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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)
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 data that are multidimensional compositional data (all dimensions sum to 1 or 100). I have learned how to use three of the variables to create a 2d ternary plot.
I would like to add a fourth dimension such that my plot looks like this.
I am willing to use python or R. I am using pyr2 to create the ternary plots in python using R right now, but just because that's an easy solution. If the ternary data could be transformed into 3d coordinates a simple wire plot could be used.
This post shows how 3d compositional data can be transformed into 2d data so that normal plotting method can be used. One solution would be to do the same thing in 3d.
Here is some sample Data:
c1 c2 c3 c4
0 0.082337 0.097583 0.048608 0.771472
1 0.116490 0.065047 0.066202 0.752261
2 0.114884 0.135018 0.073870 0.676229
3 0.071027 0.097207 0.070959 0.760807
4 0.066284 0.079842 0.103915 0.749959
5 0.016074 0.074833 0.044532 0.864561
6 0.066277 0.077837 0.058364 0.797522
7 0.055549 0.057117 0.045633 0.841701
8 0.071129 0.077620 0.049066 0.802185
9 0.089790 0.086967 0.083101 0.740142
10 0.084430 0.094489 0.039989 0.781093
Well, I solved this myself using a wikipedia article, an SO post, and some brute force. Sorry for the wall of code, but you have to draw all the plot outlines and labels and so forth.
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import axes3d, Axes3D
from itertools import combinations
import pandas as pd
def plot_ax(): #plot tetrahedral outline
verts=[[0,0,0],
[1,0,0],
[0.5,np.sqrt(3)/2,0],
[0.5,0.28867513, 0.81649658]]
lines=combinations(verts,2)
for x in lines:
line=np.transpose(np.array(x))
ax.plot3D(line[0],line[1],line[2],c='0')
def label_points(): #create labels of each vertices of the simplex
a=(np.array([1,0,0,0])) # Barycentric coordinates of vertices (A or c1)
b=(np.array([0,1,0,0])) # Barycentric coordinates of vertices (B or c2)
c=(np.array([0,0,1,0])) # Barycentric coordinates of vertices (C or c3)
d=(np.array([0,0,0,1])) # Barycentric coordinates of vertices (D or c3)
labels=['a','b','c','d']
cartesian_points=get_cartesian_array_from_barycentric([a,b,c,d])
for point,label in zip(cartesian_points,labels):
if 'a' in label:
ax.text(point[0],point[1]-0.075,point[2], label, size=16)
elif 'b' in label:
ax.text(point[0]+0.02,point[1]-0.02,point[2], label, size=16)
else:
ax.text(point[0],point[1],point[2], label, size=16)
def get_cartesian_array_from_barycentric(b): #tranform from "barycentric" composition space to cartesian coordinates
verts=[[0,0,0],
[1,0,0],
[0.5,np.sqrt(3)/2,0],
[0.5,0.28867513, 0.81649658]]
#create transformation array vis https://en.wikipedia.org/wiki/Barycentric_coordinate_system
t = np.transpose(np.array(verts))
t_array=np.array([t.dot(x) for x in b]) #apply transform to all points
return t_array
def plot_3d_tern(df,c='1'): #use function "get_cartesian_array_from_barycentric" to plot the scatter points
#args are b=dataframe to plot and c=scatter point color
bary_arr=df.values
cartesian_points=get_cartesian_array_from_barycentric(bary_arr)
ax.scatter(cartesian_points[:,0],cartesian_points[:,1],cartesian_points[:,2],c=c)
#Create Dataset 1
np.random.seed(123)
c1=np.random.normal(8,2.5,20)
c2=np.random.normal(8,2.5,20)
c3=np.random.normal(8,2.5,20)
c4=[100-x for x in c1+c2+c3] #make sur ecomponents sum to 100
#df unecessary but that is the format of my real data
df1=pd.DataFrame(data=[c1,c2,c3,c4],index=['c1','c2','c3','c4']).T
df1=df1/100
#Create Dataset 2
np.random.seed(1234)
c1=np.random.normal(16,2.5,20)
c2=np.random.normal(16,2.5,20)
c3=np.random.normal(16,2.5,20)
c4=[100-x for x in c1+c2+c3]
df2=pd.DataFrame(data=[c1,c2,c3,c4],index=['c1','c2','c3','c4']).T
df2=df2/100
#Create Dataset 3
np.random.seed(12345)
c1=np.random.normal(25,2.5,20)
c2=np.random.normal(25,2.5,20)
c3=np.random.normal(25,2.5,20)
c4=[100-x for x in c1+c2+c3]
df3=pd.DataFrame(data=[c1,c2,c3,c4],index=['c1','c2','c3','c4']).T
df3=df3/100
fig = plt.figure()
ax = Axes3D(fig) #Create a 3D plot in most recent version of matplot
plot_ax() #call function to draw tetrahedral outline
label_points() #label the vertices
plot_3d_tern(df1,'b') #call function to plot df1
plot_3d_tern(df2,'r') #...plot df2
plot_3d_tern(df3,'g') #...
The accepted answer explains how to do this in python but the question was also asking about R.
I've provided an answer in this thread on how to do this 'manually' in R.
Otherwise, you can use the klaR package directly for this:
df <- matrix(c(
0.082337, 0.097583, 0.048608, 0.771472,
0.116490, 0.065047, 0.066202, 0.752261,
0.114884, 0.135018, 0.073870, 0.676229,
0.071027, 0.097207, 0.070959, 0.760807,
0.066284, 0.079842, 0.103915, 0.749959,
0.016074, 0.074833, 0.044532, 0.864561,
0.066277, 0.077837, 0.058364, 0.797522,
0.055549, 0.057117, 0.045633, 0.841701,
0.071129, 0.077620, 0.049066, 0.802185,
0.089790, 0.086967, 0.083101, 0.740142,
0.084430, 0.094489, 0.039989, 0.781094
), byrow = TRUE, nrow = 11, ncol = 4)
# install.packages(c("klaR", "scatterplot3d"))
library(klaR)
#> Loading required package: MASS
quadplot(df)
Created on 2020-08-14 by the reprex package (v0.3.0)
I'm moving from basemap to cartopy given basemap is going to be phased out. I've previously used the basemap.interp functionality to interpolate data, e.g. say I have data at 1 degree resolution (180x360), I would run the following to interpolate to 0.5 degrees.
import numpy as np
from mpl_toolkits import basemap
Old_Lon = np.linspace(-180,180,360)
Old_Lat = np.linspace(-90,90,180)
New_Lon = np.linspace(-180,180,720)
New_Lat = np.linspace(-90,90,360)
New_Lon,New_Lat = np.meshgrid(New_Lon,New_Lat)
New_Data = basemap.interp(Old_Data,Old_Lon,Old_Lat,New_Lon,New_Lat,order=0)
order gives me options to choose from nearest neighbour, bi-linear etc. Is there an alternative that does this in as simple way? I've seen scipy has interpolation but I'm not sure how to apply it. Any help would be appreciated!
I eventually decided to take the raw code from Basemap and make it into a standalone function - I'll be recommending it to the cartopy guys to implement it as its a useful feature. Posting here as could be useful to someone else:
def Interp(datain,xin,yin,xout,yout,interpolation='NearestNeighbour'):
"""
Interpolates a 2D array onto a new grid (only works for linear grids),
with the Lat/Lon inputs of the old and new grid. Can perfom nearest
neighbour interpolation or bilinear interpolation (of order 1)'
This is an extract from the basemap module (truncated)
"""
# Mesh Coordinates so that they are both 2D arrays
xout,yout = np.meshgrid(xout,yout)
# compute grid coordinates of output grid.
delx = xin[1:]-xin[0:-1]
dely = yin[1:]-yin[0:-1]
xcoords = (len(xin)-1)*(xout-xin[0])/(xin[-1]-xin[0])
ycoords = (len(yin)-1)*(yout-yin[0])/(yin[-1]-yin[0])
xcoords = np.clip(xcoords,0,len(xin)-1)
ycoords = np.clip(ycoords,0,len(yin)-1)
# Interpolate to output grid using nearest neighbour
if interpolation == 'NearestNeighbour':
xcoordsi = np.around(xcoords).astype(np.int32)
ycoordsi = np.around(ycoords).astype(np.int32)
dataout = datain[ycoordsi,xcoordsi]
# Interpolate to output grid using bilinear interpolation.
elif interpolation == 'Bilinear':
xi = xcoords.astype(np.int32)
yi = ycoords.astype(np.int32)
xip1 = xi+1
yip1 = yi+1
xip1 = np.clip(xip1,0,len(xin)-1)
yip1 = np.clip(yip1,0,len(yin)-1)
delx = xcoords-xi.astype(np.float32)
dely = ycoords-yi.astype(np.float32)
dataout = (1.-delx)*(1.-dely)*datain[yi,xi] + \
delx*dely*datain[yip1,xip1] + \
(1.-delx)*dely*datain[yip1,xi] + \
delx*(1.-dely)*datain[yi,xip1]
return dataout
--
The SciPy interpolation routines return a function that you can call to perform an interpolation. For nearest neighbour interpolation on a regular grid, you can use scipy.interpolate.RegularGridInterpolator:
import numpy as np
from scipy.interpolate import RegularGridInterpolator
nearest_function = RegularGridInterpolator(
(old_lon, old_lat), old_data, method="nearest", bounds_error=False
)
new_data = np.array(
[[nearest_function([i, j]) for j in new_lat] for i in new_lon]
).squeeze()
That isn't perfect, though, because lon=175 are all fill values. (If I hadn't set bounds_error=False then you'd get an error there.) In that case, you need to ask how you want to wrap around the dateline. A straightforward solution would be to copy the lon=0 line to the end of the array and call it lon=180.
Should you want linear or higher order interpolation one day, which I'd recommend if your data are points rather than cells, you can use scipy.interpolate.RectBivariateSpline:
import numpy as np
from scipy.interpolate import RectBivariateSpline
old_step = 10
old_lon = np.arange(-180, 180, old_step)
old_lat = np.arange(-90, 90, old_step)
old_data = np.random.random((len(old_lon), len(old_lat)))
interp_function = RectBivariateSpline(old_lon, old_lat, old_data, kx=1, ky=1)
new_lon = np.arange(-180, 180, new_step)
new_lat = np.arange(-90, 90, new_step)
new_data = interp_function(new_lon, new_lat)
I am trying to write a simple python code for a plot of intensity vs wavelength for a given temperature, T=200K.
So far I have this...
import scipy as sp
import math
import matplotlib.pyplot as plt
import numpy as np
pi = np.pi
h = 6.626e-34
c = 3.0e+8
k = 1.38e-23
def planck(wav, T):
a = 2.0*h*pi*c**2
b = h*c/(wav*k*T)
intensity = a/ ( (wav**5)*(math.e**b - 1.0) )
return intensity
I don't know how to define wavelength(wav) and thus produce the plot of Plancks Formula. Any help would be appreciated.
Here's a basic plot. To plot using plt.plot(x, y, fmt) you need two arrays x and y of the same size, where x is the x coordinate of each point to plot and y is the y coordinate, and fmt is a string describing how to plot the numbers.
So all you need to do is create an evenly spaced array of wavelengths (an np.array which I named wavelengths). This can be done with arange(start, end, spacing) which will create an array from start to end (not inclusive) spaced at spacing apart.
Then compute the intensity using your function at each of those points in the array (which will be stored in another np.array), and then call plt.plot to plot them. Note numpy let's you do mathematical operations on arrays quickly in a vectorized form which will be computationally efficient.
import matplotlib.pyplot as plt
import numpy as np
h = 6.626e-34
c = 3.0e+8
k = 1.38e-23
def planck(wav, T):
a = 2.0*h*c**2
b = h*c/(wav*k*T)
intensity = a/ ( (wav**5) * (np.exp(b) - 1.0) )
return intensity
# generate x-axis in increments from 1nm to 3 micrometer in 1 nm increments
# starting at 1 nm to avoid wav = 0, which would result in division by zero.
wavelengths = np.arange(1e-9, 3e-6, 1e-9)
# intensity at 4000K, 5000K, 6000K, 7000K
intensity4000 = planck(wavelengths, 4000.)
intensity5000 = planck(wavelengths, 5000.)
intensity6000 = planck(wavelengths, 6000.)
intensity7000 = planck(wavelengths, 7000.)
plt.plot(wavelengths*1e9, intensity4000, 'r-')
# plot intensity4000 versus wavelength in nm as a red line
plt.plot(wavelengths*1e9, intensity5000, 'g-') # 5000K green line
plt.plot(wavelengths*1e9, intensity6000, 'b-') # 6000K blue line
plt.plot(wavelengths*1e9, intensity7000, 'k-') # 7000K black line
# show the plot
plt.show()
And you see:
You probably will want to clean up the axes labels, add a legend, plot the intensity at multiple temperatures on the same plot, among other things. Consult the relevant matplotlib documentation.
You may also want to use the RADIS library, which allows you to plot the Planck function against wavelengths, or against frequency / wavenumber, if needed !
from radis import sPlanck
sPlanck(wavelength_min=135, wavelength_max=3000, T=4000).plot()
sPlanck(wavelength_min=135, wavelength_max=3000, T=5000).plot(nfig='same')
sPlanck(wavelength_min=135, wavelength_max=3000, T=6000).plot(nfig='same')
sPlanck(wavelength_min=135, wavelength_max=3000, T=7000).plot(nfig='same')
Just want to point out that there seems to be an equivalent of what OP wants to do in astropy:
https://docs.astropy.org/en/stable/api/astropy.modeling.physical_models.BlackBody.html
Unfortunately, it is not very clear to me yet how to get wavelength vs frequency based expression.