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I have a dataframe that has 3 columns, Latitude, Longitude and Median_Income. I need to get the average median income for all points within x km of the original point into a 4th column. I need to do this for each observation.

I have tried making 3 functions which I use apply to attempt to do this quickly. However, the dataframes take forever to process (hours). I haven't seen an error yet, so it appears to be working okay.

The Haversine formula, I found on here. I am using it to calculate the distance between 2 points using lat/lon.

from math import radians, cos, sin, asin, sqrt

def haversine(lon1, lat1, lon2, lat2):

    #Calculate the great circle distance between two points 
    #on the earth (specified in decimal degrees)

    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])

    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    r = 6371 # Radius of earth in kilometers. Use 3956 for miles
    return c * r

My hav_checker function will check the distance of the current row against all other rows returning a dataframe with the haversine distance in a column.

def hav_checker(row, lon, lat):

    hav = haversine(row['longitude'], row['latitude'], lon, lat)

    return hav

My value grabber fucntion uses the frame returned by hav_checker to return the mean value from my target column (median_income).

For reference, I am using the California housing dataset to build this out.

def value_grabber(row, frame, threshold, target_col):

    frame = frame.copy()

    frame['hav'] = frame.apply(hav_checker, lon = row['longitude'], lat = row['latitude'], axis=1)

    mean_tar = frame.loc[frame.loc[:,'hav'] <= threshold, target_col].mean()

    return mean_tar

I am trying to return these 3 columns to my original dataframe for a feature engineering project within a larger class project.

df['MedianIncomeWithin3KM'] = df.apply(value_grabber, frame=df, threshold=3, target_col='median_income', axis=1)

df['MedianIncomeWithin1KM'] = df.apply(value_grabber, frame=df, threshold=1, target_col='median_income', axis=1)

df['MedianIncomeWithinHalfKM'] = df.apply(value_grabber, frame=df, threshold=.5, target_col='median_income', axis=1)

I have been able to successfully do this with looping but it is extremely time intensive and need a faster solution.

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  • \$\begingroup\$ Can you include some example data, so we can actually run the code? \$\endgroup\$ – Graipher Apr 16 at 15:29
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vectorization

You are doing all the calculations in normal python space. Try to do as much as possible in numpy space

dummy data

np.random.seed(0)
coords = (np.random.random(size=(N, dim)) - 0.5) * 360
median_income = np.random.normal(size=N) * 10000 + 5000
df = pd.DataFrame(
    {
        "lat": coords[:, 0],
        "lon": coords[:, 1],
        "median_income": np.random.normal(size=N) * 10000 + 30000,
    }
)

instead of using math.radians, usenp.radians to calculate this for the whole matrix at once:

coords_rad = np.radians(df[["lat", "lon"]].values)

select only the upper triangle

For this section, I borrowed a bit from this SO post

p1, p2 = np.triu_indices(N,k=1)        # k=1 eliminates diagonal indices

havesine distances

lat1, lon1 = coords_rad[p1].T
lat2, lon2 = coords_rad[p2].T
d_lat = lat2 - lat1
d_lon = lon2 - lon1
r = 6371
distances = 2 * r * np.arcsin(
    np.sqrt(
        np.sin(d_lat / 2) ** 2
        + np.cos(lat1) * np.cos(lat2) * np.sin(d_lon / 2) ** 2
    )
)
array([ 6318.56953693,  5685.87555152,  8221.15833653,  6489.20595509,
        8755.09024969,  7805.61189508,  6919.53162119, 15295.76892719,
        8706.83662262,  8113.95651365, 14532.71048537, 11780.39186778,
        7556.99686671, 11832.44825307,  7137.04783302,  9306.23652045,
        5446.80037496,  8740.28196777, 10242.77405649, 14237.95015622,
       12225.48901658,  2112.82250374, 11830.45390613, 13194.16431067,
        3966.47195107, 11375.98162917,  5385.20026834, 10745.8851006 ,
       15975.57051313, 13621.58550369,  7573.94148257,  2037.20795034,
       12284.11555433, 17912.47114836,  9676.18614574,  6000.06279665,
       14392.65091451, 11339.26110213,  2465.57715011, 14204.32921867,
       15974.00480201,  8347.16187191,  9820.5895048 , 12576.27804606,
        9720.35934264])

A way to minimize the memory footprint of this is by choosing the correct dtype by adding .astype("e") for example. The correct dtype for this application is the smallest one that still delivers the necessary resolution, so needs to be chosen with your data taken into consideration.

Distance matrix

You can assemble a distance matrix

distance_matrix = np.zeros((N, N))
distance_matrix [(p1, p2)] = distances 
distance_matrix [(p2, p1)] = distances 
 array([[    0.        ,  6318.56953693,  5685.87555152,  8221.15833653,  6489.20595509,  8755.09024969,  7805.61189508,  6919.53162119, 15295.76892719,  8706.83662262],
       [ 6318.56953693,     0.        ,  8113.95651365, 14532.71048537, 11780.39186778,  7556.99686671, 11832.44825307,  7137.04783302,  9306.23652045,  5446.80037496],
       [ 5685.87555152,  8113.95651365,     0.        ,  8740.28196777, 10242.77405649, 14237.95015622, 12225.48901658,  2112.82250374, 11830.45390613, 13194.16431067],
       [ 8221.15833653, 14532.71048537,  8740.28196777,     0.        ,  3966.47195107, 11375.98162917,  5385.20026834, 10745.8851006 , 15975.57051313, 13621.58550369],
       [ 6489.20595509, 11780.39186778, 10242.77405649,  3966.47195107,     0.        ,  7573.94148257,  2037.20795034, 12284.11555433, 17912.47114836,  9676.18614574],
       [ 8755.09024969,  7556.99686671, 14237.95015622, 11375.98162917,  7573.94148257,     0.        ,  6000.06279665, 14392.65091451, 11339.26110213,  2465.57715011],
       [ 7805.61189508, 11832.44825307, 12225.48901658,  5385.20026834,  2037.20795034,  6000.06279665,     0.        , 14204.32921867, 15974.00480201,  8347.16187191],
       [ 6919.53162119,  7137.04783302,  2112.82250374, 10745.8851006 , 12284.11555433, 14392.65091451, 14204.32921867,     0.        ,  9820.5895048 , 12576.27804606],
       [15295.76892719,  9306.23652045, 11830.45390613, 15975.57051313, 17912.47114836, 11339.26110213, 15974.00480201,  9820.5895048 ,     0.        ,  9720.35934264],
       [ 8706.83662262,  5446.80037496, 13194.16431067, 13621.58550369,  9676.18614574,  2465.57715011,  8347.16187191, 12576.27804606,  9720.35934264,     0.        ]])

Then you can use

close_points = pd.DataFrame(np.where((distance_matrix < d_crit) & (0 < distance_matrix)), index=["p1", "p2"]).T

To get the points which are closer than the critical distance (4km in your case, 10000km for this dummy data).

Another way to get the close points without assembling the distance_matrix is this:

point_combinations = np.array((p1, p2)).T
close_points = pd.DataFrame(
    np.concatenate(  # if A is close to B, B is close to A
        (
            point_combinations[np.ix_(close, [0, 1])],
            point_combinations[np.ix_(close, [1, 0])],   # if A is close to B, B is close to A
        )
    ),
    columns=["p1", "p2"],
)

Then get the mean of the close median incomes, you could use DataFrame.groupby

df["neighbours_mean"] = close_points.groupby("p1").apply(
    lambda x: (df.loc[x["p2"], "median_income"]).mean()
)
  lat lon median_income   neighbours_mean
0 17.57286141383691   77.468171894071 30457.58517301446   30794.78854097742
1 36.994815385791796  16.15794587888287   28128.161499741665  29640.567671359968
2 -27.484272237994304 52.5218807039962    45327.79214358458   28367.842422379927
3 -22.468603945430697 141.0382802815487   44693.58769900285   32114.24852431677
4 166.91859378037054  -41.961053222720025 31549.474256969163  32323.686056555547
5 105.02101370975925  10.402171111045613  33781.625196021734  28564.170628892793
6 24.49604199381563   153.21478978535797  21122.142523698873  34409.152403209606
7 -154.4270190487607  -148.63345210744535 10192.035317760732  32608.604330769795
8 -172.72137692148274 119.74314439725768  26520.878506738474  23294.56216951406
9 100.13643034194618  133.2043733688549   31563.489691039802  28593.31119269739

please test this against a sample set of your data


memory

If you still run into memory problems, you'll need to start calculation the distances in chunks, and then later concatenate them. An alternative is to use dask instead of pandas and numpy

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  • \$\begingroup\$ I was able to get this solution to work. It's many times faster, however I did run into some memory issues along the way. \$\endgroup\$ – krewsayder Apr 16 at 20:23

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