Related
I have three 1D arrays:
idxs: the index data
weights: the weight of each index in idxs
bins: the bin which is used to calculate minimum weight in it.
Here's my current method of using the idxs to check the data called weights in which bin, and then calculate the min/max of binned weights:
Get slices that shows which bin each idxs element belongs to.
Sort slices and weights at the same time.
Calculate the minimum of weights in each bin (slice).
numpy method
import random
import numpy as np
# create example data
out_size = int(10)
bins = np.arange(3, out_size-3)
idxs = np.arange(0, out_size)
#random.shuffle(idxs)
# set duplicated slice manually for test
idxs[4] = idxs[3]
idxs[6] = idxs[7]
weights = idxs
# get which bin idxs belong to
slices = np.digitize(idxs, bins)
# get index and weights in bins
valid = (bins.max() >= idxs) & (idxs >= bins.min())
valid_slices = slices[valid]
valid_weights = weights[valid]
# sort slice and weights
sort_index = valid_slices.argsort()
valid_slices_sort = valid_slices[sort_index]
valid_weights_sort = valid_weights[sort_index]
# get index of each first unque slices
unique_slices, unique_index = np.unique(valid_slices_sort, return_index=True)
# calculate the minimum
res_sub = np.minimum.reduceat(valid_weights_sort, unique_index)
# save results
res = np.full((out_size), np.nan)
res[unique_slices-1] = res_sub
print(res)
Results:
array([ 3., nan, 5., nan, nan, nan, nan, nan, nan, nan])
If I increase the out_size to 1e7 and shuffle the data, the speed (from np.digitize to the end) is slow:
13.5 s ± 136 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
And, here's the consumed time of each part:
np.digitize: 10.8 s ± 12.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
valid: 171 ms ± 3.78 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
argsort and slice: 2.02 s ± 33.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
unique: 9.9 ms ± 113 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
np.minimum.reduceat: 5.11 ms ± 52.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
np.digitize() costs the most of time: 10.8 s. And, the next is argsort: 2.02 s.
I also check the consumed time of calculating mean using np.histogram:
counts, _ = np.histogram(idxs, bins=out_size, range=(0, out_size))
sums, _ = np.histogram(idxs, bins=out_size, range=(0, out_size), weights = weights, density=False)
mean = sums / np.where(counts == 0, np.nan, counts)
33.2 s ± 3.47 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
This is similar to my method of calculating minimum.
scipy method
from scipy.stats import binned_statistic
statistics, _, _ = binned_statistic(idxs, weights, statistic='min', bins=bins)
print(statistics)
The result is a little different, but the speed is much slower (x6) for the longer(1e7) shuffled data:
array([ 3., nan, 5.])
1min 20s ± 6.93 s per loop (mean ± std. dev. of 7 runs, 1 loop each)
Summary
I want to figure out a quicker method. If the method is also suitable for dask, that would be excellent!
User Case
Here's how my real data (1D) look like:
SultanOrazbayev showed a quick approach; I'll add a cool one.
mask = bins[:, None] == idxs[None, :]
result = np.nanmin(np.where(mask, weights, np.nan), axis=-1)
# Note: may produce (expected) runtime warning if bin has no values
of course, you can also do np.nanmax, np.nanmean, etc.
The above assumes that your bins are indeed single values. If they are ranges, you need slightly little more work to construct the mask
lower_mask = idxs[None, :] >= bins[:, None]
upper_mask = np.empty_like(lower_mask)
upper_mask[:-1, ...] = idxs[None, :] < bins[1:, None]
upper_mask[-1, ...] = False
mask = lower_mask & upper_mask
at which point you can use np.nanmin like above.
Ofc np.where and the broadcast to create a mask will create new arrays of shape (len(bins), len(idxs)) with their respective datatype. If this is of no concern to you, then the above is probably good enough.
If it is a problem (because you are pressed for RAM), then my first suggestion is to buy more RAM. If - for some stupid reason (say, money) - that doesn't work, you can avoid the copy of weights by using a masked array over a manually re-strided view into weights
import numpy.ma as ma
mask = ...
restrided_weights = np.lib.stride_tricks.as_strided(weights, shape=(bins.size, idxs.size), strides=(0, idxs.strides[0]))
masked = ma.masked_array(restrided_weights, mask=~mask, fill_value=np.nan, dtype=np.float64)
result = masked.min(axis=-1).filled(np.nan)
this avoids both, a copy of weights and the above-mentioned runtime warning.
If you don't even have enough memory to construct mask, then you can try processing the data in chunks.
Last I checked, Dask used to have funny behavior when fed with manually strided arrays. There was some work on this though, so you may want to double-check if that is resolved, in which case you can happily parallelize the above.
Update based on your further comments to this answer and the other:
You can do this computation in chunks to avoid memory issues due to your large number of bins (1e4 in magnitude). Putting the concrete numbers into a full example and adding a progress bar indicates <40 seconds runtime on a single core.
import numpy.ma as ma
from tqdm import trange
import numpy as np
import random
# create example data
out_size = int(1.5e5)
#bins = np.arange(3, out_size-3)
bins = np.arange(3, int(3.8e4-3), dtype=np.int64)
idxs = np.arange(0, out_size)
random.shuffle(idxs)
# set duplicated slice manually for test
idxs[4] = idxs[3]
idxs[6] = idxs[7]
weights = idxs
chunk_size = 100
# preallocate buffers to avoid array creation in main loop
extended_bins = np.empty(len(bins) + 1, dtype=bins.dtype)
extended_bins[:-1] = bins
extended_bins[-1] = np.iinfo(bins.dtype).max # last bin goes to infinity
mask_buffer = np.empty((chunk_size, len(idxs)), dtype=bool)
result = np.empty_like(bins, dtype=np.float64)
for low in trange(0, len(bins), chunk_size):
high = min(low + chunk_size, len(bins))
chunk_size = high - low
mask_buffer[:chunk_size, ...] = ~((bins[low:high, None] <= idxs[None, :]) & (extended_bins[low+1:high+1, None] > idxs[None, :]))
mask = mask_buffer[:chunk_size, ...]
restrided_weights = np.lib.stride_tricks.as_strided(weights, shape=mask.shape, strides=(0, idxs.strides[0]))
masked = ma.masked_array(restrided_weights, mask=mask, fill_value=np.nan, dtype=np.float64)
result[low:high] = masked.min(axis=-1).filled(np.nan)
Bonus: For min and max only there is a cool trick that you can use: sort idxs and weights based on weights (ascending for min, descending for max). This way, you can immediately look up the min/max value and can avoid the masked array and the custom strides altogether. This relies on some not so well documented behavior of np.argmax, which takes a fast-pass for boolean arrays and doesn't search the full array.
It only works for these two cases, and you'd have to fall back to the above for more sophisticated things (mean), but for those two it shaves off another ~70% and a run on a single core clocks in at <12 seconds.
# fast min/max
from tqdm import trange
import numpy as np
# create example data
out_size = int(1.5e5)
#bins = np.arange(3, out_size-3)
bins = np.arange(3, int(3.8e4-3), dtype=np.int64)
idxs = np.arange(0, out_size)
random.shuffle(idxs)
# set duplicated slice manually for test
idxs[4] = idxs[3]
idxs[6] = idxs[7]
weights = idxs
order = np.argsort(weights)
weights_sorted = np.empty((len(weights) + 1), dtype=np.float64)
weights_sorted[:-1] = weights[order]
weights_sorted[-1] = np.nan
idxs_sorted = idxs[order]
extended_bins = np.empty(len(bins) + 1, dtype=bins.dtype)
extended_bins[:-1] = bins
extended_bins[-1] = np.iinfo(bins.dtype).max # last bin goes to infinity
# preallocate buffers to avoid array creation in main loop
chunk_size = 1000
mask_buffer = np.empty((chunk_size, len(idxs) + 1), dtype=bool)
mask_buffer[:, -1] = True
result = np.empty_like(bins, dtype=np.float64)
for low in trange(0, len(bins), chunk_size):
high = min(low + chunk_size, len(bins))
chunk_size = high - low
mask_buffer[:chunk_size, :-1] = (bins[low:high, None] <= idxs_sorted[None, :]) & (extended_bins[low+1:high+1, None] > idxs_sorted[None, :])
mask = mask_buffer[:chunk_size, ...]
weight_idx = np.argmax(mask, axis=-1)
result[low:high] = weights_sorted[weight_idx]
A quick approach to achieve this is with dask.dataframe and pd.cut, I first show the pandas version:
import numpy as np
from scipy.stats import binned_statistic as bs
import pandas as pd
nrows=10**7
df = pd.DataFrame(np.random.rand(nrows, 2), columns=['x', 'val'])
bins = np.linspace(df['x'].min(), df['x'].max(), 10)
df['binned_x'] = pd.cut(df['x'], bins=bins, right=False)
result_pandas = df.groupby('binned_x')['val'].min().values
result_scipy = bs(df['x'], df['val'], 'min', bins=bins)[0]
print(np.isclose(result_pandas, result_scipy))
# [ True True True True True True True True True]
Now to go from pandas to dask, you will need to make sure that bins are consistent across partitions, so take a look here. Once every partition is binned consistently, you want to apply the desired operation (min/max/sum/count):
import dask.dataframe as dd
ddf = dd.from_pandas(df, npartitions=10)
def f(df, bins):
df = df.copy()
df['binned_x'] = pd.cut(df['x'], bins=bins, right=False)
result = df.groupby('binned_x', as_index=False)['val'].min()
return result
result_dask = ddf.map_partitions(f, bins).groupby('binned_x')['val'].min().compute()
print(np.isclose(result_pandas, result_dask))
# [ True True True True True True True True True]
On my laptop, the first code take about 7 3 seconds, the second code is about 10 times faster (forgot that I am double-counting both pandas and scipy performing the same operation). There is scope for playing around with partitioning, but that's context-dependent, so something you can try optimizing on your data/hardware.
Update: note that this approach will work for min/max, but for mean you will want to calculate sum and count and then divide them. There is probably a good way of keeping track of weights in doing this calculation in one go, but it might not be worth the added code complexity.
I want to calculate a rolling quantile of a large 2D matrix with dimensions (1e6, 1e5), column wise. I am looking for the fastest way, since I need to perform this operation thousands of times, and it's very computationally expensive. For experiments window=1000 and q=0.1 is used.
import numpy as np
import pandas as pd
import multiprocessing as mp
from functools import partial
import numba as nb
X = np.random.random((10000,1000)) # Original array has dimensions of about (1e6, 1e5)
My current approaches:
Pandas: %timeit: 5.8 s ± 15.5 ms per loop
def pd_rolling_quantile(X, window, q):
return pd.DataFrame(X).rolling(window).quantile(quantile=q)
Numpy Strided: %timeit: 2min 42s ± 3.29 s per loop
def strided_app(a, L, S):
nrows = ((a.size-L)//S)+1
n = a.strides[0]
return np.lib.stride_tricks.as_strided(a, shape=(nrows,L), strides=(S*n,n))
def np_1d(x, window, q):
return np.pad(np.percentile(strided_app(x, window, 1), q*100, axis=-1), (window-1, 0) , mode='constant')
def np_rolling_quantile(X, window, q):
results = []
for i in np.arange(X.shape[1]):
results.append(np_1d(X[:,i], window, q))
return np.column_stack(results)
Multiprocessing: %timeit: 1.13 s ± 27.6 ms per loop
def mp_rolling_quantile(X, window, q):
pool = mp.Pool(processes=12)
results = pool.map(partial(pd_rolling_quantile, window=window, q=q), [X[:,i] for i in np.arange(X.shape[1])])
pool.close()
pool.join()
return np.column_stack(results)
Numba: %timeit: 2min 28s ± 182 ms per loop
#nb.njit
def nb_1d(x, window, q):
out = np.zeros(x.shape[0])
for i in np.arange(x.shape[0]-window+1)+window:
out[i-1] = np.quantile(x[i-window:i], q=q)
return out
def nb_rolling_quantile(X, window, q):
results = []
for i in np.arange(X.shape[1]):
results.append(nb_1d(X[:,i], window, q))
return np.column_stack(results)
The timings are not great, and ideally I would target an improvement of 10-50x by speed. I would appreciate any suggestions, how to speed it up. Maybe someone has ideas on using lower level languages (Cython), or other ways to speed it up with Numpy/Numba/Tensorflow based methods. Thanks!
I would recommend the new rolling-quantiles package.
To demonstrate, even the somewhat naive approach of constructing a separate filter for each column outperforms the above single-threaded pandas experiment:
pipes = [rq.Pipeline(rq.LowPass(window=1000, quantile=0.1)) for i in range(1000)]
%timeit [pipe.feed(X[:, i]) for i, pipe in enumerate(pipes)]
1.34 s ± 7.76 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
versus
df = pd.DataFrame(X)
%timeit df.rolling(1000).quantile(0.1)
5.63 s ± 27 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Both can be trivially parallelized by means of multiprocessing, as you showed.
My goal is to convert a list of pixels from RGB to Hex as quickly as possible. The format is a Numpy dimensional array (rgb colorspace) and ideally it would be converted from RGB to Hex and maintain it's shape.
My attempt at doing this uses list comprehension and with the exception of performance, it solves it. Performance wise, adding the ravel and list comprehension really slows this down. Unfortunately I just don't know enough math to know the solution of how to to speed this up:
Edited: Updated code to reflex most recent changes. Current running around 24ms on 35,000 pixel image.
def np_array_to_hex(array):
array = np.asarray(array, dtype='uint32')
array = (1 << 24) + ((array[:, :, 0]<<16) + (array[:, :, 1]<<8) + array[:, :, 2])
return [hex(x)[-6:] for x in array.ravel()]
>>> np_array_to_hex(img)
['afb3bc', 'abaeb5', 'b3b4b9', ..., '8b9dab', '92a4b2', '9caebc']
I tried it with a LUT ("Look Up Table") - it takes a few seconds to initialise and it uses 100MB (0.1GB) of RAM, but that's a small price to pay amortised over a million images:
#!/usr/bin/env python3
import numpy as np
def np_array_to_hex1(array):
array = np.asarray(array, dtype='uint32')
array = ((array[:, :, 0]<<16) + (array[:, :, 1]<<8) + array[:, :, 2])
return array
def np_array_to_hex2(array):
array = np.asarray(array, dtype='uint32')
array = (1 << 24) + ((array[:, :, 0]<<16) + (array[:, :, 1]<<8) + array[:, :, 2])
return [hex(x)[-6:] for x in array.ravel()]
def me(array, LUT):
h, w, d = array.shape
# Reshape to a color vector
z = np.reshape(array,(-1,3))
# Make array and fill with 32-bit colour number
y = np.zeros((h*w),dtype=np.uint32)
y = z[:,0]*65536 + z[:,1]*256 + z[:,2]
return LUT[y]
# Define dummy image of 35,000 RGB pixels
w,h = 175, 200
im = np.random.randint(0,256,(h,w,3),dtype=np.uint8)
# %timeit np_array_to_hex1(im)
# 112 µs ± 1.1 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
# %timeit np_array_to_hex2(im)
# 8.42 ms ± 136 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
# This may take time to set up, but amortize that over a million images...
LUT = np.zeros((256*256*256),dtype='a6')
for i in range(256*256*256):
h = hex(i)[2:].zfill(6)
LUT[i] = h
# %timeit me(im,LUT)
# 499 µs ± 8.15 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
So that appears to be 4x slower than your fastest which doesn't work, and 17x faster that your slowest which does work.
My next suggestion is to use multi-threading or multi-processing so all your CPU cores get busy in parallel and reduce your overall time by a factor of 4 or more assuming you have a reasonably modern 4+ core CPU.
The following data, represent 2 given histograms split into 13 bins:
key 0 1-9 10-18 19-27 28-36 37-45 46-54 55-63 64-72 73-81 82-90 91-99 100
A 1.274580708 2.466224824 5.045757621 7.413716262 8.958855646 10.41325305 11.14150951 10.91949012 11.29095648 10.95054297 10.10976255 8.128781795 1.886568472
B 0 1.700493692 4.059243006 5.320899616 6.747120132 7.899067471 9.434997257 11.24520022 12.94569391 12.83598464 12.6165661 10.80636314 4.388370817
I'm trying to follow this article in order to calculate the intersection between those 2 histograms, using this method:
def histogram_intersection(h1, h2, bins):
bins = numpy.diff(bins)
sm = 0
for i in range(len(bins)):
sm += min(bins[i]*h1[i], bins[i]*h2[i])
return sm
Since my data is already calculated as a histogram, I can't use numpy built-in function, so I'm failing to provide the function the necessary data.
How can I process my data to fit the algorithm?
Since you have the same number of bins for both of the histograms you can use:
def histogram_intersection(h1, h2):
sm = 0
for i in range(13):
sm += min(h1[i], h2[i])
return sm
You can calculate it faster and more simply with Numpy:
#!/usr/bin/env python3
import numpy as np
A = np.array([1.274580708,2.466224824,5.045757621,7.413716262,8.958855646,10.41325305,11.14150951,10.91949012,11.29095648,10.95054297,10.10976255,8.128781795,1.886568472])
B = np.array([0,1.700493692,4.059243006,5.320899616,6.747120132,7.899067471,9.434997257,11.24520022,12.94569391,12.83598464,12.6165661,10.80636314,4.388370817])
def histogram_intersection(h1, h2):
sm = 0
for i in range(13):
sm += min(h1[i], h2[i])
return sm
print(histogram_intersection(A,B))
print(np.sum(np.minimum(A,B)))
Output
88.44792356099998
88.447923561
But if you time it, Numpy only takes 60% of the time:
%timeit histogram_intersection(A,B)
5.02 µs ± 65.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit np.sum(np.minimum(A,B))
3.22 µs ± 11.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Some caveats first : in your data bins are ranges, in your algorithm they are numbers. You must redefine bins for that.
Furthermore, min(bins[i]*h1[i], bins[i]*h2[i]) is bins[i]*min(h1[i], h2[i]), so the result can be obtained by :
hists=pandas.read_clipboard(index_col=0) # your data
bins=arange(-4,112,9) # try for bins but edges are different here
mins=hists.min('rows')
intersection=dot(mins,bins)
Is there a way to calculate many histograms along an axis of an nD-array? The method I currently have uses a for loop to iterate over all other axes and calculate a numpy.histogram() for each resulting 1D array:
import numpy
import itertools
data = numpy.random.rand(4, 5, 6)
# axis=-1, place `200001` and `[slice(None)]` on any other position to process along other axes
out = numpy.zeros((4, 5, 200001), dtype="int64")
indices = [
numpy.arange(4), numpy.arange(5), [slice(None)]
]
# Iterate over all axes, calculate histogram for each cell
for idx in itertools.product(*indices):
out[idx] = numpy.histogram(
data[idx],
bins=2 * 100000 + 1,
range=(-100000 - 0.5, 100000 + 0.5),
)[0]
out.shape # (4, 5, 200001)
Needless to say this is very slow, however I couldn't find a way to solve this using numpy.histogram, numpy.histogram2d or numpy.histogramdd.
Here's a vectorized approach making use of the efficient tools np.searchsorted and np.bincount. searchsorted gives us the loactions where each element is to be placed based on the bins and bincount does the counting for us.
Implementation -
def hist_laxis(data, n_bins, range_limits):
# Setup bins and determine the bin location for each element for the bins
R = range_limits
N = data.shape[-1]
bins = np.linspace(R[0],R[1],n_bins+1)
data2D = data.reshape(-1,N)
idx = np.searchsorted(bins, data2D,'right')-1
# Some elements would be off limits, so get a mask for those
bad_mask = (idx==-1) | (idx==n_bins)
# We need to use bincount to get bin based counts. To have unique IDs for
# each row and not get confused by the ones from other rows, we need to
# offset each row by a scale (using row length for this).
scaled_idx = n_bins*np.arange(data2D.shape[0])[:,None] + idx
# Set the bad ones to be last possible index+1 : n_bins*data2D.shape[0]
limit = n_bins*data2D.shape[0]
scaled_idx[bad_mask] = limit
# Get the counts and reshape to multi-dim
counts = np.bincount(scaled_idx.ravel(),minlength=limit+1)[:-1]
counts.shape = data.shape[:-1] + (n_bins,)
return counts
Runtime test
Original approach -
def org_app(data, n_bins, range_limits):
R = range_limits
m,n = data.shape[:2]
out = np.zeros((m, n, n_bins), dtype="int64")
indices = [
np.arange(m), np.arange(n), [slice(None)]
]
# Iterate over all axes, calculate histogram for each cell
for idx in itertools.product(*indices):
out[idx] = np.histogram(
data[idx],
bins=n_bins,
range=(R[0], R[1]),
)[0]
return out
Timings and verification -
In [2]: data = np.random.randn(4, 5, 6)
...: out1 = org_app(data, n_bins=200001, range_limits=(- 2.5, 2.5))
...: out2 = hist_laxis(data, n_bins=200001, range_limits=(- 2.5, 2.5))
...: print np.allclose(out1, out2)
...:
True
In [3]: %timeit org_app(data, n_bins=200001, range_limits=(- 2.5, 2.5))
10 loops, best of 3: 39.3 ms per loop
In [4]: %timeit hist_laxis(data, n_bins=200001, range_limits=(- 2.5, 2.5))
100 loops, best of 3: 3.17 ms per loop
Since, in the loopy solution, we are looping through the first two axes. So, let's increase their lengths as that would show us how good is the vectorized one -
In [59]: data = np.random.randn(400, 500, 6)
In [60]: %timeit org_app(data, n_bins=21, range_limits=(- 2.5, 2.5))
1 loops, best of 3: 9.59 s per loop
In [61]: %timeit hist_laxis(data, n_bins=21, range_limits=(- 2.5, 2.5))
10 loops, best of 3: 44.2 ms per loop
In [62]: 9590/44.2 # Speedup number
Out[62]: 216.9683257918552
The first solution provided a nice short idiom which uses numpy sortedsearch which has the cost of a sort and many searches. But numpy has a fast route in its source code which is done in Python in fact, which can deal with equal bin edge range mathematically. This solution uses only a vectorized subtraction and multiplication and some comparisons instead.
This solution will follow numpy code for the search sorted, type imputations, and handles weights as well as complex numbers. It is basically the first solution combined with numpy histogram fast route, and some extra type, and iteration details, etc.
_range = range
def hist_np_laxis(a, bins=10, range=None, weights=None):
# Initialize empty histogram
N = a.shape[-1]
data2D = a.reshape(-1,N)
limit = bins*data2D.shape[0]
# gh-10322 means that type resolution rules are dependent on array
# shapes. To avoid this causing problems, we pick a type now and stick
# with it throughout.
bin_type = np.result_type(range[0], range[1], a)
if np.issubdtype(bin_type, np.integer):
bin_type = np.result_type(bin_type, float)
bin_edges = np.linspace(range[0],range[1],bins+1, endpoint=True, dtype=bin_type)
# Histogram is an integer or a float array depending on the weights.
if weights is None:
ntype = np.dtype(np.intp)
else:
ntype = weights.dtype
n = np.zeros(limit, ntype)
# Pre-compute histogram scaling factor
norm = bins / (range[1] - range[0])
# We set a block size, as this allows us to iterate over chunks when
# computing histograms, to minimize memory usage.
BLOCK = 65536
# We iterate over blocks here for two reasons: the first is that for
# large arrays, it is actually faster (for example for a 10^8 array it
# is 2x as fast) and it results in a memory footprint 3x lower in the
# limit of large arrays.
for i in _range(0, data2D.shape[0], BLOCK):
tmp_a = data2D[i:i+BLOCK]
block_size = tmp_a.shape[0]
if weights is None:
tmp_w = None
else:
tmp_w = weights[i:i + BLOCK]
# Only include values in the right range
keep = (tmp_a >= range[0])
keep &= (tmp_a <= range[1])
if not np.logical_and.reduce(np.logical_and.reduce(keep)):
tmp_a = tmp_a[keep]
if tmp_w is not None:
tmp_w = tmp_w[keep]
# This cast ensures no type promotions occur below, which gh-10322
# make unpredictable. Getting it wrong leads to precision errors
# like gh-8123.
tmp_a = tmp_a.astype(bin_edges.dtype, copy=False)
# Compute the bin indices, and for values that lie exactly on
# last_edge we need to subtract one
f_indices = (tmp_a - range[0]) * norm
indices = f_indices.astype(np.intp)
indices[indices == bins] -= 1
# The index computation is not guaranteed to give exactly
# consistent results within ~1 ULP of the bin edges.
decrement = tmp_a < bin_edges[indices]
indices[decrement] -= 1
# The last bin includes the right edge. The other bins do not.
increment = ((tmp_a >= bin_edges[indices + 1])
& (indices != bins - 1))
indices[increment] += 1
((bins*np.arange(i, i+block_size)[:,None] * keep)[keep].reshape(indices.shape) + indices).reshape(-1)
#indices = scaled_idx.reshape(-1)
# We now compute the histogram using bincount
if ntype.kind == 'c':
n.real += np.bincount(indices, weights=tmp_w.real,
minlength=limit)
n.imag += np.bincount(indices, weights=tmp_w.imag,
minlength=limit)
else:
n += np.bincount(indices, weights=tmp_w,
minlength=limit).astype(ntype)
n.shape = a.shape[:-1] + (bins,)
return n
data = np.random.randn(4, 5, 6)
out1 = hist_laxis(data, n_bins=200001, range_limits=(- 2.5, 2.5))
out2 = hist_np_laxis(data, bins=200001, range=(- 2.5, 2.5))
print(np.allclose(out1, out2))
True
%timeit hist_np_laxis(data, bins=21, range=(- 2.5, 2.5))
92.1 µs ± 504 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit hist_laxis(data, n_bins=21, range_limits=(- 2.5, 2.5))
55.1 µs ± 3.66 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Although the first solution is faster in the small example and even the larger example:
data = np.random.randn(400, 500, 6)
%timeit hist_np_laxis(data, bins=21, range=(- 2.5, 2.5))
264 ms ± 2.68 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
%timeit hist_laxis(data, n_bins=21, range_limits=(- 2.5, 2.5))
71.6 ms ± 377 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
It is not ALWAYS faster:
data = np.random.randn(400, 6, 500)
%timeit hist_np_laxis(data, bins=101, range=(- 2.5, 2.5))
71.5 ms ± 128 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit hist_laxis(data, n_bins=101, range_limits=(- 2.5, 2.5))
76.9 ms ± 137 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
However, the numpy variation is only faster when the last axis is large. And its a very slight increase. In all other cases I tried, the first solution is much faster regardless of bin count and size of the first 2 dimensions. The only important line ((bins*np.arange(i, i+block_size)[:,None] * keep)[keep].reshape(indices.shape) + indices).reshape(-1) might be more optimizable though I have not found a faster method yet.
This would also imply the sheer number of vectorized operations of O(n) is outdoing the O(n log n) of the sort and repeated incremental searches.
However, realistic use cases will have a last axis with a lot of data and the prior axes with few. So in reality the samples in the first solution are too contrived to fit the desired performance.
Axis addition for histogram is noted as an issue in the numpy repo: https://github.com/numpy/numpy/issues/13166.
An xhistogram library has sought to solve this problem as well: https://xhistogram.readthedocs.io/en/latest/