方法#1
corr2_coeff_rowwise 列出了如何在行之间进行元素关联。我们可以将其分解为两列之间元素相关的情况。所以，我们最终会得到一个使用corr2_coeff_rowwise 的循环。然后，我们将尝试对其进行矢量化，并查看其中是否有可以矢量化的片段：
- 使用
mean 获取平均值。这可以使用统一过滤器进行矢量化。
- 接下来是获取这些平均值与输入数组中的滑动元素之间的差异。要移植到矢量化的，我们将使用
broadcasting。
Rest 保持不变，以从 pearsonr 的两个输出中获取第一个。
要获得第二个输出，我们返回source code。考虑到第一个系数输出，这应该是直截了当的。
因此，考虑到这些，我们最终会得到这样的结果 -
import scipy.special as special
from scipy.ndimage import uniform_filter
def sliding_corr1(a,b,W):
# a,b are input arrays; W is window length
am = uniform_filter(a.astype(float),W)
bm = uniform_filter(b.astype(float),W)
amc = am[W//2:-W//2+1]
bmc = bm[W//2:-W//2+1]
da = a[:,None]-amc
db = b[:,None]-bmc
# Get sliding mask of valid windows
m,n = da.shape
mask1 = np.arange(m)[:,None] >= np.arange(n)
mask2 = np.arange(m)[:,None] < np.arange(n)+W
mask = mask1 & mask2
dam = (da*mask)
dbm = (db*mask)
ssAs = np.einsum('ij,ij->j',dam,dam)
ssBs = np.einsum('ij,ij->j',dbm,dbm)
D = np.einsum('ij,ij->j',dam,dbm)
coeff = D/np.sqrt(ssAs*ssBs)
n = W
ab = n/2 - 1
pval = 2*special.btdtr(ab, ab, 0.5*(1 - abs(np.float64(coeff))))
return coeff,pval
因此，从 pandas 系列的输入中获得最终输出 -
out = sliding_corr1(df['x'].to_numpy(copy=False),df['y'].to_numpy(copy=False),50)
方法#2
与Approach #1 非常相似，但我们将使用numba 来提高内存效率，以取代之前方法中的步骤#2。
from numba import njit
import math
@njit(parallel=True)
def sliding_corr2_coeff(a,b,amc,bmc):
L = len(a)-W+1
out00 = np.empty(L)
for i in range(L):
out_a = 0
out_b = 0
out_D = 0
for j in range(W):
d_a = a[i+j]-amc[i]
d_b = b[i+j]-bmc[i]
out_D += d_a*d_b
out_a += d_a**2
out_b += d_b**2
out00[i] = out_D/math.sqrt(out_a*out_b)
return out00
def sliding_corr2(a,b,W):
am = uniform_filter(a.astype(float),W)
bm = uniform_filter(b.astype(float),W)
amc = am[W//2:-W//2+1]
bmc = bm[W//2:-W//2+1]
coeff = sliding_corr2_coeff(a,b,amc,bmc)
ab = W/2 - 1
pval = 2*special.btdtr(ab, ab, 0.5*(1 - abs(np.float64(coeff))))
return coeff,pval
方法#3
与上一个非常相似，除了我们将所有系数工作推到numba -
@njit(parallel=True)
def sliding_corr3_coeff(a,b,W):
L = len(a)-W+1
out00 = np.empty(L)
for i in range(L):
a_mean = 0.0
b_mean = 0.0
for j in range(W):
a_mean += a[i+j]
b_mean += b[i+j]
a_mean /= W
b_mean /= W
out_a = 0
out_b = 0
out_D = 0
for j in range(W):
d_a = a[i+j]-a_mean
d_b = b[i+j]-b_mean
out_D += d_a*d_b
out_a += d_a*d_a
out_b += d_b*d_b
out00[i] = out_D/math.sqrt(out_a*out_b)
return out00
def sliding_corr3(a,b,W):
coeff = sliding_corr3_coeff(a,b,W)
ab = W/2 - 1
pval = 2*special.btdtr(ab, ab, 0.5*(1 - np.abs(coeff)))
return coeff,pval
时间安排 -
In [181]: df = pd.DataFrame({'x': np.random.rand(10000), 'y': np.random.rand(10000)})
In [182]: %timeit sliding_corr2(df['x'].to_numpy(copy=False),df['y'].to_numpy(copy=False),50)
100 loops, best of 3: 5.05 ms per loop
In [183]: %timeit sliding_corr3(df['x'].to_numpy(copy=False),df['y'].to_numpy(copy=False),50)
100 loops, best of 3: 5.51 ms per loop
注意：