Interpolating within a grid in python

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I have a number of height values across a grid, specified in a series of lists:

[3, 1, -2, -3, -3] # x = 0m 

[2, -7, -14, -30, -39] # x = 10m

[46, 22, 5, -2, -8] # x = 20m

The example above shows a 40m x 20m grid in the form

[x=0y=0, x=0y=10...][x=10y=0, x=10y=10...] etc.

I need to calculate the height at a specific x,y coordinate. I have looked at various interpolate functions and example but can't make any sense out of them - please help!

An example would be the height at x = 5m, y = 5m in teh grid above, which would be in the middle of the values 3, 1 2 and -7 (-1ish?)

Thanks

asked Jan 3 at 12:26

tp503

Is it what you need? docs.scipy.org/doc/scipy-0.14.0/reference/generated/…

– dyukha
Jan 3 at 12:34

You may want to check this post

– Cedric Zoppolo
Jan 3 at 17:49

add a comment |

I have a number of height values across a grid, specified in a series of lists:

[3, 1, -2, -3, -3] # x = 0m 

[2, -7, -14, -30, -39] # x = 10m

[46, 22, 5, -2, -8] # x = 20m

The example above shows a 40m x 20m grid in the form

[x=0y=0, x=0y=10...][x=10y=0, x=10y=10...] etc.

I need to calculate the height at a specific x,y coordinate. I have looked at various interpolate functions and example but can't make any sense out of them - please help!

An example would be the height at x = 5m, y = 5m in teh grid above, which would be in the middle of the values 3, 1 2 and -7 (-1ish?)

Thanks

asked Jan 3 at 12:26

tp503

Is it what you need? docs.scipy.org/doc/scipy-0.14.0/reference/generated/…

– dyukha
Jan 3 at 12:34

You may want to check this post

– Cedric Zoppolo
Jan 3 at 17:49

add a comment |

I have a number of height values across a grid, specified in a series of lists:

[3, 1, -2, -3, -3] # x = 0m 

[2, -7, -14, -30, -39] # x = 10m

[46, 22, 5, -2, -8] # x = 20m

The example above shows a 40m x 20m grid in the form

[x=0y=0, x=0y=10...][x=10y=0, x=10y=10...] etc.

I need to calculate the height at a specific x,y coordinate. I have looked at various interpolate functions and example but can't make any sense out of them - please help!

An example would be the height at x = 5m, y = 5m in teh grid above, which would be in the middle of the values 3, 1 2 and -7 (-1ish?)

Thanks

asked Jan 3 at 12:26

tp503

I have a number of height values across a grid, specified in a series of lists:

[3, 1, -2, -3, -3] # x = 0m 

[2, -7, -14, -30, -39] # x = 10m

[46, 22, 5, -2, -8] # x = 20m

The example above shows a 40m x 20m grid in the form

[x=0y=0, x=0y=10...][x=10y=0, x=10y=10...] etc.

I need to calculate the height at a specific x,y coordinate. I have looked at various interpolate functions and example but can't make any sense out of them - please help!

An example would be the height at x = 5m, y = 5m in teh grid above, which would be in the middle of the values 3, 1 2 and -7 (-1ish?)

Thanks

python interpolation

asked Jan 3 at 12:26

tp503

asked Jan 3 at 12:26

tp503

asked Jan 3 at 12:26

tp503

asked Jan 3 at 12:26

tp503

asked Jan 3 at 12:26

tp503

Is it what you need? docs.scipy.org/doc/scipy-0.14.0/reference/generated/…

– dyukha
Jan 3 at 12:34

You may want to check this post

– Cedric Zoppolo
Jan 3 at 17:49

add a comment |

Is it what you need? docs.scipy.org/doc/scipy-0.14.0/reference/generated/…

– dyukha
Jan 3 at 12:34

You may want to check this post

– Cedric Zoppolo
Jan 3 at 17:49

Is it what you need? docs.scipy.org/doc/scipy-0.14.0/reference/generated/…

– dyukha
Jan 3 at 12:34

You may want to check this post

– Cedric Zoppolo
Jan 3 at 17:49

add a comment |

3 Answers
3

active

oldest

votes

You can use scipy.interpolate.interp2d. The code below should do the job:

import numpy as np

from scipy.interpolate import interp2d



# Original data (e.g. measurements)

a = [3, 1, -2, -3, -3]

b = [2, -7, -14, -30, -39]

c = [46, 22, 5, -2, -8]



x = [0, 10, 20]             # x-coordinates

y = [0, 10, 20, 30, 40]     # y-coordinates



# Organise data in matrix

z = np.vstack([a, b, c]).T



# Create interpolation function

f_z = interp2d(x, y, z)



# Desired x/y values

x_interp = 5

y_interp = 5



# Collect interpolated z-value

z_interp = f_z(x_interp, y_interp)

print(z_interp)  # (result: [-0.25])

answered Jan 3 at 14:02

MPA

84711429

This works perfectly, thank you!

– tp503
Jan 3 at 16:20

add a comment |

Solution: Transform into 2D array

import numpy as np



a = [3, 1, -2, -3, -3] # x = 0m 

b = [2, -7, -14, -30, -39] # x = 10m

c = [46, 22, 5, -2, -8] # x = 20m



all_lists = [a,b,c]





grid = np.vstack(all_lists) # Edit: thanks @jochen for better code



>>print(grid.shape)

(3, 5)

Now you can access grid via x,y coordinates

grid[1,1] and etc

Based on this, you can create logic.

If x = 5, y = 5, then you can create square corners indexes by this:

indexes: x/5 +0, x/5+1, y/5,y/5+1 ---> by combining them you get [0,0],[0,1],[1,0],[1,1]

At the end you just feed grid with those indexes

SUM = grid[0,0] + grid[0,1] + grid[1,0] + grid[1,1]

edited Jan 3 at 14:50

answered Jan 3 at 12:44

Martin

1,5081516

The construction of your grid is unnecessary complex. It could be grid = np.vstack([a, b, c]) Your code actually constructs and allocates memory for 3 grids, discarding the first two.

– jochen
Jan 3 at 14:10

thanks. I tried vstack and it just extended array, but with your code, it does it better

– Martin
Jan 3 at 14:46

add a comment |

You can achieve this with interpolate.gridddata function from scipy.

With below code you can get the any interpolation you want from your grid.

import itertools

import numpy as np

from scipy.interpolate import griddata



dataX0 = [3, 1, -2, -3, -3] # x = 0m

dataX10 = [2, -7, -14, -30, -39] # x = 10m

dataX20 = [46, 22, 5, -2, -8] # x = 20m

data = dataX0 + dataX10 + dataX20

points = list(itertools.product(range(0,30,10),range(0,50,10)))



outputPoint = (5,5)

outputValue = griddata(points=points,values=data, xi=outputPoint, method="cubic")

print outputValue

The example above give you the interpolation at ouputPoint (5,5). And the output would give:

>>> 

-2.78976054957

edited Jan 3 at 15:03

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

You may want to check this post

– Cedric Zoppolo
Jan 3 at 15:01

1

The post you linked applies to most answers given here. Perhaps you could add it as a comment to the original question, so that it is more likely to get noticed by other readers?

– MPA
Jan 3 at 15:19

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

You can use scipy.interpolate.interp2d. The code below should do the job:

import numpy as np

from scipy.interpolate import interp2d



# Original data (e.g. measurements)

a = [3, 1, -2, -3, -3]

b = [2, -7, -14, -30, -39]

c = [46, 22, 5, -2, -8]



x = [0, 10, 20]             # x-coordinates

y = [0, 10, 20, 30, 40]     # y-coordinates



# Organise data in matrix

z = np.vstack([a, b, c]).T



# Create interpolation function

f_z = interp2d(x, y, z)



# Desired x/y values

x_interp = 5

y_interp = 5



# Collect interpolated z-value

z_interp = f_z(x_interp, y_interp)

print(z_interp)  # (result: [-0.25])

answered Jan 3 at 14:02

MPA

84711429

This works perfectly, thank you!

– tp503
Jan 3 at 16:20

add a comment |

You can use scipy.interpolate.interp2d. The code below should do the job:

import numpy as np

from scipy.interpolate import interp2d



# Original data (e.g. measurements)

a = [3, 1, -2, -3, -3]

b = [2, -7, -14, -30, -39]

c = [46, 22, 5, -2, -8]



x = [0, 10, 20]             # x-coordinates

y = [0, 10, 20, 30, 40]     # y-coordinates



# Organise data in matrix

z = np.vstack([a, b, c]).T



# Create interpolation function

f_z = interp2d(x, y, z)



# Desired x/y values

x_interp = 5

y_interp = 5



# Collect interpolated z-value

z_interp = f_z(x_interp, y_interp)

print(z_interp)  # (result: [-0.25])

answered Jan 3 at 14:02

MPA

84711429

This works perfectly, thank you!

– tp503
Jan 3 at 16:20

add a comment |

You can use scipy.interpolate.interp2d. The code below should do the job:

import numpy as np

from scipy.interpolate import interp2d



# Original data (e.g. measurements)

a = [3, 1, -2, -3, -3]

b = [2, -7, -14, -30, -39]

c = [46, 22, 5, -2, -8]



x = [0, 10, 20]             # x-coordinates

y = [0, 10, 20, 30, 40]     # y-coordinates



# Organise data in matrix

z = np.vstack([a, b, c]).T



# Create interpolation function

f_z = interp2d(x, y, z)



# Desired x/y values

x_interp = 5

y_interp = 5



# Collect interpolated z-value

z_interp = f_z(x_interp, y_interp)

print(z_interp)  # (result: [-0.25])

answered Jan 3 at 14:02

MPA

84711429

You can use scipy.interpolate.interp2d. The code below should do the job:

import numpy as np

from scipy.interpolate import interp2d



# Original data (e.g. measurements)

a = [3, 1, -2, -3, -3]

b = [2, -7, -14, -30, -39]

c = [46, 22, 5, -2, -8]



x = [0, 10, 20]             # x-coordinates

y = [0, 10, 20, 30, 40]     # y-coordinates



# Organise data in matrix

z = np.vstack([a, b, c]).T



# Create interpolation function

f_z = interp2d(x, y, z)



# Desired x/y values

x_interp = 5

y_interp = 5



# Collect interpolated z-value

z_interp = f_z(x_interp, y_interp)

print(z_interp)  # (result: [-0.25])

answered Jan 3 at 14:02

MPA

84711429

answered Jan 3 at 14:02

MPA

84711429

answered Jan 3 at 14:02

MPA

84711429

answered Jan 3 at 14:02

MPA

84711429

This works perfectly, thank you!

– tp503
Jan 3 at 16:20

add a comment |

This works perfectly, thank you!

– tp503
Jan 3 at 16:20

This works perfectly, thank you!

– tp503
Jan 3 at 16:20

add a comment |

Solution: Transform into 2D array

import numpy as np



a = [3, 1, -2, -3, -3] # x = 0m 

b = [2, -7, -14, -30, -39] # x = 10m

c = [46, 22, 5, -2, -8] # x = 20m



all_lists = [a,b,c]





grid = np.vstack(all_lists) # Edit: thanks @jochen for better code



>>print(grid.shape)

(3, 5)

Now you can access grid via x,y coordinates

grid[1,1] and etc

Based on this, you can create logic.

If x = 5, y = 5, then you can create square corners indexes by this:

indexes: x/5 +0, x/5+1, y/5,y/5+1 ---> by combining them you get [0,0],[0,1],[1,0],[1,1]

At the end you just feed grid with those indexes

SUM = grid[0,0] + grid[0,1] + grid[1,0] + grid[1,1]

edited Jan 3 at 14:50

answered Jan 3 at 12:44

Martin

1,5081516

The construction of your grid is unnecessary complex. It could be grid = np.vstack([a, b, c]) Your code actually constructs and allocates memory for 3 grids, discarding the first two.

– jochen
Jan 3 at 14:10

thanks. I tried vstack and it just extended array, but with your code, it does it better

– Martin
Jan 3 at 14:46

add a comment |

Solution: Transform into 2D array

import numpy as np



a = [3, 1, -2, -3, -3] # x = 0m 

b = [2, -7, -14, -30, -39] # x = 10m

c = [46, 22, 5, -2, -8] # x = 20m



all_lists = [a,b,c]





grid = np.vstack(all_lists) # Edit: thanks @jochen for better code



>>print(grid.shape)

(3, 5)

Now you can access grid via x,y coordinates

grid[1,1] and etc

Based on this, you can create logic.

If x = 5, y = 5, then you can create square corners indexes by this:

indexes: x/5 +0, x/5+1, y/5,y/5+1 ---> by combining them you get [0,0],[0,1],[1,0],[1,1]

At the end you just feed grid with those indexes

SUM = grid[0,0] + grid[0,1] + grid[1,0] + grid[1,1]

edited Jan 3 at 14:50

answered Jan 3 at 12:44

Martin

1,5081516

The construction of your grid is unnecessary complex. It could be grid = np.vstack([a, b, c]) Your code actually constructs and allocates memory for 3 grids, discarding the first two.

– jochen
Jan 3 at 14:10

thanks. I tried vstack and it just extended array, but with your code, it does it better

– Martin
Jan 3 at 14:46

add a comment |

Solution: Transform into 2D array

import numpy as np



a = [3, 1, -2, -3, -3] # x = 0m 

b = [2, -7, -14, -30, -39] # x = 10m

c = [46, 22, 5, -2, -8] # x = 20m



all_lists = [a,b,c]





grid = np.vstack(all_lists) # Edit: thanks @jochen for better code



>>print(grid.shape)

(3, 5)

Now you can access grid via x,y coordinates

grid[1,1] and etc

Based on this, you can create logic.

If x = 5, y = 5, then you can create square corners indexes by this:

indexes: x/5 +0, x/5+1, y/5,y/5+1 ---> by combining them you get [0,0],[0,1],[1,0],[1,1]

At the end you just feed grid with those indexes

SUM = grid[0,0] + grid[0,1] + grid[1,0] + grid[1,1]

edited Jan 3 at 14:50

answered Jan 3 at 12:44

Martin

1,5081516

Solution: Transform into 2D array

import numpy as np



a = [3, 1, -2, -3, -3] # x = 0m 

b = [2, -7, -14, -30, -39] # x = 10m

c = [46, 22, 5, -2, -8] # x = 20m



all_lists = [a,b,c]





grid = np.vstack(all_lists) # Edit: thanks @jochen for better code



>>print(grid.shape)

(3, 5)

Now you can access grid via x,y coordinates

grid[1,1] and etc

Based on this, you can create logic.

If x = 5, y = 5, then you can create square corners indexes by this:

indexes: x/5 +0, x/5+1, y/5,y/5+1 ---> by combining them you get [0,0],[0,1],[1,0],[1,1]

At the end you just feed grid with those indexes

SUM = grid[0,0] + grid[0,1] + grid[1,0] + grid[1,1]

edited Jan 3 at 14:50

answered Jan 3 at 12:44

Martin

1,5081516

edited Jan 3 at 14:50

answered Jan 3 at 12:44

Martin

1,5081516

answered Jan 3 at 12:44

Martin

1,5081516

answered Jan 3 at 12:44

Martin

1,5081516

The construction of your grid is unnecessary complex. It could be grid = np.vstack([a, b, c]) Your code actually constructs and allocates memory for 3 grids, discarding the first two.

– jochen
Jan 3 at 14:10

thanks. I tried vstack and it just extended array, but with your code, it does it better

– Martin
Jan 3 at 14:46

add a comment |

The construction of your grid is unnecessary complex. It could be grid = np.vstack([a, b, c]) Your code actually constructs and allocates memory for 3 grids, discarding the first two.

– jochen
Jan 3 at 14:10

thanks. I tried vstack and it just extended array, but with your code, it does it better

– Martin
Jan 3 at 14:46

The construction of your grid is unnecessary complex. It could be grid = np.vstack([a, b, c]) Your code actually constructs and allocates memory for 3 grids, discarding the first two.

– jochen
Jan 3 at 14:10

thanks. I tried vstack and it just extended array, but with your code, it does it better

– Martin
Jan 3 at 14:46

add a comment |

You can achieve this with interpolate.gridddata function from scipy.

With below code you can get the any interpolation you want from your grid.

import itertools

import numpy as np

from scipy.interpolate import griddata



dataX0 = [3, 1, -2, -3, -3] # x = 0m

dataX10 = [2, -7, -14, -30, -39] # x = 10m

dataX20 = [46, 22, 5, -2, -8] # x = 20m

data = dataX0 + dataX10 + dataX20

points = list(itertools.product(range(0,30,10),range(0,50,10)))



outputPoint = (5,5)

outputValue = griddata(points=points,values=data, xi=outputPoint, method="cubic")

print outputValue

The example above give you the interpolation at ouputPoint (5,5). And the output would give:

>>> 

-2.78976054957

edited Jan 3 at 15:03

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

You may want to check this post

– Cedric Zoppolo
Jan 3 at 15:01

1

The post you linked applies to most answers given here. Perhaps you could add it as a comment to the original question, so that it is more likely to get noticed by other readers?

– MPA
Jan 3 at 15:19

add a comment |

You can achieve this with interpolate.gridddata function from scipy.

With below code you can get the any interpolation you want from your grid.

import itertools

import numpy as np

from scipy.interpolate import griddata



dataX0 = [3, 1, -2, -3, -3] # x = 0m

dataX10 = [2, -7, -14, -30, -39] # x = 10m

dataX20 = [46, 22, 5, -2, -8] # x = 20m

data = dataX0 + dataX10 + dataX20

points = list(itertools.product(range(0,30,10),range(0,50,10)))



outputPoint = (5,5)

outputValue = griddata(points=points,values=data, xi=outputPoint, method="cubic")

print outputValue

The example above give you the interpolation at ouputPoint (5,5). And the output would give:

>>> 

-2.78976054957

edited Jan 3 at 15:03

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

You may want to check this post

– Cedric Zoppolo
Jan 3 at 15:01

1

The post you linked applies to most answers given here. Perhaps you could add it as a comment to the original question, so that it is more likely to get noticed by other readers?

– MPA
Jan 3 at 15:19

add a comment |

You can achieve this with interpolate.gridddata function from scipy.

With below code you can get the any interpolation you want from your grid.

import itertools

import numpy as np

from scipy.interpolate import griddata



dataX0 = [3, 1, -2, -3, -3] # x = 0m

dataX10 = [2, -7, -14, -30, -39] # x = 10m

dataX20 = [46, 22, 5, -2, -8] # x = 20m

data = dataX0 + dataX10 + dataX20

points = list(itertools.product(range(0,30,10),range(0,50,10)))



outputPoint = (5,5)

outputValue = griddata(points=points,values=data, xi=outputPoint, method="cubic")

print outputValue

The example above give you the interpolation at ouputPoint (5,5). And the output would give:

>>> 

-2.78976054957

edited Jan 3 at 15:03

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

You can achieve this with interpolate.gridddata function from scipy.

With below code you can get the any interpolation you want from your grid.

import itertools

import numpy as np

from scipy.interpolate import griddata



dataX0 = [3, 1, -2, -3, -3] # x = 0m

dataX10 = [2, -7, -14, -30, -39] # x = 10m

dataX20 = [46, 22, 5, -2, -8] # x = 20m

data = dataX0 + dataX10 + dataX20

points = list(itertools.product(range(0,30,10),range(0,50,10)))



outputPoint = (5,5)

outputValue = griddata(points=points,values=data, xi=outputPoint, method="cubic")

print outputValue

The example above give you the interpolation at ouputPoint (5,5). And the output would give:

>>> 

-2.78976054957

edited Jan 3 at 15:03

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

edited Jan 3 at 15:03

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

answered Jan 3 at 14:48

Cedric Zoppolo

1,36211529

You may want to check this post

– Cedric Zoppolo
Jan 3 at 15:01

1

The post you linked applies to most answers given here. Perhaps you could add it as a comment to the original question, so that it is more likely to get noticed by other readers?

– MPA
Jan 3 at 15:19

add a comment |

You may want to check this post

– Cedric Zoppolo
Jan 3 at 15:01

1

The post you linked applies to most answers given here. Perhaps you could add it as a comment to the original question, so that it is more likely to get noticed by other readers?

– MPA
Jan 3 at 15:19

You may want to check this post

– Cedric Zoppolo
Jan 3 at 15:01

The post you linked applies to most answers given here. Perhaps you could add it as a comment to the original question, so that it is more likely to get noticed by other readers?

– MPA
Jan 3 at 15:19

add a comment |

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