平均重叠婚姻？

Question

假设你有一个形状为(n, m)的矩阵。你知道吗

此外，还有k较小的形状矩阵(s, m)。你知道吗

这些k矩阵可能是通过将较大的矩阵分片生成的，例如：

stride = z
ranges = [] # will contain sub lists of start / end positions
for i in range(0, n, stride):
    if i + s > n:
        ranges.append([n-s, n]) #<-- if not evenly divisible include last ragged bit
        break
    else:
        ranges.append([i, i+s])

# k = len(ranges)

for a, b in ranges:
    submat = mat[a:b] # <--- produces submats of shape (s, m)
    # not necessarily where submats come from, just for 
    # simple example purpose, feel free to add random noise to each submat

在numpy中，如何连接这些k重叠的子矩阵，平均重叠区域？你知道吗

我们的目标是，然后采取这些潜艇和改革原来的垫子，例如：

blank = np.zeros((n, m))
for i in range(len(submats)):
    a, b = ranges[i]
    blank[a:b] += submats[i] #<--- doesn't account for different amounts of overlapping regions

混凝土编号：

n = 693
m = 10
# so mat has shape (693, 10)

s = 500
stride = 50

ranges = [[0, 500], [50, 550], [100, 600], [150, 650], [193, 693]]
# notice that the range (0,50) doesn't need to be averaged

k = 5 # len(ranges)

# so we have k submats of shape (500, 10)

我目前正在做：

def count_overlap(max_len, ranges): # from example 693, and [[0, 500], ...]
    tally = np.zeros(max_len)
    for i in range(max_len):
        for a, b in ranges:
            if a <= i and i < b:
                tally[i] += 1
    return tally


olap = count_overlap(693, ranges)
olap[:55]
# ([1., 1., ..., 1., 2., 2., 2., 2., 2.])
olap[-50:]
# ([2., 2., 2., 2., 2., 2., 2., 2., 1., 1., ..., 1., 1., 1.])

知道如何划分垫子的每个索引

Answer 1

在我自己的问题中，我提供了另一种（不那么优雅的）方法，但我并没有试图回答我自己的问题。相反，我只是简单地包装了相关的功能和解决方案，供其他人使用：

助手

def shard_rng(maxlen, sublen, stride):
    ranges = [] 
    for i in range(0, maxlen, stride):
        if i + sublen > maxlen:
            ranges.append([maxlen-sublen, maxlen])
            break
        else:
            ranges.append([i, i+sublen])
    return ranges

# for testing. stitched_mat - mat should be 1
def split_mat(mat, ranges):
    submats = []
    for a,b in ranges:
        submats.append(mat[a:b] + 1)
    return submats

# part of solution 1
def weight_rngs(ranges):
    n = ranges[-1][-1]
    bins = map(np.bincount,np.array(ranges).T,(None,None),(n+1,n+1))
    vals = np.subtract(*bins).cumsum()
    weights = 1 / vals[:n,None]
    return weights

解决方案

由@Paul Panzer提供

# solution 1
def stitch_mats(shape, submats, ranges):    
    stitched = np.zeros(shape)
    weights = weight_rngs(ranges)
    for submat, (start, stop) in zip(submats, ranges):
        stitched[start:stop] += weights[start:stop] * submat        
    return stitched

# solution 2
def stitch_mats2(shape, submats, ranges):
    ranges = np.array(ranges)
    ro = ranges.ravel().argsort(kind='stable')

    # put 1 for starting and -1 for ending, take cumsum
    cnts = (1-((ro&1)<<1)).cumsum()

    stitched = np.zeros((n,m))
    # add slices
    for submat, (start, stop) in zip(submats,ranges):
        stitched[start:stop] += submat

    rs = ranges.ravel()[ro]
    # divide by overlap
    for start, stop, count in zip(rs[:-1],rs[1:],cnts[:-1]):
        stitched[start:stop] /= count
    return stitched

试验

n = 693
m = 10
s = 500 # sublen
stride = 50

mat = np.random.randint(0,10,(n,m))
ranges = shard_rng(n, s, stride)
submats = split_mat(mat, ranges)


stitched_1 = stitch_mats(mat.shape, submats, ranges)
stitched_2 = stitch_mats2(mat.shape, submats, ranges)

np.unique(stitched_1-mat-1.), np.unique(stitched_2-mat-1.)
# array([-8.8817842e-16,  0.0000000e+00,  4.4408921e-16,  8.8817842e-16]), array([0.])

Answer 2

下面是使用bincount+cumsum逐行计算重叠的方法：

更新：添加了另一个只使用切片的方法。我希望这会更快。你知道吗

import numpy as np

n = 693
m = 10
# so mat has shape (693, 10)

s = 500
stride = 50

ranges = [[0, 500], [50, 550], [100, 600], [150, 650], [193, 693]]
# notice that the range (0,50) doesn't need to be averaged

k = 5 # len(ranges)

mat = np.random.randint(0,10,(n,m))
submats = []
for a, b in ranges:
    submats.append(mat[a:b])


ranges = np.asarray(ranges)
out = np.zeros((n,m))
# put a 1 at every start and a -1 at every stop
# then take the cumsum this will assign to each row the
# number of intervals it is in
# finally, take the reciprocal
weight = 1 / np.subtract(*map(np.bincount,ranges.T,(None,None),(n+1,n+1))).cumsum()[:n,None]
for sm,(a,b) in zip(submats,ranges):
    out[a:b] += weight[a:b] * sm


# method 2

# sort range ends
ro = ranges.ravel().argsort(kind='stable')
# put 1 for starting and -1 for ending, take cumsum
cnts = (1-((ro&1)<<1)).cumsum()
out = np.zeros((n,m))
# add slices
for sm,(a,b) in zip(submats,ranges):
    out[a:b] += sm
rs = ranges.ravel()[ro]
# divide by overlap
for a,b,c in zip(rs[:-1],rs[1:],cnts[:-1]):
    out[a:b] /= c