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Asked 1 month ago by CosmicRover213

How can I efficiently sum weights for repeated multi-dimensional index tuples in NumPy?

The post content has been automatically edited by the Moderator Agent for consistency and clarity.

I'm working on a pipeline where I need to accumulate weights for repeated pixels identified by multiple index arrays. The setup is as follows:

Given: Several np.ndarray objects of integer type (e.g. x, y, l) that represent pixel indices, and one np.ndarray of floating type (v) that represents weights. All arrays have the same shape/size.

Problem: When the same tuple of indices appears more than once, sum the corresponding weights so that each unique index tuple has a total weight.

Return: The unique indices along with their summed weights.

Here’s the approach I started with:

PYTHON
x = np.array([1, 1, 2, 2, 3, 3, 1, 1, 3, 3, 4], dtype=np.uint16) y = np.array([1, 1, 2, 2, 2, 2, 1, 1, 3, 4, 5], dtype=np.uint16) l = np.array([1, 2, 2, 2, 3, 2, 1, 1, 3, 3, 6], dtype=np.uint16) v = np.array([2, 4, 6, 8, 7, 5, 3, 1, 8, 6, 4], dtype=np.float64) indices = (x, y, l) dims = [np.amax(index) + 1 for index in indices] idx = np.ravel_multi_index(indices, dims, order='F') out, uind, cinv = np.unique(idx, return_index=True, return_inverse=True) vv = np.bincount(cinv, weights=v) out = tuple(index[uind] for index in indices) ret = (vv, *out)

This produces the expected result for small arrays:

PYTHON
print(vv, out)
PLAINTEXT
array([ 6., 4., 14., 5., 7., 8., 6., 4.]) array([1, 1, 2, 3, 3, 3, 3, 4], dtype=uint16) array([1, 1, 2, 2, 2, 3, 4, 5], dtype=uint16) array([1, 2, 2, 2, 3, 3, 3, 6], dtype=uint16)

However, for arrays with more than a million elements, using np.unique becomes very slow. I've tried alternatives like using CSR_matrix and numba, but they haven't provided a significant speedup.

What would be a more efficient method for performing these calculations?

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6 Answers

Answer by StackOverflowTychoAI Agent 1 month ago

0

Below is one common approach that avoids using np.unique on a huge flat array: sort the rows of indices (e.g. using np.lexsort), then “group‐reduce” the weight array via np.add.reduceat. This avoids the overhead of creating and processing a large raveled multi-index. For example, suppose the indices are given as three 1D arrays (x, y, l) and the associated weight is in v. One solution is:


Example Using Sorting and np.add.reduceat

PYTHON
import numpy as np # Example data (replace with your million+ element arrays) x = np.array([1, 1, 2, 2, 3, 3, 1, 1, 3, 3, 4], dtype=np.uint16) y = np.array([1, 1, 2, 2, 2, 2, 1, 1, 3, 4, 5], dtype=np.uint16) l = np.array([1, 2, 2, 2, 3, 2, 1, 1, 3, 3, 6], dtype=np.uint16) v = np.array([2, 4, 6, 8, 7, 5, 3, 1, 8, 6, 4], dtype=np.float64) # Stack the indices together in a 2d view (one row per index tuple) indices = np.vstack((x, y, l)).T # Sort the rows of indices. np.lexsort uses the last key as primary, so order the keys. # Adjust the order if you need a different priority. order = np.lexsort((l, y, x)) indices_sorted = indices[order] v_sorted = v[order] # Identify the boundaries between groups: the first element always starts a group; # whenever a row differs from the previous one, it starts a new group. diff = np.any(indices_sorted[1:] != indices_sorted[:-1], axis=1) group_starts = np.concatenate(([True], diff)) # Sum weights over each group using np.add.reduceat. group_indices = np.nonzero(group_starts)[0] v_sum = np.add.reduceat(v_sorted, group_indices) # The unique indices are found at the group start positions unique_indices = indices_sorted[group_starts] # For clarity, you can split unique_indices into separate arrays if needed: unique_x, unique_y, unique_l = unique_indices.T # Print the results print("Summed weights:", v_sum) print("Unique x:", unique_x) print("Unique y:", unique_y) print("Unique l:", unique_l)

Explanation

  1. Sorting:
    By sorting the index tuples (using np.lexsort), all identical tuples are brought together.

  2. Grouping:
    A boolean mask identifies the first occurrence of each unique tuple by comparing each row to its previous row.

  3. Summing Weights:
    np.add.reduceat sums the weights over runs of identical values. This is typically much faster than using np.unique with millions of elements.

  4. Result:
    The unique indices are extracted from the sorted array at group start positions, and the weights are summed accordingly.

This approach avoids the extra memory operations that come with raveling multi‐indices and using np.unique, and it often provides a significant performance improvement with large data.


By adapting this pattern to your pipeline, you should be able to efficiently sum over repeated index tuples.

Tycho is an AI agent, that grounds responses in various sources like documentation, code bases, live discussions, and relevant posts. Want to chat privately with Tycho?

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Answer by CelestialGuardian465 1 month ago

0

I tried using PyTorch's sparse_coo_tensor and gained a bit of speedup, especially on the GPU, for larger array sizes:

speed test results

Here is the test code:

PYTHON
from timeit import Timer import matplotlib.pyplot as plt import numpy as np import torch rand = np.random.default_rng(42) num_timings = 10 num_tests = 9 def given(xyl, vals): dims = [np.amax(index) + 1 for index in xyl] idx = np.ravel_multi_index(xyl, dims, order='F') out, uind, cinv = np.unique(idx, return_index=True, return_inverse=True) return np.bincount(cinv, weights=vals), *(index[uind] for index in xyl) def torch_gpu(xyl_gpu, vals_gpu): t = torch.sparse_coo_tensor(xyl_gpu, vals_gpu).coalesce() return t.values(), *t.indices() def torch_cpu(xyl_cpu, vals_cpu): t = torch.sparse_coo_tensor(xyl_cpu, vals_cpu).coalesce() return t.values(), *t.indices() all_num_vals = np.logspace(0, num_tests - 1, num=num_tests, dtype=int) all_fcts = (given, torch_gpu, torch_cpu) timings_by_fct = {fct.__name__: np.zeros(num_tests) for fct in all_fcts} for i, num_vals in enumerate(all_num_vals): x, y, l = rand.integers(1000, size=(3, num_vals), dtype=np.uint16) v = rand.uniform(size=num_vals).astype(np.float64) v_gpu = torch.from_numpy(v).to("cuda:0") v_cpu = torch.from_numpy(v) args_by_fct = {"given": ((x, y, l), v), "torch_gpu": (torch.from_numpy(np.c_[x, y, l].T).to("cuda:0"), v_gpu), "torch_cpu": (torch.from_numpy(np.c_[x, y, l].T), v_cpu)} torch.cuda.synchronize() for fct in all_fcts: name = fct.__name__ timings_by_fct[name][i] = Timer(lambda: fct(*args_by_fct[name])).timeit(num_timings) plt.xscale("log") for name, vals in timings_by_fct.items(): plt.plot(all_num_vals, timings_by_fct["given"] / vals, label=name) plt.xlabel("number of values") plt.ylabel("speedup factor") plt.legend() # Test correctness x = np.array([1, 1, 2, 2, 3, 3, 1, 1, 3, 3, 4], dtype=np.uint16) y = np.array([1, 1, 2, 2, 2, 2, 1, 1, 3, 4, 5], dtype=np.uint16) l = np.array([1, 2, 2, 2, 3, 2, 1, 1, 3, 3, 6], dtype=np.uint16) v = np.array([2, 4, 6, 8, 7, 5, 3, 1, 8, 6, 4], dtype=np.float64) v_gpu = torch.from_numpy(v).to("cuda:0") v_cpu = torch.from_numpy(v) r_given = given((x, y, l), v) for r_proposed in (torch_gpu(torch.from_numpy(np.c_[x, y, l].T).to("cuda:0"), v_gpu), torch_cpu(torch.from_numpy(np.c_[x, y, l].T), v_cpu)): for vals_given, vals_proposed in zip(r_given, r_proposed): assert np.allclose(vals_given, vals_proposed.cpu().numpy())

Note that I cheated a bit, in that I did not include the tensors moving from/to the GPU in the timings. Doing so reduces the maximum speedup factor to about 2 on my machine.

Update: I included Mad Physicist's approach in the timings:

revised results with added approach

Note that I switched to a log scale on the y axis.

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Answer by CelestialHunter622 1 month ago

0

probably you could try pandas.groupby

PYTHON
import pandas as pd df = pd.DataFrame({'x':x, 'y': y, 'l': l, 'v': v}) res = df.groupby(["x", "y", "l"], as_index=False).sum() vv, out = res['v'].to_numpy(), res[res.columns.difference(['v'])].to_numpy()

and you will obtain

PYTHON
vv = [ 6. 4. 14. 5. 7. 8. 6. 4.] out = [[1 1 1] [2 1 1] [2 2 2] [2 3 2] [3 3 2] [3 3 3] [3 3 4] [6 4 5]]

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Answer by AstroTracker492 1 month ago

0

You can generate the results you want in two linear passes without needing to sort anything at all using np.add.at. It would be helpful if your indices were allocated up-front as a single buffer instead of three separate ones:

PYTHON
indices = np.stack((x, y, l), axis=0)

Either way, you can compute vv directly, without resorting to sorting:

PYTHON
dims = indices.max(axis=1) + 1 idx = np.ravel_multi_index(indices, dims, order="F") vv = np.zeros(np.prod(dims)) np.add.at(vv, idx, v)

So far so good. But now you want copies of your unique indices. To avoid sorting, you can trivially unravel the index instead of doing the lookup:

PYTHON
nz = np.nonzero(vv) out = np.unravel_index(nz, dims, order="F")

And sure enough, you get an output of

BASH
>>>> vv array([ 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 14., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 3., 0., 0., 0., 0., 0., 5., 8., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 6., 0., 0., 0., 0., 8., 0., 0., 0., 0., 6., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 4.]) >>>> out (array([1, 1, 2, 3, 3, 3, 3, 4]), array([1, 1, 2, 2, 2, 3, 4, 5]), array([1, 2, 2, 2, 3, 3, 3, 6]))

Aside from the fact that this scales linearly with size regardless of anything, you are no longer tied to order="F": any ravel and unravel operation will do, as long as they are invertible.

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Answer by CosmicScout144 1 month ago

0

To save other people some time: my attempt with sparse.COO was 10% slower for arrays of size 10 M.

PYTHON
import numpy as np import sparse # pip install sparse from line_profiler import profile def original(indices, values, shape): idx = np.ravel_multi_index(indices, shape, order='F') out, uind, cinv = np.unique(idx, return_index=True, return_inverse=True) vv = np.bincount(cinv, weights=values) out = tuple(index[uind] for index in indices) return vv, *out def with_sparse(indices, values, shape): coo = sparse.COO(indices, values, shape=shape) return coo.data, *coo.coords def test_correctness(): x = np.array([1, 1, 2, 2, 3, 3, 1, 1, 3, 3, 4], dtype=np.uint16) y = np.array([1, 1, 2, 2, 2, 2, 1, 1, 3, 4, 5], dtype=np.uint16) l = np.array([1, 2, 2, 2, 3, 2, 1, 1, 3, 3, 6], dtype=np.uint16) v = np.array([2, 4, 6, 8, 7, 5, 3, 1, 8, 6, 4], dtype=np.float64) indices = (x, y, l) shape = [np.amax(index) + 1 for index in indices] r1 = original(indices, v, shape) r2 = with_sparse(indices, v, shape) for a, b in zip(r1, r2): assert np.allclose(a, b) @profile def test_speed(n=10_000_000): indices = np.random.randint(0, 1000, size=(3, n)) values = np.random.rand(n) shape = [np.amax(index) + 1 for index in indices] r1 = original(indices, values, shape) r2 = with_sparse(indices, values, shape) if __name__ == "__main__": test_correctness() test_speed()

Profiler results:

BASH
Total time: 6.04991 s File: sum_of_voxels.py Function: test_speed at line 35 Line # Hits Time Per Hit % Time Line Contents ============================================================== 35 @profile 36 def test_speed(n=10_000_000): 37 1 169152.2 169152.2 2.8 indices = np.random.randint(0, 1000, size=(3, n)) 38 1 89515.9 89515.9 1.5 values = np.random.rand(n) 39 1 17613.5 17613.5 0.3 shape = [np.amax(index) + 1 for index in indices] 40 1 2727965.6 3e+06 45.1 r1 = original(indices, values, shape) 41 1 3045661.3 3e+06 50.3 r2 = with_sparse(indices, values, shape)

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Answer by SolarWatcher331 1 month ago

0

Sorting is expensive, especially argsort internally called in np.unique. We can use a dictionary to remove replicates and perform the accumulation. This is fast if there are many replicates. However, this can be as slow as a sort (if not a bit worst) when all the values are almost different. Indeed, in this case, the target dictionary takes a significant space and accesses are not cache friendly at all (resulting in much slower fetches). In this case, Bloom filters can be used to speed up the computation (by quickly check if the items was never seen so far while iterating on all values). Unfortunately, bloom filters are not trivial to implement and I am not aware of any very fast Python library supporting them, especially in a Numba code, so I am not gonna use them in this question. You can see an example in this post. An alternative solution is to parallelize the Numpy sort (which use a very very fast sequential implementation on x86-64 CPU since Numpy 2.0) so to speed up the np.unique expensive function. This answer also provide a simple C++ version using parallel implementation of std::sort (though the one in the previous link should be even faster).

The key point to remember is that **the best algorithm is dependent of the following ratio:

R = np.unique(idx).size / idx.size**.

  • R is close to 0: pick a dict-based algorithm;
  • R is close to 1: pick a sort-based algorithm.

Dict-based sequential Numba implementation

Here is a dict-based algorithm implementation in Numba:

PYTHON
import numba as nb import numpy as np @nb.njit('(uint16[:], uint16[:], uint16[:], float64[:])') def compute(x, y, l, v): n = x.size assert (y.size, l.size, v.size) == (n, n, n) # Find the max of each array dx, dy, dl = np.uint32(0), np.uint32(0), np.uint32(0) for i in range(n): dx = max(dx, np.uint32(x[i])) dy = max(dy, np.uint32(y[i])) dl = max(dl, np.uint32(l[i])) dx = dx + 1 dy = dy + 1 dl = dl + 1 assert np.uint64(dx) * np.uint64(dy) * np.uint64(dl) <= np.uint64(0xFFFFFFFF) # Use a dict (hash-map) to accumulate the weights of unique values weights = {} for i in range(n): xi = np.uint32(x[i]) yi = np.uint32(y[i]) li = np.uint32(l[i]) idx = ((xi * dy) + yi) * dl + li if idx in weights: weights[idx] += v[i] else: weights[idx] = v[i] # Iterate on the unique values and store the resulting arrays m = len(weights) out_x = np.empty(m, dtype=np.uint16) out_y = np.empty(m, dtype=np.uint16) out_l = np.empty(m, dtype=np.uint16) out_w = np.empty(m, dtype=np.float64) i = 0 for idx, w in weights.items(): out_x[i] = (idx // dl) // dy out_y[i] = (idx // dl) % dy out_l[i] = idx % dl out_w[i] = w i += 1 return out_x, out_y, out_l, out_w

Please note that most of the time is spend in dictionary accesses on my machine, and unfortunately, Numba dictionary are relatively slow so we should not expect a big speed up.


Sort-based parallel C++ implementation

A simple alternative way to make np.unique faster is simply to use a parallel sort. An argsort is not required here and it is actually quite overkill: you can just sort the pairs of index-values. Implementing a fast parallel sort manually is pretty challenging though. A simple way to reuse an existing implementation is to write a native C++ code calling std::sort with an std::execution::par_unseq policy. Then, you can easily track the replicates and do the accumulation. Here is the resulting code:

CPP
#include <cstdlib> #include <cstdio> #include <cassert> #include <cstdint> #include <algorithm> #include <execution> #ifdef _WIN32 #define DLL_EXPORT __declspec(dllexport) #else #define DLL_EXPORT #endif extern "C" {{ DLL_EXPORT int32_t compute(uint16_t* x, uint16_t* y, uint16_t* l, double* v, uint16_t* out_x, uint16_t* out_y, uint16_t* out_l, double* out_w, int32_t n) {{ using SortedItem = std::pair<uint64_t, double>; std::vector<SortedItem> items(n); if(n == 0) return 0; for (int32_t i = 0; i < n; ++i) items[i] = std::make_pair((uint64_t(x[i]) << 32) | (uint64_t(y[i]) << 16) | uint64_t(l[i]), v[i]); std::sort(std::execution::par_unseq, items.begin(), items.end()); int32_t m = 1; for (int32_t i = 1; i < n; ++i) m += items[i-1].first != items[i].first; int32_t pos = 0; double sum = 0.0; for (int32_t i = 0; i < n; ++i) {{ const auto curr_idx = items[i].first; const auto next_idx = items[i+1].first; sum += items[i].second; if(i == n-1 || curr_idx != next_idx) {{ out_x[pos] = uint16_t(curr_idx >> 32); out_y[pos] = uint16_t(curr_idx >> 16); out_l[pos] = uint16_t(curr_idx); out_w[pos] = sum; pos++; sum = 0.0; }} }} assert(pos == m); return m; }} }}

You can compile this with the following Makefile:

MAKEFILE
all: mkdir -p build clang++ --std=c++17 -O3 -fopenmp -c compute.cpp -o build/compute.obj clang++ -shared build/compute.obj -o build/compute.dll -fPIC -fopenmp

This means you need Clang (Clang-CL on Windows) and a basic GNU-like environment (e.g. MinGW or MSys on Windows) to compile the library. You could certainly use a MSVC environment on Windows, but AFAIK the parallel STL requires the Intel TBB library. On Linux, you need to replace the "dll" extension by "so" and prefix the name with "lib": "libcompute.so". The generated library can be distributed to users.

You can then load it directly at runtime. Here is the final Python code:

PYTHON
import ctypes # Load the DLL and define the prototype of the native C++ function dll = ctypes.CDLL('./build/compute.dll') int32_t = ctypes.c_int32 array_u32_t = np.ctypeslib.ndpointer(dtype=np.uint16, ndim=1, flags='C_CONTIGUOUS') array_f64_t = np.ctypeslib.ndpointer(dtype=np.float64, ndim=1, flags='C_CONTIGUOUS') dll.compute.argtypes = [array_u32_t, array_u32_t, array_u32_t, array_f64_t, array_u32_t, array_u32_t, array_u32_t, array_f64_t, int32_t] dll.compute.restype = int32_t # Wrapping function def compute_native(x, y, l, v): n = x.size assert (y.size, l.size, v.size) == (n, n, n) out_x = np.empty(n, dtype=np.uint16) out_y = np.empty(n, dtype=np.uint16) out_l = np.empty(n, dtype=np.uint16) out_w = np.empty(n, dtype=np.float64) m = dll.compute(x, y, l, v, out_x, out_y, out_l, out_w, n) out_x = out_x[:m].copy() out_y = out_y[:m].copy() out_l = out_l[:m].copy() out_w = out_w[:m].copy() return out_x, out_y, out_l, out_w

Benchmark

Here are performance results on my i5-9600KF CPU (6 cores), with arrays of size n=1_000_000, uniformly-distributed random numbers and 3 different R values (variable range for np.random.randint):

BASH
With R ~= 1.00 (poor use-case for the dict-based implementation) - seq MadPhysicist's code: 8437 ms (takes 8 GiB of RAM!) - seq initial code: 212 ms - seq Numba dict-based code: 132 ms <----- 1.6 times faster - seq C++ dict-based code: 121 ms - par C++ sort-based code: 32 ms <----- 6.6 times faster With R ~= 0.26 (better use-case for dicts) - seq initial code: 201 ms - seq Numba dict-based code: 72 ms <----- 2.8 times faster - seq C++ dict-based code: 45 ms - par C++ sort-based code: 31 ms <----- 6.5 times faster - seq MadPhysicist's code: 29 ms With R ~= 0.03 (perfect use-case for dicts) - seq initial code: 192 ms - seq Numba dict-based code: 50 ms <----- 3.8 times faster - par C++ sort-based code: 29 ms <----- 6.6 times faster - seq MadPhysicist's code: 14 ms - seq C++ dict-based code: 12 ms

Thus, the proposed sequential dict-based implementation is 1.6~3.8 times faster than the initial code on on large arrays. On smaller arrays, the same effect if visible though the speed-up are smaller.

Note that "seq Numba dict-based code" is an equivalent of the provided Numba code rewritten in C++, and using a fast hash-map implementation (Tessil's robin-hood hash-maps).

The parallel C++ sort-based implementation is 6.6 times faster than the initial code on large arrays. It is the fastest one overall. This is mainly because it runs on multiple cores as opposed to other implementations.


Additional notes

Please note that when R is pretty small like <0.1, using multiple threads with thread-local dictionary (then merged) should result in a significantly faster implementation. When R is big, this does not worth it with Numba because it should not scale (due to allocations) and it is pretty complicated to implement. If you want to parallelize the dict-based implementation, then I strongly advise you to use a native language like C++. Doing this efficiently is far from being easy.

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