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I am working with a dataset that has images in shape (105,105,3). Since this raw image gives us a range of (0 to 105) on the x axis and (105 to 0) on the y axis, I am converting the image into (x,y) datapoints by using the code below where example is the raw image:
example = x_train[21]
x = []
y = []
i1 = np.argwhere(example < 1)
H = i1.shape[1]
i1[:1] = [H-1]-i1[:1]
for i in i1:
x.append(i[0])
y.append(i[1])
plt.scatter(x,y)
xy = np.column_stack((x,y))
This code changes
to this:
Obviously i am doing something wrong and there is some random fricken dot in the bottom left corner that is driving me crazy. Is there a better less stupid way to do this? Ideally a nice mapping with the proper orientation and size would be great.
I can't understand the code starting from #LIST DEST PIXEL INDICES, I don't get why the and x and y
are the way they are, why aren't they starting from 0,0 instead of DIM//2. From my understanding, this function transforms the image by "inverse" transforming the destination pixels and plugging the original pixels in the destination pixels positions.
get_mat(*args) returns a 3x3 matrix that does various transformations, pretend it is a rotation matrix
take IMAGE_SIZE[0] as 224
this is part of Chris Deotte Kaggel notebook https://www.kaggle.com/cdeotte/rotation-augmentation-gpu-tpu-0-96
def transform(image,label):
# input image - is one image of size [dim,dim,3] not a batch of [b,dim,dim,3]
# output - image randomly rotated, sheared, zoomed, and shifted
DIM = IMAGE_SIZE[0]
XDIM = DIM%2 #fix for size 331
rot = 15. * tf.random.normal([1],dtype='float32')
shr = 5. * tf.random.normal([1],dtype='float32')
h_zoom = 1.0 + tf.random.normal([1],dtype='float32')/10.
w_zoom = 1.0 + tf.random.normal([1],dtype='float32')/10.
h_shift = 16. * tf.random.normal([1],dtype='float32')
w_shift = 16. * tf.random.normal([1],dtype='float32')
# GET TRANSFORMATION MATRIX
m = get_mat(rot,shr,h_zoom,w_zoom,h_shift,w_shift)
# LIST DESTINATION PIXEL INDICES
x = tf.repeat( tf.range(DIM//2,-DIM//2,-1), DIM )
y = tf.tile( tf.range(-DIM//2,DIM//2),[DIM] )
z = tf.ones([DIM*DIM],dtype='int32')
idx = tf.stack( [x,y,z] )
# ROTATE DESTINATION PIXELS ONTO ORIGIN PIXELS
idx2 = K.dot(m,tf.cast(idx,dtype='float32'))
idx2 = K.cast(idx2,dtype='int32')
idx2 = K.clip(idx2,-DIM//2+XDIM+1,DIM//2)
# FIND ORIGIN PIXEL VALUES
idx3 = tf.stack( [DIM//2-idx2[0,], DIM//2-1+idx2[1,]] )
d = tf.gather_nd(image,tf.transpose(idx3))
return tf.reshape(d,[DIM,DIM,3]),label
This seems like an affine transform of an image done the simple way(without multisampling), i.e. for all of destination pixels find corresponding pixel in source array and take its value.
destination pixel coorinates are put on a list idx, which is the transformed with matrix m, then they are cast into integers, and culled to image size.
idx3 is assembled from reverse-offset components (x,y) of source coorinates and used to index original image
why aren't they starting from 0,0 instead of DIM//2
(0,0) is top left corner of an image data array. When thinking of matrix transform you usually think that (0,0) point lies at the center of an image, and thus an offset
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 -
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.
I have thousands of polygons stored in a table format (given their 4 corner coordinates) which represent small regions of the earth. In addition, each polygon has a data value.
The file looks for example like this:
lat1, lat2, lat3, lat4, lon1, lon2, lon3, lon4, data
57.27, 57.72, 57.68, 58.1, 151.58, 152.06, 150.27, 150.72, 13.45
56.96, 57.41, 57.36, 57.79, 151.24, 151.72, 149.95, 150.39, 56.24
57.33, 57.75, 57.69, 58.1, 150.06, 150.51, 148.82, 149.23, 24.52
56.65, 57.09, 57.05, 57.47, 150.91, 151.38, 149.63, 150.06, 38.24
57.01, 57.44, 57.38, 57.78, 149.74, 150.18, 148.5, 148.91, 84.25
...
Many of the polygons intersect or overlap. Now I would like to create a n*m matrix ranging from -90° to 90° latitude and -180° to 180° longitude in steps of, for instance, 0.25°x0.25° to store the (area-weighted) mean data value of all polygons that fall within each pixel.
So, one pixel in the regular mesh shall get the mean value of one or more polygons (or none if no polygon overlaps with the pixel). Each polygon should contribute to this mean value depending on its area fraction within this pixel.
Basically the regular mesh and the polygons look like this:
If you look at pixel 2, you see that two polygons are inside this pixel. Thus, I have to take the mean data value of both polygons considering their area fractions. The result should be then stored in the regular mesh pixel.
I looked around the web and found no satisfactory approach for this so far. Since I am using Python/Numpy for daily work I would like to stick to it. Is this possible? The package shapely looks promising but I don't know where to begin with...
Porting everything to a postgis database is an awful amount of effort and I guess there will be quite a few obstacles in my way.
There are plenty of ways to do it, but yes, Shapely can help. It appears that your polygons are quadrilateral, but the approach I'll sketch doesn't count on that. You won't need anything other than box() and Polygon() from shapely.geometry.
For each pixel, find the polygons that approximately overlap with it by comparing the pixels bounds to the minimum bounding box of each polygon.
from shapely.geometry import box, Polygon
for pixel in pixels:
# say the pixel has llx, lly, urx, ury values.
pixel_shape = box(llx, lly, urx, ury)
for polygon in approximately_overlapping:
# say the polygon has a ``value`` and a 2-D array of coordinates
# [[x0,y0],...] named ``xy``.
polygon_shape = Polygon(xy)
pixel_value += polygon_shape.intersection(pixel_shape).area * value
If the pixel and polygon don't intersect, the area of their intersection will be 0 and the contribution of that polygon to that pixel vanishes.
I added a couple of things to my initial question, but this is a working solution so far. Do you have any ideas to speed things up? It is still quite slow. As input, I have over 100000 polygons and the meshgrid has 720*1440 grid cells. That is also why I changed the order, because there are a lot of grid cells with no intersecting polygons. Furthermore, when there is only one polygon that intersects with a grid cell, the grid cell receives the whole data value of the polygon.
In addition, since I have to store the area fraction and the data value for the "post-processing" part, I set the possible number of intersections to 10.
from shapely.geometry import box, Polygon
import h5py
import numpy as np
f = h5py.File('data.he5','r')
geo = f['geo'][:] #10 columns: 4xlat, lat center, 4xlon, lon center
product = f['product'][:]
f.close()
#prepare the regular meshgrid
delta = 0.25
darea = delta**-2
llx, lly = np.meshgrid( np.arange(-180, 180, delta), np.arange(-90, 90, delta) )
urx, ury = np.meshgrid( np.arange(-179.75, 180.25, delta), np.arange(-89.75, 90.25, delta) )
lly = np.flipud(lly)
ury = np.flipud(ury)
llx = llx.flatten()
lly = lly.flatten()
urx = urx.flatten()
ury = ury.flatten()
#initialize the data structures
data = np.zeros(len(llx),'f2')+np.nan
counter = np.zeros(len(llx),'f2')
fraction = np.zeros( (len(llx),10),'f2')
value = np.zeros( (len(llx),10),'f2')
#go through all polygons
for ii in np.arange(1000):#len(hcho)):
percent = (float(ii)/float(len(hcho)))*100
print("Polygon: %i (%0.3f %%)" % (ii, percent))
xy = [ [geo[ii,5],geo[ii,0]], [geo[ii,7],geo[ii,2]], [geo[ii,8],geo[ii,3]], [geo[ii,6],geo[ii,1]] ]
polygon_shape = Polygon(xy)
# only go through grid cells which might intersect with the polygon
minx = np.min( geo[ii,5:9] )
miny = np.min( geo[ii,:3] )
maxx = np.max( geo[ii,5:9] )
maxy = np.max( geo[ii,:3] )
mask = np.argwhere( (lly>=miny) & (lly<=maxy) & (llx>=minx) & (llx<=maxx) )
if mask.size:
cc = 0
for mm in mask:
cc = int(counter[mm])
pixel_shape = box(llx[mm], lly[mm], urx[mm], ury[mm])
fraction[mm,cc] = polygon_shape.intersection(pixel_shape).area * darea
value[mm,cc] = hcho[ii]
counter[mm] += 1
print("post-processing")
mask = np.argwhere(counter>0)
for mm in mask:
for cc in np.arange(counter[mm]):
maxfraction = np.sum(fraction[mm,:])
value[mm,cc] = (fraction[mm,cc]/maxfraction) * value[mm,cc]
data[mm] = np.mean(value[mm,:int(counter[mm])])
data = data.reshape( 720, 1440 )