Efficiently find neighbors on multiple dimensions and calculate

ghz 9hours ago ⋅ 2 views

Efficiently find neighbors on multiple dimensions and calculate sum of values based on proximity

I am tasked with finding the total value of all elements within a variable distance of a central element. The elements are arranged using 3 dimensions (columns in my data). Each element has a unique location given the 3 dimensions (and has a unique-id).

I have a working version that does what I want, however it is terribly slow. I am using itertuples, finding the value per tuple using a subset dataframe, apply(np.isclose), and I set the value with .at (see code below).

The problem is not so much the function of my code as it is the scalability. Since I want to set a variable distance to measure, and I want to calculate this value for each row, it ends up iterating nrows x ndistances, and currently each iteration takes 1.7 seconds (my data has >25,000 rows, I estimated ~12 hours per each distance I try).

import pandas as pd
import numpy as np

Example of data structure:

df = pd.DataFrame({'id':[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19], 
                          'x':[-2,-2,-2,-1,-1,-1,-1,0,0,0,0,0,1,1,1,1,2,2,2], 
                          'y':[2,1,0,2,1,0,-1,2,1,0,-1,-2,1,0,-1,-2,0,-1,-2], 
                          'z':[0,1,2,-1,0,1,2,-2,-1,0,1,2,-2,-1,0,1,-2,-1,0], 
                          'val':[0,0,0,1,0,0,6,3,7,11,0,0,14,18,10,4,20,15,2]})
df.set_index('id', inplace=True)
# The 'val' column can have any non-negative whole number, I've just picked some randomly.

'Working' code so far:

n = 0  #Initial distance
while n < 3:  #This part allows me to set my distance range
    df['n{0}'.format(n)] = np.nan  #create a column for the new values
    for row in df.itertuples():
        valsum = df[(df['x'].apply(np.isclose, b=row.x, atol=n)) & 
                    (df['y'].apply(np.isclose, b=row.y, atol=n)) & 
                    (df['z'].apply(np.isclose, b=row.z, atol=n))].val.sum()
        df.at[row.Index, 'n{0}'.format(n)] = valsum
    n += 1

Current / Desired output:

    x   y   z   val n0  n1  n2
id                          
1   -2  2   0   0   0   1   22
2   -2  1   1   0   0   0   25
3   -2  0   2   0   0   6   17
4   -1  2   -1  1   1   11  54
5   -1  1   0   0   0   19  70
6   -1  0   1   0   0   17  57
7   -1  -1  2   6   6   6   31
8   0   2   -2  3   3   25  74
9   0   1   -1  7   7   54  99
10  0   0   0   11  11  46  111
11  0   -1  1   0   0   31  73
12  0   -2  2   0   0   10  33
13  1   1   -2  14  14  62  99
14  1   0   -1  18  18  95  105
15  1   -1  0   10  10  60  107
16  1   -2  1   4   4   16  66
17  2   0   -2  20  20  67  100
18  2   -1  -1  15  15  65  101
19  2   -2  0   2   2   31  80

I know that having the 'n0' column is equal to 'val' column, because the search distance is 0, but I wanted to hopefully show what I am looking for. The sum of all the items in the val column is 111, which is the same when (x,y,z) = (0,0,0). This is because (0,0,0) is the center of my data in this example, and therefore having a distance of 2 captures all of the elements. I'd like to do this for a bandwidth of distances, say, 5-10.

My ultimate question is: How can I do this but faster / more efficiently?

Answer

To improve the performance of your code and make it scalable, we need to take a few steps that avoid the expensive itertuples loop and the repeated calculation of the distance for each row. The main issues with the existing code are the following:

  1. Iterating over rows (itertuples): This results in multiple passes over the data, and for each row, you're filtering the DataFrame and performing an expensive operation.

  2. Using apply with np.isclose for each row: This is inefficient because apply iterates over all rows to perform the comparisons, which could be optimized.

  3. Accessing and modifying the DataFrame using .at: This adds overhead for each modification, which can be slow with large datasets.

Steps for Optimization

To speed up the computation, we can:

  1. Use NumPy for vectorized distance calculations: Instead of applying np.isclose to each row, we can calculate distances for all rows at once using NumPy. This allows for more efficient handling of large data.

  2. Pre-calculate distance matrices: For each distance n, calculate the Euclidean distance between each point and the others, and then sum the values where the distance is within the given range.

  3. Avoid repeated filtering: Use matrix operations (such as np.meshgrid and broadcasting) to avoid repeatedly filtering the DataFrame.

Optimized Approach

Here’s a more efficient approach using NumPy for the distance calculation and avoiding the repeated filtering of rows:

import pandas as pd
import numpy as np

# Example DataFrame
df = pd.DataFrame({
    'id': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19], 
    'x': [-2, -2, -2, -1, -1, -1, -1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2], 
    'y': [2, 1, 0, 2, 1, 0, -1, 2, 1, 0, -1, -2, 1, 0, -1, -2, 0, -1, -2], 
    'z': [0, 1, 2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, -2, -1, 0], 
    'val': [0, 0, 0, 1, 0, 0, 6, 3, 7, 11, 0, 0, 14, 18, 10, 4, 20, 15, 2]
})
df.set_index('id', inplace=True)

# Define the optimized function to calculate the sum of values within a certain distance range
def compute_nearby_sums(df, max_distance):
    # Extract coordinate columns and value column
    coords = df[['x', 'y', 'z']].values
    values = df['val'].values
    
    # Compute pairwise squared Euclidean distances (avoid sqrt for efficiency)
    diff = coords[:, None, :] - coords[None, :, :]
    squared_distances = np.sum(diff**2, axis=-1)
    
    # Create an empty DataFrame to store the results
    nearby_sums = pd.DataFrame(index=df.index)
    
    for dist in range(0, max_distance+1):  # Loop over the distance ranges
        # Get a mask for distances within the current range (inclusive)
        mask = (squared_distances <= dist**2)  # Compare squared distances to avoid sqrt
        nearby_sums[f'n{dist}'] = np.sum(values[mask], axis=1)  # Sum the values for each point within distance
    
    return nearby_sums

# Set the maximum distance you want to compute for
max_distance = 2

# Compute the sums for distances 0 to max_distance
nearby_sums = compute_nearby_sums(df, max_distance)

# Display the result
result_df = df.join(nearby_sums)
print(result_df)

Explanation:

  1. Euclidean Distance Calculation:

    • We calculate the squared Euclidean distance between all pairs of points in a vectorized manner using NumPy. This avoids the need for expensive np.isclose checks for each individual row.
    • We subtract all coordinates from each other and compute the squared sum, which gives us a distance matrix where each element (i, j) represents the squared distance between the i-th and j-th points.

  2. Efficient Masking:

    • For each distance n, we create a boolean mask (mask) where the squared distances are less than or equal to n^2. This allows us to sum the val column values only for points within the distance range without having to repeatedly filter the DataFrame.

  3. Precompute Sums:

    • We compute the sums for all points within the given distance range (0 <= dist <= max_distance) at once using np.sum. This eliminates the need to loop through each row individually.

  4. Result:

    • The nearby_sums DataFrame contains the sums of the val column for each row within each distance range. We then join this DataFrame with the original DataFrame to get the final result.

Performance Gains:

  • Vectorized Operations: By leveraging NumPy’s vectorized operations, we avoid the Python-level for-loops and apply calls, which significantly improves performance.
  • Avoiding Repeated Filtering: Instead of filtering the DataFrame repeatedly for each row, we calculate all pairwise distances once and then use a mask to sum the values, which is far more efficient.

Scaling:

This approach should scale much better than your original method because it reduces the complexity from nrows * ndistances operations to something more manageable, using matrix operations and broadcasting.

Let me know if you need further optimizations or have any additional questions!

#py