wise order depends on the choice of a pivot point. From your question, one good choice of the pivot point is the center of mass.
Something like this:
# Find the Center of Mass: data is a numpy array of shape (Npoints, 2)
mean = np.mean(data, axis=0)
# Compute angles
angles = np.arctan2((data-mean)[:, 1], (data-mean)[:, 0])
# Transform angles from [-pi,pi] -> [0, 2*pi]
angles[angles < 0] = angles[angles < 0] + 2 * np.pi
# Sort
sorting_indices = np.argsort(angles)
sorted_data = data[sorting_indices]

Not really a python question I think, but still I think you could try sorting by - sign(y) * x doing something like:
def counter_clockwise_sort(points):
return sorted(points, key=lambda point: point['x'] * (-1 if point['y'] >= 0 else 1))
should work fine, assuming you read your points properly into a list of dicts of format {'x': 0.12312, 'y': 0.912}
EDIT: This will work as long as you cross the X axis only twice, like in your example.

If:
the shape is arbitrarily complex and
the point spacing is ~random
then I think this is a really hard problem.
For what it's worth, I have faced a similar problem in the past, and I used a traveling salesman solver. In particular, I used the LKH solver. I see there is a Python repo for solving the problem, LKH-TSP. Once you have an order to the points, I don't think it will be too hard to decide on a clockwise vs clockwise ordering.

If we want to answer your specific problem, we need to pick a pivot point.
Since you want to sort according to the starting point you picked, I would take a pivot in the middle (x=4,y=0 will do).
Since we're sorting counterclockwise, we'll take arctan2(-(y-pivot_y),-(x-center_x)) (we're flipping the x axis).
We get the following, with a gradient colored scatter to prove correctness (fyi I removed the first line of the dat file after downloading):
import numpy as np
import matplotlib.pyplot as plt
points = np.loadtxt('points.dat')
#oneliner for ordering points (transform, adjust for 0 to 2pi, argsort, index at points)
ordered_points = points[np.argsort(np.apply_along_axis(lambda x: np.arctan2(-x[1],-x[0]+4) + np.pi*2, axis=1,arr=points)),:]
#color coding 0-1 as str for gray colormap in matplotlib
plt.scatter(ordered_points[:,0], ordered_points[:,1],c=[str(x) for x in np.arange(len(ordered_points)) / len(ordered_points)],cmap='gray')
Result (in the colormap 1 is white and 0 is black), they're numbered in the 0-1 range by order:

For points with comparable distances between their neighbouring pts, we can use KDTree to get two closest pts for each pt. Then draw lines connecting those to give us a closed shape contour. Then, we will make use of OpenCV's findContours to get contour traced always in counter-clockwise manner. Now, since OpenCV works on images, we need to sample data from the provided float format to uint8 image format. Given, comparable distances between two pts, that should be pretty safe. Also, OpenCV handles it well to make sure it traces even sharp corners in curvatures, i.e. smooth or not-smooth data would work just fine. And, there's no pivot requirement, etc. As such all kinds of shapes would be good to work with.
Here'e the implementation -
import numpy as np
import matplotlib.pyplot as plt
from scipy.spatial.distance import pdist
from scipy.spatial import cKDTree
import cv2
from scipy.ndimage.morphology import binary_fill_holes
def counter_clockwise_order(a, DEBUG_PLOT=False):
b = a-a.min(0)
d = pdist(b).min()
c = np.round(2*b/d).astype(int)
img = np.zeros(c.max(0)[::-1]+1, dtype=np.uint8)
d1,d2 = cKDTree(c).query(c,k=3)
b = c[d2]
p1,p2,p3 = b[:,0],b[:,1],b[:,2]
for i in range(len(b)):
cv2.line(img,tuple(p1[i]),tuple(p2[i]),255,1)
cv2.line(img,tuple(p1[i]),tuple(p3[i]),255,1)
img = (binary_fill_holes(img==255)*255).astype(np.uint8)
if int(cv2.__version__.split('.')[0])>=3:
_,contours,hierarchy = cv2.findContours(img.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
else:
contours,hierarchy = cv2.findContours(img.copy(),cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
cont = contours[0][:,0]
f1,f2 = cKDTree(cont).query(c,k=1)
ordered_points = a[f2.argsort()[::-1]]
if DEBUG_PLOT==1:
NPOINTS = len(ordered_points)
for i in range(NPOINTS):
plt.plot(ordered_points[i:i+2,0],ordered_points[i:i+2,1],alpha=float(i)/(NPOINTS-1),color='k')
plt.show()
return ordered_points
Sample run -
# Load data in a 2D array with 2 columns
a = np.loadtxt('random_shape.csv',delimiter=' ')
ordered_a = counter_clockwise_order(a, DEBUG_PLOT=1)
Output -

3d coordinates x,y,z to 3d numpy array

I have a 3d mask which is an ellipsoid. I have extracted the coordinates of the mask using np.argwhere. The coordinates can be assigned as x, y, z as in the example code. My question is how can I get my mask back (in the form of 3d numpy or boolean array of the same shape) from the coordinates x, y, z ?
import numpy as np
import scipy
import skimage
from skimage import draw
mask = skimage.draw.ellipsoid(10,12,18)
print mask.shape
coord = np.argwhere(mask)
x = coord[:,0]
y = coord[:,1]
z = coord[:,2]
The above code gives me boolean mask of the shape (23, 27, 39) and now I want to construct the same mask of exactly same shape using x, y, z coordinates. How can it be done?
I would like to modify the question above a bit. Now if I rotate my coordinates using quaternion which will give me new set of coordinates and then with new coordinates x1,y1,z1 I want to construct my boolean mask of shape (23,27,39) as that of original mask ? How can that be done ?
import quaternion
angle1 = 90
rotation = np.exp(quaternion.quaternion(0,0, 1) * angle1*(np.pi/180) / 2)
coord_rotd = quaternion.rotate_vectors(rotation, coord)
x1 = coord_rotd[:,0]
y1 = coord_rotd[:,1]
z1 = coord_rotd[:,2]

You can use directly x, y and z to reconstruct your mask. First, use a new array with the same shape as your mask. I pre-filled everything with zeros (i.e. False). Next, set each coordinate defined by x, y and z to True:
new_mask = np.zeros_like(mask)
new_mask[x,y,z] = True
# Check if mask and new_mask is the same
np.allclose(mask, new_mask)
# True
If you ask, if you can reconstruct your mask only knowing x, y and z, this is not possible. Because you loose information of what is not filled. Just imagine having your ellipsoid at a corner of a huge cube. How would you know (only knowing how the ellipsoid looks), how large the cube is?
Regarding your second question:
You have to fix your coordinates, because they can be out of your scenery. So I defined a function that takes care of this:
def fixCoordinates(coord, shape):
# move to the positive edge
# remove negative indices
# you can also add now +1 to
# have a margin around your ellipse
coord -= coord.min(0)
# trim coordinates outside of scene
for i, s in enumerate(shape):
coord[coord[:,i] >= s] = s-1
# Return coordinates and change dtype
return coord.astype(np.int)
And if you modify your code slightly, you can use the same strategy as before:
# your code
import quaternion
angle1 = 90
rotation = np.exp(quaternion.quaternion(0,0, 1) * angle1*(np.pi/180) / 2)
coord_rotd = quaternion.rotate_vectors(rotation, coord_rotd)
# Create new mask
new_mask2 = np.zeros_like(new_mask)
# Fix coordinates
coord_rotd = fixCoordinates(coord_rotd, mask.shape)
x1 = coord_rotd[:,0]
y1 = coord_rotd[:,1]
z1 = coord_rotd[:,2]
# create new mask, similar as before
new_mask2[x1, y1, z1] = True
Given your example rotation, you can now plot both masks (that have the same shape), side by side:

If you know the shape of your old mask, try this:
new_mask = np.full(old_mask_shape, True) # Fill new_mask with True everywhere
new_mask[x,y,z] = False # Set False for the ellipsoid part alone
Note:
old_mask_shape should be the same as shape of the image on which you intend to apply the mask.
If you want a True mask rather than a False one (if you want the ellipsoid part to be True and everywhere else False) just interchange True and False in the above two lines of code.

Looking for a way to adjust the values of one array based on another array?

I started with a set of bivariate data. My goal is to first find points in that data set for which the y-values are outliers. Then, I wanted to create a new data set that included not only the outlier points, but also any points with an x-value of within 0.01 of any given outlier point.
Then (if possible) I want to subtract the original outlier x-values from the new x-set, so that I have a group of points with x-values of between -0.01 and 0.01, with x-value now indicating distance from an original outlier x-value.
I have this code:
import numpy as np
mean = np.mean(y)
SD = np.std(y)
x_indices = [i for i in range(len(y)) if ((y[i]) > ((2*SD)+mean))]
expanded_indices = [i for i in range(len(x)) if np.any((abs(x[i] - x[x_indices])) < 0.01)]
This worked great, and now I can call (and plot) x and y using the indices:
plt.plot(x[expanded_indices],y[expanded_indices])
However, I have no idea how to subtract the original "x_indices" values to get an x range of -0.01 to 0.01, since everything I tried failed.
I want to do something like what I have below, except I know that I can't subtract two arrays of different sizes, and I'm worried I can't use np.any in this context either.
x_values = [(x[expanded_indices] - x[indices]) if np.any((abs(x[expanded_indices] - x[indices])) < 0.01)]
Any ideas? I'm sorry this is so long -- I'm very new at this and pretty lost. I've been giving it a go for the last few hours and any assistance would be appreciated. Thanks!
sample data could be as follows:
x =[0,0.994,0.995,0.996,0.997,0.998,1.134,1.245,1.459,1.499,1.500,1.501,2.103,2.104,2.105,2.106]
y =
[1.5,1.6,1.5,1.6,10,1.5,1.5,1.5,1.6,1.6,1.5,1.6,1.5,11,1.6,1.5]

Once you have the set with y-outliers values and the set with the expanded values, you can go over the whole second set with a for loop and subtract the corresponding 1st set value using 2 For() loops:
import numpy as np
x =np.array([0,0.994,0.995,0.996,0.997,0.998,1.134,1.245,1.459,1.499,1.500,1.501,2.103,2.104,2.105,2.106])
y = np.array([1.5,1.6,1.5,1.6,10,1.5,1.5,1.5,1.6,1.6,1.5,1.6,1.5,11,1.6,1.5])
mean = np.mean(y)
SD = np.std(y)
# elements with y-element outside defined region
indices = [i for i in range(len(y)) if ((y[i]) > ((2*SD)+mean))]
my_1st_set = x[indices]
# Set with values within 0.01 difference with 1st set points
expanded_indices = [i for i in range(len(x)) if np.any((abs(x[i] - x[x_indices])) < 0.01)]
my_2nd_set = x[expanded_indices]
# A final set with the subtracted values from the 2nd set
my_final_set = my_2nd_set
for i in range(my_final_set.size):
for j in range(my_1st_set.size):
if abs(my_final_set[i] - my_1st_set[j]) < 0.01:
my_final_set[i] = x[i] - my_1st_set[j]
break
my_final_set is a numpy array with the resulting values of subtracting the original expanded_indices values with their corresponding value of the first set

Let's see if I understood you correctly. This code should find the outliers, and put an array into res for each outlier.
import numpy as np
mean = np.mean(y)
SD = np.std(y)
x = np.array([0,0.994,0.995,0.996,0.997,0.998,1.134,1.245,1.459,1.499,1.500,1.501,2.103,2.104,2.105,2.106])
y = np.array([1.5,1.6,1.5,1.6,10,1.5,1.5,1.5,1.6,1.6,1.5,1.6,1.5,11,1.6,1.5])
outlier_indices = np.abs(y - mean) > 2*SD
res = []
for x_at_outlier in x[np.flatnonzero(outlier_indices)]:
part_res = x[np.abs(x - x_at_outlier) < 0.01]
part_res -= np.mean(part_res)
res.append(part_res)
res is now a list of arrays, with each array containing the values around one outlier. Perhaps it is easier to continue working with the data in this format?
If you want all of them in one numpy array:
res = np.hstack(res)