7

I would like to interpolate 2D array "test" whose dimensions are 4x4 (just as example, in reality close to 1000x1000) with a grid of shape 8x8.

import numpy as np

X = np.arange(0,4,1)
Y = np.arange(0,4,1)

points = np.vstack((X,Y))
points = points.T #my coordinates

#my values as a 2D array
test = np.array([[ 1.2514318 ,  1.25145821,  1.25148472,  1.25151133],
       [ 1.25087456,  1.25090105,  1.25092764,  1.25095435],
       [ 1.25031581,  1.25034238,  1.25036907,  1.25039586],
       [ 1.24975557,  1.24978222,  1.24980898,  1.24983587]])

I try with griddata but it seems work only 1D isnt it? as the errors tells me i have "different number of values and points" Do i make a mistake? 

from scipy.interpolate import griddata
grid_x, grid_y = np.mgrid[0:4:8j, 0:4:8j]
grid_z0 = griddata(points, test, (grid_x, grid_y), method='linear')
user3601754
  • 3,792
  • 11
  • 43
  • 77

2 Answers2

7

you can do this with scipy.interpolate.interp2d and numpy.meshgrid.

You need to make sure your new X and Y ranges go over the same range as the old ones, just with a smaller stepsize. This is easy with np.linspace:

import numpy as np
from scipy import interpolate

mymin,mymax = 0,3
X = np.linspace(mymin,mymax,4)
Y = np.linspace(mymin,mymax,4)

x,y = np.meshgrid(X,Y)

test = np.array([[ 1.2514318 ,  1.25145821,  1.25148472,  1.25151133],
       [ 1.25087456,  1.25090105,  1.25092764,  1.25095435],
       [ 1.25031581,  1.25034238,  1.25036907,  1.25039586],
       [ 1.24975557,  1.24978222,  1.24980898,  1.24983587]])

f = interpolate.interp2d(x,y,test,kind='cubic')

# use linspace so your new range also goes from 0 to 3, with 8 intervals
Xnew = np.linspace(mymin,mymax,8)
Ynew = np.linspace(mymin,mymax,8)

test8x8 = f(Xnew,Ynew)

print test8x8
>>> [[ 1.2514318   1.25144311  1.25145443  1.25146577  1.25147714  1.25148852  1.25149991  1.25151133]
     [ 1.25119317  1.25120449  1.25121583  1.25122719  1.25123856  1.25124995  1.25126137  1.25127281]
     [ 1.25095426  1.2509656   1.25097695  1.25098832  1.25099971  1.25101112  1.25102255  1.25103401]
     [ 1.25071507  1.25072642  1.25073779  1.25074918  1.25076059  1.25077201  1.25078346  1.25079494]
     [ 1.25047561  1.25048697  1.25049835  1.25050976  1.25052119  1.25053263  1.2505441   1.25055558]
     [ 1.25023587  1.25024724  1.25025864  1.25027007  1.25028151  1.25029297  1.25030446  1.25031595]
     [ 1.24999585  1.25000724  1.25001866  1.2500301   1.25004156  1.25005304  1.25006453  1.25007605]
     [ 1.24975557  1.24976698  1.24977841  1.24978985  1.24980132  1.24981281  1.24982433  1.24983587]]
tmdavison
  • 64,360
  • 12
  • 187
  • 165
  • That works! ;) but for my case, it seems to be that there it "Too many data points to interpolate" as error :( – user3601754 Oct 21 '15 at 15:35
  • how big is your new array? maybe you could split it into sub-arrays to perform the interpolation, and then stitch them together afterwards? – tmdavison Oct 21 '15 at 17:00
  • 1
    You can reduce your memory overhead a bit by creating an open grid for your `x` and `y` coordinates, i.e. an `(nx, 1)` array and an `(1, ny)` rather than two `(nx, ny)` arrays. For example, `x, y = np.meshgrid(X, Y, sparse=True)` or `x, y = np.ix_(X, Y)`. – ali_m Oct 21 '15 at 18:39
  • 1
    Shouldn't scipy.interpolate.RegularGridInterpolator do the job much more efficiently? – fdermishin Nov 22 '17 at 22:17
  • Why are you picking 3 as `mymax` here? I'm unsure how that choice affects the final result. – curious_cosmo Aug 06 '18 at 17:10
1

A convenient and fast way to do this would be by using skimage.transform.resize. It worked well for me also on large meshgrids:

import numpy as np
from skimage.transform import resize

test = np.random.rand(1000,1000)

dim1, dim2 = 8, 8

test_resized = resize(test,(dim1,dim2))

print(test_resized.shape)
>>> (8, 8)
MKo
  • 56
  • 3