I have dataframe which contains coordinates and measurements, something similar to this (this is fake):

id   lat          long        mes
0     -14.1309    -52.4561    0.1
1     -14.1312    -52.5327    0.05
2     -14.1308    -52.3324    0.07
3     -14.1302    -52.3323    0.03
2     -14.1302    -52.3312    0.01

I want to interpolate this data points. I want my final raster to have size that I have already defined (3586, 2284). I have tried to do something similar to this post :

xt,yt = df['long'].values, df['lat'].values
zt =  df['mes'].values

from scipy.interpolate import griddata
CONC = griddata((xt,yt), zt, method='cubic')

But then it says I'm missing the xi argument:

TypeError: griddata() missing 1 required positional argument: 'xi'

My end goal is to interpolate these points to get raster with the given dimensions (3586, 2284) with the correct coordinates. I'm also open to use other libraries, but seems like scipy is the best one.

Edit: I have tried @snowman2 solution, however it return empty raster:

geo_grid = make_geocube(
    resolution=(-0.1, 0.1),
    rasterize_function=partial(rasterize_points_griddata, method="cubic"),

enter image description here

Then I have tried also using xarray to interpolate , I could plot my data but could not interpolate:


idx = pd.MultiIndex.from_arrays(arrays=[Latitude,Longitude], names=["Latitude","Longitude"])
s = pd.Series(data=data, index=idx)
# use from_series method
da = xr.DataArray.from_series(s)

enter image description here

this can be plotted: enter image description here

But when it is interpolated, I get only nan:

dsi = da.interp(Latitude=Latitude, Longitude=Longitude,method='linear')

enter image description here

  • Suggest you read the help file for this tool?
    – Hornbydd
    Commented Aug 19, 2021 at 12:02
  • @Hornbydd yes, just not sure how to define the xi, I know what is the size e of the result raster- e.g the shape of the new array, I know the values, but I don't understand the xi, what does it mean point which to interpolate the data? is confusing me because I have already xt,xy , and I don't understand from the example in the original post what is it and how it was determined.
    – ReutKeller
    Commented Aug 19, 2021 at 13:21
  • 2
    xi is the coordinates at which you want to sample, so that would be the coordinates of your target raster's cell centers. You could create these with np.meshgrid(), see the examples here
    – mikewatt
    Commented Aug 19, 2021 at 16:42
  • @mikewatt thank you for your respond. There is still soemthing I don't understand - xi should be the coordinates of the final raster? of the full raster? i'm confused as in the post I based on, the write defined xx and yy with np.linspace(110, 120, 40) (and different numbrs for yy) and i'm not sure where it comes from. I have tried to put my xt and yt but that priduces array with no shape. soemthing here still confusing me.
    – ReutKeller
    Commented Aug 23, 2021 at 8:36
  • 2
    Check this one: gis.stackexchange.com/a/305894/28714
    – dmh126
    Commented Aug 26, 2021 at 11:30

3 Answers 3



import pandas

df = pandas.DataFrame({
    "lat": [-14.1309, -14.1312, -14.1308, -14.1302, -14.1302],
    "long": [-52.4561, -52.5327, -52.3324, -52.3323, -52.3312],
    "mes": [0.1, 0.05, 0.07, 0.03, 0.01],

Step 1: Convert to geodataframe


import geopandas

gdf = geopandas.GeoDataFrame(
    df, geometry=geopandas.points_from_xy(df['long'], df['lat']), crs="EPSG:4326",

Step 2: Convert to raster

from functools import partial

from geocube.api.core import make_geocube
from geocube.rasterize import rasterize_points_griddata

geo_grid_cubic = make_geocube(
    resolution=(-0.00001, 0.001),
    rasterize_function=partial(rasterize_points_griddata, method="cubic"),

Cubic interpolation

You can also fill in the missing data:

geo_grid_cubic = make_geocube(
    resolution=(-0.00001, 0.001),
    rasterize_function=partial(rasterize_points_griddata, method="cubic"),

Cubic interpolation with nearest fill

And you can use linear interpolation as well:

geo_grid_linear = make_geocube(
    resolution=(-0.00001, 0.001),
    rasterize_function=partial(rasterize_points_griddata, method="linear"),

Linear interpolation with nearest fill With radial interpolation:

geo_radial = make_geocube(
    resolution=(-0.00001, 0.001),

radial interpolation Default griddata interpolation:

geo_griddata = make_geocube(
    resolution=(-0.00001, 0.001),

griddata interpolation

  • I have tried your solution but got blank raster (see the original post, the edit section).
    – ReutKeller
    Commented Aug 30, 2021 at 8:20
  • I would try making the resolution smaller. Maybe 0.001?
    – snowman2
    Commented Aug 30, 2021 at 14:28
  • I updated the response with a working example.
    – snowman2
    Commented Aug 31, 2021 at 1:11

How about QGIS 3 and its IDW Interpolation tool?

enter image description here

Step-by-step instructions:

  1. Convert the text data into a vector layer (use Add Delimited Text Layer with Ctrl+ Shift +T key combination).
  2. [Optional] Convert a degree coordinate system (such as WGS 84) to a meter coordinate system (such as UTM 30N) with the Reproject layer tool.
  3. Use the IDW Interpolation tool to interpolate the data with the desired parameters (see picture above).

P.S. Your example data lies almost on the same line, so you can't use it as an example. Add more points to make it look more like a square.

  • I did not want to use QGIS, the whole point was for me to try solve it outside QGIS , but it's indeed good solution as well.
    – ReutKeller
    Commented Sep 5, 2021 at 6:54

Just an update for another solution I have found : using the Scikit GStat library . This solution worked for me the best. It allowed me to use Kriking interpolation in convinient wat with my data.

example for how I run kriging with this library:

df, geometry=gpd.points_from_xy(df.Longitude, df.Latitude))

#this is function from sentinel hub that I used but you can get the bbox with other libraries

totalPointsArray = np.zeros([df.shape[0],3])

for index, point in pdf.iterrows():
     pointArray = np.array([point.geometry.coords.xy[0][0],point.geometry.coords.xy[1][0],point['Data']])
        totalPointsArray[index] = pointArray
xmin = cords_bbox[0][0]
xmax = cords_bbox[2][0]
ymin = cords_bbox[2][1]
ymax = cords_bbox[0][1]
# number of pixels with 1m resolution
nx = (int(xmax - xmin + bbox_size[0]))
ny = (int(ymax - ymin + bbox_size[1]))
xi = np.linspace(xmin, xmax, nx)
yi = np.linspace(ymin, ymax, ny)
xi, yi = np.meshgrid(xi, yi)     



V = Variogram(data[['x', 'y']].values, data.z.values, normalize=False, maxlag=60, n_lags=15)
fig = V.plot(show=False)
print('Sample variance: %.2f   Variogram sill: %.2f' % (data.z.var(), V.describe()['sill']))
ok = OrdinaryKriging(V, min_points=5, max_points=15, mode='exact')
field = ok.transform(xi.flatten(), yi.flatten()).reshape(xi.shape)
s2 = ok.sigma.reshape(xi.shape)

plt.title('Kriging Interpolation')

Using Kriging produced better interpolation in this case and I found this library very nice and easy to use. They also ahve very nice and detailed tutorials in their documentation.

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