I want iterate through the for loop to return three points

points = array



def get_nearest_point(x, y, n_points):



    dist = 

    val = 

    j = (x, y)

    for z in range(n_points):

        for i in range(len(points)):

            dist += [Distance(points[i][0],points[i][1],x,y)]

            a = dist.index(min(dist))

        val.append(points[a])

        dist.remove(min(dist))

        return val

Output :

get_nearest_point(50, 50,3)

[[54, 57]]

I want return something like [[54,57],[56,78],[78,90]]

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

asked Jan 3 at 13:30

Lodwin Moloto

3

Do you have a question?

– jpp
Jan 3 at 13:32

add a comment |

points = array



def get_nearest_point(x, y, n_points):



    dist = 

    val = 

    j = (x, y)

    for z in range(n_points):

        for i in range(len(points)):

            dist += [Distance(points[i][0],points[i][1],x,y)]

            a = dist.index(min(dist))

        val.append(points[a])

        dist.remove(min(dist))

        return val

Output :

get_nearest_point(50, 50,3)

[[54, 57]]

I want return something like [[54,57],[56,78],[78,90]]

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

asked Jan 3 at 13:30

Lodwin Moloto

3

Do you have a question?

– jpp
Jan 3 at 13:32

add a comment |

points = array



def get_nearest_point(x, y, n_points):



    dist = 

    val = 

    j = (x, y)

    for z in range(n_points):

        for i in range(len(points)):

            dist += [Distance(points[i][0],points[i][1],x,y)]

            a = dist.index(min(dist))

        val.append(points[a])

        dist.remove(min(dist))

        return val

Output :

get_nearest_point(50, 50,3)

[[54, 57]]

I want return something like [[54,57],[56,78],[78,90]]

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

asked Jan 3 at 13:30

Lodwin Moloto

points = array



def get_nearest_point(x, y, n_points):



    dist = 

    val = 

    j = (x, y)

    for z in range(n_points):

        for i in range(len(points)):

            dist += [Distance(points[i][0],points[i][1],x,y)]

            a = dist.index(min(dist))

        val.append(points[a])

        dist.remove(min(dist))

        return val

Output :

get_nearest_point(50, 50,3)

[[54, 57]]

I want return something like [[54,57],[56,78],[78,90]]

python

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

asked Jan 3 at 13:30

Lodwin Moloto

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

asked Jan 3 at 13:30

Lodwin Moloto

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

edited Jan 3 at 16:01

Akshay L Aradhya

1,05811128

asked Jan 3 at 13:30

Lodwin Moloto

asked Jan 3 at 13:30

Lodwin Moloto

asked Jan 3 at 13:30

Lodwin Moloto

3

Do you have a question?

– jpp
Jan 3 at 13:32

add a comment |

3

Do you have a question?

– jpp
Jan 3 at 13:32

Do you have a question?

– jpp
Jan 3 at 13:32

add a comment |

3 Answers
3

active

oldest

votes

What you are trying to do is essentially find the indices of the smallest 3 elements in the given array. You can easily do this with numpy.argsort method

Anyways here is your code with numpy :

import numpy as np



def get_nearest_point(x, y, n):



    dist = [Distance(point[0], point[1], x, y) for point in points]

    indices = np.argsort(dist)

    return [points[i] for i in indices[:n]]

edited Jan 3 at 16:16

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

add a comment |

welcome to the site!

The issue with your code as written is that the return statement is inside the loop- instead, move it outside. That way you append all three values to val before you return it.

answered Jan 3 at 16:10

Paul Becotte

4,90921630

There are still a lot more errors in his code.

– Akshay L Aradhya
Jan 3 at 16:14

Maybe so, but I still find it useful to answer the question as asked, especially for a new question asker.

– Paul Becotte
Jan 3 at 16:17

add a comment |

Here we calculate distance for every point, sort it by distance and get first n points.

def get_nearest_point(x, y, n_points):

    points_with_dist =   # store here pairs (dist, point)

    for point in points:

        points_with_dist.append((

            Distance(point[0], point[1], x, y),

            point

        ))

    result_with_dist = sorted(points_with_dist, key=lambda struct: struct[0])[:n_points]

    result = [p[1] for p in result_with_dist]  # we need only points

    return result

answered Jan 3 at 17:03

Danyla Hulchuk

1365

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

What you are trying to do is essentially find the indices of the smallest 3 elements in the given array. You can easily do this with numpy.argsort method

Anyways here is your code with numpy :

import numpy as np



def get_nearest_point(x, y, n):



    dist = [Distance(point[0], point[1], x, y) for point in points]

    indices = np.argsort(dist)

    return [points[i] for i in indices[:n]]

edited Jan 3 at 16:16

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

add a comment |

What you are trying to do is essentially find the indices of the smallest 3 elements in the given array. You can easily do this with numpy.argsort method

Anyways here is your code with numpy :

import numpy as np



def get_nearest_point(x, y, n):



    dist = [Distance(point[0], point[1], x, y) for point in points]

    indices = np.argsort(dist)

    return [points[i] for i in indices[:n]]

edited Jan 3 at 16:16

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

add a comment |

What you are trying to do is essentially find the indices of the smallest 3 elements in the given array. You can easily do this with numpy.argsort method

Anyways here is your code with numpy :

import numpy as np



def get_nearest_point(x, y, n):



    dist = [Distance(point[0], point[1], x, y) for point in points]

    indices = np.argsort(dist)

    return [points[i] for i in indices[:n]]

edited Jan 3 at 16:16

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

What you are trying to do is essentially find the indices of the smallest 3 elements in the given array. You can easily do this with numpy.argsort method

Anyways here is your code with numpy :

import numpy as np



def get_nearest_point(x, y, n):



    dist = [Distance(point[0], point[1], x, y) for point in points]

    indices = np.argsort(dist)

    return [points[i] for i in indices[:n]]

edited Jan 3 at 16:16

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

edited Jan 3 at 16:16

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

answered Jan 3 at 16:07

Akshay L Aradhya

1,05811128

add a comment |

welcome to the site!

The issue with your code as written is that the return statement is inside the loop- instead, move it outside. That way you append all three values to val before you return it.

answered Jan 3 at 16:10

Paul Becotte

4,90921630

There are still a lot more errors in his code.

– Akshay L Aradhya
Jan 3 at 16:14

Maybe so, but I still find it useful to answer the question as asked, especially for a new question asker.

– Paul Becotte
Jan 3 at 16:17

add a comment |

welcome to the site!

The issue with your code as written is that the return statement is inside the loop- instead, move it outside. That way you append all three values to val before you return it.

answered Jan 3 at 16:10

Paul Becotte

4,90921630

There are still a lot more errors in his code.

– Akshay L Aradhya
Jan 3 at 16:14

Maybe so, but I still find it useful to answer the question as asked, especially for a new question asker.

– Paul Becotte
Jan 3 at 16:17

add a comment |

welcome to the site!

The issue with your code as written is that the return statement is inside the loop- instead, move it outside. That way you append all three values to val before you return it.

answered Jan 3 at 16:10

Paul Becotte

4,90921630

welcome to the site!

The issue with your code as written is that the return statement is inside the loop- instead, move it outside. That way you append all three values to val before you return it.

answered Jan 3 at 16:10

Paul Becotte

4,90921630

answered Jan 3 at 16:10

Paul Becotte

4,90921630

answered Jan 3 at 16:10

Paul Becotte

4,90921630

answered Jan 3 at 16:10

Paul Becotte

4,90921630

There are still a lot more errors in his code.

– Akshay L Aradhya
Jan 3 at 16:14

Maybe so, but I still find it useful to answer the question as asked, especially for a new question asker.

– Paul Becotte
Jan 3 at 16:17

add a comment |

There are still a lot more errors in his code.

– Akshay L Aradhya
Jan 3 at 16:14

Maybe so, but I still find it useful to answer the question as asked, especially for a new question asker.

– Paul Becotte
Jan 3 at 16:17

There are still a lot more errors in his code.

– Akshay L Aradhya
Jan 3 at 16:14

Maybe so, but I still find it useful to answer the question as asked, especially for a new question asker.

– Paul Becotte
Jan 3 at 16:17

add a comment |

Here we calculate distance for every point, sort it by distance and get first n points.

def get_nearest_point(x, y, n_points):

    points_with_dist =   # store here pairs (dist, point)

    for point in points:

        points_with_dist.append((

            Distance(point[0], point[1], x, y),

            point

        ))

    result_with_dist = sorted(points_with_dist, key=lambda struct: struct[0])[:n_points]

    result = [p[1] for p in result_with_dist]  # we need only points

    return result

answered Jan 3 at 17:03

Danyla Hulchuk

1365

add a comment |

Here we calculate distance for every point, sort it by distance and get first n points.

def get_nearest_point(x, y, n_points):

    points_with_dist =   # store here pairs (dist, point)

    for point in points:

        points_with_dist.append((

            Distance(point[0], point[1], x, y),

            point

        ))

    result_with_dist = sorted(points_with_dist, key=lambda struct: struct[0])[:n_points]

    result = [p[1] for p in result_with_dist]  # we need only points

    return result

answered Jan 3 at 17:03

Danyla Hulchuk

1365

add a comment |

Here we calculate distance for every point, sort it by distance and get first n points.

def get_nearest_point(x, y, n_points):

    points_with_dist =   # store here pairs (dist, point)

    for point in points:

        points_with_dist.append((

            Distance(point[0], point[1], x, y),

            point

        ))

    result_with_dist = sorted(points_with_dist, key=lambda struct: struct[0])[:n_points]

    result = [p[1] for p in result_with_dist]  # we need only points

    return result

answered Jan 3 at 17:03

Danyla Hulchuk

1365

Here we calculate distance for every point, sort it by distance and get first n points.

def get_nearest_point(x, y, n_points):

    points_with_dist =   # store here pairs (dist, point)

    for point in points:

        points_with_dist.append((

            Distance(point[0], point[1], x, y),

            point

        ))

    result_with_dist = sorted(points_with_dist, key=lambda struct: struct[0])[:n_points]

    result = [p[1] for p in result_with_dist]  # we need only points

    return result

answered Jan 3 at 17:03

Danyla Hulchuk

1365

answered Jan 3 at 17:03

Danyla Hulchuk

1365

answered Jan 3 at 17:03

Danyla Hulchuk

1365

answered Jan 3 at 17:03

Danyla Hulchuk

1365

add a comment |

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