Skip to content
New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

"most common" Aggregator with Dask #263

Open
BSchilperoort opened this issue Sep 21, 2023 · 7 comments
Open

"most common" Aggregator with Dask #263

BSchilperoort opened this issue Sep 21, 2023 · 7 comments

Comments

@BSchilperoort
Copy link

Hi! Thanks for developing Flox, it has been quite useful in our workflows.

We have been working on a method to reduce categorical data using flox, using a "most common" strategy.

    most_common = Aggregation(
        name="most_common",
        numpy=_custom_grouped_reduction,
        chunk=None,
        combine=None,
    )

    result = flox.xarray.xarray_reduce(
        data,
        *coords,
        func=most_common,
        expected_groups=bounds,
    )

This works very well for in-memory datasets, however, not on Dask data (and I have not been able to get it to work with Dask data).

If the input is dask data, the following error is returned

NotImplementedError: Aggregation 'most_common' is only implemented for dask arrays when method='blockwise'.

If method="blockwise" is set, then the following error is returned.

InvalidIndexError: Reindexing only valid with uniquely valued Index objects

I have tried to implement chunk= and combine=, but I don't think that is possible with this method.

Memory limit of xarray_reduce

Additionally, because of the size of our input dataset the group_idx array does not fit in memory. This is due to the size being dim1 * dim2 * ... * size(np.int64).

Our current workaround is to apply the reduction on subset slices of the source data and target data, and then merging them. This works OK but is quite ugly and slow.

Is there another way to avoid this limit?


Most common implementation
def _most_common_label(neighbors: np.ndarray) -> int:
    """Find the most common label in a neighborhood.

    Note that if more than one labels have the same frequency which is the highest,
    then the first label in the list will be picked.
    """
    unique_labels, counts = np.unique(neighbors, return_counts=True)
    return unique_labels[np.argmax(counts)]


def _custom_grouped_reduction(
    group_idx: np.ndarray,
    array: np.ndarray,
    *,
    axis: int = -1,
    size: int | None = None,
    fill_value=None,
    dtype=None
) -> np.ndarray:
    """Custom grouped reduction for flox.Aggregation to get most common label.

    Args:
        group_idx : integer codes for group labels (1D)
        array : values to reduce (nD)
        axis : axis of array along which to reduce.
            Requires array.shape[axis] == len(group_idx)
        size : expected number of groups. If none,
            output.shape[-1] == number of uniques in group_idx
        fill_value : fill_value for when number groups in group_idx is less than size
        dtype : dtype of output

    Returns:
        np.ndarray with array.shape[-1] == size, containing a single value per group
    """
    return npg.aggregate_numpy.aggregate(
        group_idx,
        array,
        func=_most_common_label,
        axis=axis,
        size=size,
        fill_value=fill_value,
        dtype=dtype,
    )
@dcherian
Copy link
Collaborator

dcherian commented Sep 21, 2023

Ah, now this is interesting!

The general solution is hard and approximate (I think). You'll need to implement something like count min sketch (I'm sure there are others)

The easier way: You'll have to decompose the problem and compute each unique item and associated count in chunk (like you already do), merge those intermediates together in combine and then pick the top-most in finalize.

Are there useful properties for the field you're collapsing or the field you're grouping by (example plots of both these fields would be useful)? For examlke

  • are there only a small number of unique labels?
  • method="blockwise" only works

@BSchilperoort
Copy link
Author

BSchilperoort commented Sep 22, 2023

Hi Deepak, thank you for your reply.

The easier way: You'll have to decompose the problem and compute each unique item and associated count in chunk (like you already do), merge those intermediates together in combine and then pick the top-most in finalize.

I attempted to do this, however I am unsure of how to exactly make this work. Below is the code, but I am not sure of how to wrap the individual functions in such a way that they're compatible with Aggregation.

I could not make much sense of the documentation/docstrings beyond the very simple examples which are available. I also received some errors relating to setting a element with a sequence (as each chunk will return an array with unique values, instead of a single value).

Code
def unique_labels(a: np.ndarray) -> np.ndarray:
    labels = np.unique(a)
    return labels


def unique_counts(a: np.ndarray) -> np.ndarray:
    _, counts = np.unique(a, return_counts=True)
    return counts


def most_common_chunked(multi_values: np.ndarray, multi_counts: np.ndarray, **kwargs):
    all_values, index = np.unique(multi_values, return_inverse=True)
    all_counts = np.zeros(all_values.size, np.int64)
    np.add.at(all_counts, index, multi_counts.ravel())  # inplace
    return all_values[all_counts.argmax()]


most_common = Aggregation(
    name="most_common",
    numpy=_custom_grouped_reduction,
    chunk=(unique_labels, unique_counts),  # first compute blockwise
    combine=(wrap_stack, wrap_stack),  # stack these intermediate results
    finalize=most_common_chunked,  # get most common value from the combined result
    fill_value=0,
)

Are there useful properties for the field you're collapsing or the field you're grouping by (example plots of both these fields would be useful)? For example are there only a small number of unique labels?

Our specific use case is a high resolution land cover dataset, which we want to be able to regrid easily. Of course once the "most common" strategy works, knowing the 2nd (or n-th) most common is also interesting.

Due to the nature of our dataset it will have a limited number of unique labels.

@dcherian
Copy link
Collaborator

Unfortunately, this will need some major thinking. You'll have to handle the unique and count intermediates together, similar to how the argreduction is run. This is not trivial.

Is it possible to rechunk so that a blockwise solution works? Can you describe your problem precisely? A reproducible example would be best...

Due to the nature of our dataset it will have a limited number of unique labels.

OK that's good for the exact nature of this solution.

@dcherian
Copy link
Collaborator

Also np.unique flattens the array. Is that OK for your purposes?

@dcherian
Copy link
Collaborator

dcherian commented Oct 4, 2023

#269 should fix your blockwise problems. I'm thinking I can just add support for mode applied blockwise. Would that solve your use case?

@BSchilperoort
Copy link
Author

BSchilperoort commented Oct 4, 2023

Hi Deepak, thank you for your replies and the modifications to the code. I am currently too busy with other projects to focus on this, but I will get back to it and try it out as soon as I can make some time!


Also np.unique flattens the array. Is that OK for your purposes?

Yes that is fine.

A reproducible example would be best...

I do have a demo notebook on our repository where I show a basic use case. It is not easily reproducible as it requires a very large dataset at the moment, but it might give you a better idea of what we are trying to achieve.

@dcherian
Copy link
Collaborator

dcherian commented Apr 30, 2024

Ah I've been dense. We can simply compute the histogram and then get the mode:

import flox
import numpy as np
import pandas as pd

array = np.array([[0, 0, 0, 1, 1, 5, 4, 5, 5, 3], [0, 0, 0, 1, 1, 3, 4, 5, 5, 3]])
array[1, :] += 4
by = np.array([0, 0, 0, 1, 1, 2, 2, 2, 3, 3])

The default only works with numpy and problems that can be executed "blockwise"

flox.groupby_reduce(array, by, func="mode", axis=-1)
(array([[0, 1, 5, 3],
        [4, 5, 7, 7]]),
 array([0, 1, 2, 3]))

This is the more general implementation that should always work regardless of chunking

result, uniques, groups = flox.groupby_reduce(array, array, np.broadcast_to(by, array.shape), func="count", fill_value=0, axis=-1)
uniques[result.argmax(axis=-2)], groups
(array([[0, 1, 5, 3],
        [4, 5, 7, 7]]),
 array([0, 1, 2, 3]))

With dask, you'll need to provide the unique values of array in expected_groups (this could be relaxed).

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment
Labels
None yet
Projects
None yet
Development

No branches or pull requests

2 participants