Python中内存高效的大块numpy数组

Question

I need to sort a VERY large genomic dataset using numpy. I have an array of 2.6 billion floats, dimensions = (868940742, 3) which takes up about 20GB of memory on my machine once loaded and just sitting there. I have an early 2015 13' MacBook Pro with 16GB of RAM, 500GB solid state HD and an 3.1 GHz intel i7 processor. Just loading the array overflows to virtual memory but not to the point where my machine suffers or I have to stop everything else I am doing.

I build this VERY large array step by step from 22 smaller (N, 2) subarrays.

Function FUN_1 generates 2 new (N, 1) arrays using each of the 22 subarrays which I call sub_arr .

The first output of FUN_1 is generated by interpolating values from sub_arr[:,0] on array b = array([X, F(X)]) and the second output is generated by placing sub_arr[:, 0] into bins using array r = array([X, BIN(X)]) . I call these outputs b_arr and rate_arr , respectively. The function returns a 3-tuple of (N, 1) arrays:

import numpy as np

def FUN_1(sub_arr):
    """interpolate b values and rates based on position in sub_arr"""

    b = np.load(bfile)
    r = np.load(rfile)

    b_arr = np.interp(sub_arr[:,0], b[:,0], b[:,1])
    rate_arr = np.searchsorted(r[:,0], sub_arr[:,0])  # HUGE efficiency gain over np.digitize...

    return r[rate_r, 1], b_arr, sub_arr[:,1] 

I call the function 22 times in a for-loop and fill a pre-allocated array of zeros full_arr = numpy.zeros([868940742, 3]) with the values:

full_arr[:,0], full_arr[:,1], full_arr[:,2] = FUN_1

In terms of saving memory at this step, I think this is the best I can do, but I'm open to suggestions. Either way, I don't run into problems up through this point and it only takes about 2 minutes.

Here is the sorting routine (there are two consecutive sorts)

for idx in range(2):
    sort_idx = numpy.argsort(full_arr[:,idx])
    full_arr = full_arr[sort_idx]
    # ...
    # <additional processing, return small (1000, 3) array of stats>

Now this sort had been working, albeit slowly (takes about 10 minutes). However, I recently started using a larger, more fine resolution table of [X, F(X)] values for the interpolation step above in FUN_1 that returns b_arr and now the SORT really slows down, although everything else remains the same.

Interestingly, I am not even sorting on the interpolated values at the step where the sort is now lagging. Here are some snippets of the different interpolation files - the smaller one is about 30% smaller in each case and far more uniform in terms of values in the second column; the slower one has a higher resolution and many more unique values, so the results of interpolation are likely more unique, but I'm not sure if this should have any kind of effect...?

bigger, slower file:

17399307    99.4
17493652    98.8
17570460    98.2
17575180    97.6
17577127    97
17578255    96.4
17580576    95.8
17583028    95.2
17583699    94.6
17584172    94

smaller, more uniform regular file:

1       24  
1001    24  
2001    24  
3001    24  
4001    24  
5001    24
6001    24
7001    24

I'm not sure what could be causing this issue and I would be interested in any suggestions or just general input about sorting in this type of memory limiting case!

Answer 1

At the moment each call to np.argsort is generating a (868940742, 1) array of int64 indices, which will take up ~7 GB just by itself. Additionally, when you use these indices to sort the columns of full_arr you are generating another (868940742, 1) array of floats, since fancy indexing always returns a copy rather than a view .

One fairly obvious improvement would be to sort full_arr in place using its .sort() method . Unfortunately, .sort() does not allow you to directly specify a row or column to sort by. However, you can specify a field to sort by for a structured array. You can therefore force an inplace sort over one of the three columns by getting a view onto your array as a structured array with three float fields, then sorting by one of these fields:

full_arr.view('f8, f8, f8').sort(order=['f0'], axis=0)

In this case I'm sorting full_arr in place by the 0th field, which corresponds to the first column. Note that I've assumed that there are three float64 columns ( 'f8' ) - you should change this accordingly if your dtype is different. This also requires that your array is contiguous and in row-major format, ie full_arr.flags.C_CONTIGUOUS == True .

Credit for this method should go to Joe Kington for his answer here .

---

Although it requires less memory, sorting a structured array by field is unfortunately much slower compared with using np.argsort to generate an index array, as you mentioned in the comments below (see this previous question ). If you use np.argsort to obtain a set of indices to sort by, you might see a modest performance gain by using np.take rather than direct indexing to get the sorted array:

 %%timeit -n 1 -r 100 x = np.random.randn(10000, 2); idx = x[:, 0].argsort()
x[idx]
# 1 loops, best of 100: 148 µs per loop

 %%timeit -n 1 -r 100 x = np.random.randn(10000, 2); idx = x[:, 0].argsort()
np.take(x, idx, axis=0)
# 1 loops, best of 100: 42.9 µs per loop

However I wouldn't expect to see any difference in terms of memory usage, since both methods will generate a copy.

---

Regarding your question about why sorting the second array is faster - yes, you should expect any reasonable sorting algorithm to be faster when there are fewer unique values in the array because on average there's less work for it to do. Suppose I have a random sequence of digits between 1 and 10:

5  1  4  8  10  2  6  9  7  3

There are 10! = 3628800 possible ways to arrange these digits, but only one in which they are in ascending order. Now suppose there are just 5 unique digits:

4  4  3  2  3  1  2  5  1  5

Now there are 2⁵ = 32 ways to arrange these digits in ascending order, since I could swap any pair of identical digits in the sorted vector without breaking the ordering.

By default, np.ndarray.sort() uses Quicksort . The qsort variant of this algorithm works by recursively selecting a 'pivot' element in the array, then reordering the array such that all the elements less than the pivot value are placed before it, and all of the elements greater than the pivot value are placed after it. Values that are equal to the pivot are already sorted. Having fewer unique values means that, on average, more values will be equal to the pivot value on any given sweep, and therefore fewer sweeps are needed to fully sort the array.

For example:

%%timeit -n 1 -r 100 x = np.random.random_integers(0, 10, 100000)
x.sort()
# 1 loops, best of 100: 2.3 ms per loop

%%timeit -n 1 -r 100 x = np.random.random_integers(0, 1000, 100000)
x.sort()
# 1 loops, best of 100: 4.62 ms per loop

In this example the dtypes of the two arrays are the same. If your smaller array has a smaller item size compared with the larger array then the cost of copying it due to the fancy indexing will also be smaller.

Answer 2

EDIT: In case anyone new to programming and numpy comes across this post, I want to point out the importance of considering the np.dtype that you are using. In my case, I was actually able to get away with using half-precision floating point, ie np.float16 , which reduced a 20GB object in memory to 5GB and made sorting much more manageable. The default used by numpy is np.float64 , which is a lot of precision that you may not need. Check out the doc here, which describes the capacity of the different data types. Thanks to @ali_m for pointing this out in the comments.

I did a bad job explaining this question but I have discovered some helpful workarounds that I think would be useful to share for anyone who needs to sort a truly massive numpy array.

I am building a very large numpy array from 22 "sub-arrays" of human genome data containing the elements [position, value] . Ultimately, the final array must be numerically sorted "in place" based on the values in a particular column and without shuffling the values within rows.

The sub-array dimensions follow the form:

arr1.shape = (N1, 2)
...
arr22.shape = (N22, 2)

sum([N1..N2]) = 868940742 ie there are close to 1BN positions to sort.

First I process the 22 sub-arrays with the function process_sub_arrs , which returns a 3-tuple of 1D arrays the same length as the input. I stack the 1D arrays into a new (N, 3) array and insert them into an np.zeros array initialized for the full dataset:

    full_arr = np.zeros([868940742, 3])
    i, j = 0, 0

    for arr in list(arr1..arr22):  
        # indices (i, j) incremented at each loop based on sub-array size
        j += len(arr)
        full_arr[i:j, :] = np.column_stack( process_sub_arrs(arr) )
        i = j

    return full_arr

EDIT: Since I realized my dataset could be represented with half-precision floats, I now initialize full_arr as follows: full_arr = np.zeros([868940742, 3], dtype=np.float16) , which is only 1/4 the size and much easier to sort.

Result is a massive 20GB array:

full_arr.nbytes = 20854577808

As @ali_m pointed out in his detailed post, my earlier routine was inefficient:

sort_idx = np.argsort(full_arr[:,idx])
full_arr = full_arr[sort_idx]

the array sort_idx , which is 33% the size of full_arr , hangs around and wastes memory after sorting full_arr . This sort supposedly generates a copy of full_arr due to "fancy" indexing, potentially pushing memory use to 233% of what is already used to hold the massive array! This is the slow step, lasting about ten minutes and relying heavily on virtual memory.

I'm not sure the "fancy" sort makes a persistent copy however. Watching the memory usage on my machine, it seems that full_arr = full_arr[sort_idx] deletes the reference to the unsorted original, because after about 1 second all that is left is the memory used by the sorted array and the index, even if there is a transient copy.

A more compact usage of argsort() to save memory is this one:

    full_arr = full_arr[full_arr[:,idx].argsort()]

This still causes a spike at the time of the assignment, where both a transient index array and a transient copy are made, but the memory is almost instantly freed again.

@ali_m pointed out a nice trick (credited to Joe Kington) for generating a de facto structured array with a view on full_arr . The benefit is that these may be sorted "in place", maintaining stable row order:

full_arr.view('f8, f8, f8').sort(order=['f0'], axis=0)

Views work great for performing mathematical array operations, but for sorting it is far too inefficient for even a single sub-array from my dataset. In general, structured arrays just don't seem to scale very well even though they have really useful properties. If anyone has any idea why this is I would be interested to know.

One good option to minimize memory consumption and improve performance with very large arrays is to build a pipeline of small, simple functions. Functions clear local variables once they have completed so if intermediate data structures are building up and sapping memory this can be a good solution.

This a sketch of the pipeline I've used to speed up the massive array sort:

def process_sub_arrs(arr):
    """process a sub-array and return a 3-tuple of 1D values arrays"""

    return values1, values2, values3

def build_arr():
    """build the initial array by joining processed sub-arrays"""

    full_arr = np.zeros([868940742, 3])
    i, j = 0, 0

    for arr in list(arr1..arr22):  
        # indices (i, j) incremented at each loop based on sub-array size
        j += len(arr)
        full_arr[i:j, :] = np.column_stack( process_sub_arrs(arr) )
        i = j

    return full_arr

def sort_arr():
    """return full_arr and sort_idx"""

    full_arr = build_arr()
    sort_idx = np.argsort(full_arr[:, index])

    return full_arr[sort_idx]

def get_sorted_arr():
    """call through nested functions to return the sorted array"""

    sorted_arr = sort_arr()
    <process sorted_arr>

    return statistics

call stack: get_sorted_arr --> sort_arr --> build_arr --> process_sub_arrs

Once each inner function is completed get_sorted_arr() finally just holds the sorted array and then returns a small array of statistics.

EDIT: It is also worth pointing out here that even if you are able to use a more compact dtype to represent your huge array, you will want to use higher precision for summary calculations. For example, since full_arr.dtype = np.float16 , the command np.mean(full_arr[:,idx]) tries to calculate the mean in half-precision floating point, but this quickly overflows when summing over a massive array. Using np.mean(full_arr[:,idx], dtype=np.float64) will prevent the overflow.

I posted this question initially because I was puzzled by the fact that a dataset of identical size suddenly began choking up my system memory, although there was a big difference in the proportion of unique values in the new "slow" set. @ali_m pointed out that, indeed, more uniform data with fewer unique values is easier to sort:

The qsort variant of Quicksort works by recursively selecting a 'pivot' element in the array, then reordering the array such that all the elements less than the pivot value are placed before it, and all of the elements greater than the pivot value are placed after it. Values that are equal to the pivot are already sorted, so intuitively, the fewer unique values there are in the array, the smaller the number of swaps there are that need to be made.

On that note, the final change I ended up making to attempt to resolve this issue was to round the newer dataset in advance, since there was an unnecessarily high level of decimal precision leftover from an interpolation step. This ultimately had an even bigger effect than the other memory saving steps, showing that the sort algorithm itself was the limiting factor in this case.

Look forward to other comments or suggestions anyone might have on this topic, and I almost certainly misspoke about some technical issues so I would be glad to hear back :-)