问题
I'd like to compare each value x of an array with a rolling window of the n previous values. More precisely I'd like to see at which percentile this new value x would be, if we added it to the previous window:
import numpy as np
A = np.array([1, 4, 9, 28, 28.5, 2, 283, 3.2, 7, 15])
print A
n = 4 # window width
for i in range(len(A)-n):
W = A[i:i+n]
x = A[i+n]
q = sum(W <= x) * 1.0 / n
print 'Value:', x, ' Window before this value:', W, ' Quantile:', q
[ 1. 4. 9. 28. 28.5 2. 283. 3.2 7. 15. ]
Value: 28.5 Window before this value: [ 1. 4. 9. 28.] Quantile: 1.0
Value: 2.0 Window before this value: [ 4. 9. 28. 28.5] Quantile: 0.0
Value: 283.0 Window before this value: [ 9. 28. 28.5 2. ] Quantile: 1.0
Value: 3.2 Window before this value: [ 28. 28.5 2. 283. ] Quantile: 0.25
Value: 7.0 Window before this value: [ 28.5 2. 283. 3.2] Quantile: 0.5
Value: 15.0 Window before this value: [ 2. 283. 3.2 7. ] Quantile: 0.75
Question: What is the name of this computation? Is there a clever numpy way to compute this more efficiently on arrays of millions of items (with n that can be ~5000)?
Note: here is a simulation for 1M items and n=5000 but it would take ~ 2 hours:
import numpy as np
A = np.random.random(1000*1000) # the following is not very interesting with a [0,1]
n = 5000 # uniform random variable, but anyway...
Q = np.zeros(len(A)-n)
for i in range(len(Q)):
Q[i] = sum(A[i:i+n] <= A[i+n]) * 1.0 / n
if i % 100 == 0:
print "%.2f %% already done. " % (i * 100.0 / len(A))
print Q
Note: this is not similar to How to compute moving (or rolling, if you will) percentile/quantile for a 1d array in numpy?
回答1:
Your code is so slow because you're using Python's own sum() instead of numpy.sum() or numpy.array.sum(); Python's sum() has to convert all the raw values to Python objects before doing the calculations, which is really slow. Just by changing sum(...) to np.sum(...) or (...).sum(), the runtime drops to under 20 seconds.
回答2:
you can use np.lib.stride_tricks.as_strided as in the accepted answer of the question you linked. With the first example you give, it is pretty easy to understand:
A = np.array([1, 4, 9, 28, 28.5, 2, 283, 3.2, 7, 15])
n=4
print (np.lib.stride_tricks.as_strided(A, shape=(A.size-n,n),
strides=(A.itemsize,A.itemsize)))
# you get the A.size-n columns of the n rolling elements
array([[ 1. , 4. , 9. , 28. , 28.5, 2. ],
[ 4. , 9. , 28. , 28.5, 2. , 283. ],
[ 9. , 28. , 28.5, 2. , 283. , 3.2],
[ 28. , 28.5, 2. , 283. , 3.2, 7. ]])
Now to do the calculation, you can compare this array to A[n:], sum over the rows and divide by n:
print ((np.lib.stride_tricks.as_strided(A, shape=(n,A.size-n),
strides=(A.itemsize,A.itemsize))
<= A[n:]).sum(0)/(1.*n))
[1. 0. 1. 0.25 0.5 0.75] # same anwser
Now the problem is the size of you data (several M and n around 5000), not sure you can use directly this method. One way could be to chunk the data. Let's define a function
def compare_strides (arr, n):
return (np.lib.stride_tricks.as_strided(arr, shape=(n,arr.size-n),
strides=(arr.itemsize,arr.itemsize))
<= arr[n:]).sum(0)
and do the chunk, with np.concatenate and don't forget to divide by n:
nb_chunk = 1000 #this number depends on the capacity of you computer,
# not sure how to optimize it
Q = np.concatenate([compare_strides(A[chunk*nb_chunk:(chunk+1)*nb_chunk+n],n)
for chunk in range(0,A[n:].size/nb_chunk+1)])/(1.*n)
I can't do the 1M - 5000 test, but on a 5000 - 100, see the difference in timeit:
A = np.random.random(5000)
n = 100
%%timeit
Q = np.zeros(len(A)-n)
for i in range(len(Q)):
Q[i] = sum(A[i:i+n] <= A[i+n]) * 1.0 / n
#1 loop, best of 3: 6.75 s per loop
%%timeit
nb_chunk = 100
Q1 = np.concatenate([compare_strides(A[chunk*nb_chunk:(chunk+1)*nb_chunk+n],n)
for chunk in range(0,A[n:].size/nb_chunk+1)])/(1.*n)
#100 loops, best of 3: 7.84 ms per loop
#check for egality
print ((Q == Q1).all())
Out[33]: True
See the difference in time from 6750 ms to 7.84 ms. Hope it works on bigger data
回答3:
Using np.sum instead of sum was already mentioned, so my only suggestion left is to additionally consider using pandas and its rolling window function, which you can apply any arbitrary function to:
import numpy as np
import pandas as pd
A = np.random.random(1000*1000)
df = pd.DataFrame(A)
n = 5000
def fct(x):
return np.sum(x[:-1] <= x[-1]) * 1.0 / (len(x)-1)
percentiles = df.rolling(n+1).apply(fct)
print(percentiles)
回答4:
Additional benchmark: comparison between this solution and this solution:
import numpy as np, time
A = np.random.random(1000*1000)
n = 5000
def compare_strides (arr, n):
return (np.lib.stride_tricks.as_strided(arr, shape=(n,arr.size-n), strides=(arr.itemsize,arr.itemsize)) <= arr[n:]).sum(0)
# Test #1: with strides ===> 11.0 seconds
t0 = time.time()
nb_chunk = 10*1000
Q = np.concatenate([compare_strides(A[chunk*nb_chunk:(chunk+1)*nb_chunk+n],n) for chunk in range(0,A[n:].size/nb_chunk+1)])/(1.*n)
print time.time() - t0, Q
# Test #2: with just np.sum ===> 18.0 seconds
t0 = time.time()
Q2 = np.zeros(len(A)-n)
for i in range(len(Q2)):
Q2[i] = np.sum(A[i:i+n] <= A[i+n])
Q2 *= 1.0 / n # here the multiplication is vectorized; if instead, we move this multiplication to the previous line: np.sum(A[i:i+n] <= A[i+n]) * 1.0 / n, it is 6 seconds slower
print time.time() - t0, Q2
print all(Q == Q2)
There's also another (better) way, with numba and @jit decorator. Then it is much faster: only 5.4 seconds!
from numba import jit
import numpy as np
@jit # if you remove this line, it is much slower (similar to Test #2 above)
def doit():
A = np.random.random(1000*1000)
n = 5000
Q2 = np.zeros(len(A)-n)
for i in range(len(Q2)):
Q2[i] = np.sum(A[i:i+n] <= A[i+n])
Q2 *= 1.0/n
print(Q2)
doit()
When adding numba parallelization, it's even faster: 1.8 seconds!
import numpy as np
from numba import jit, prange
@jit(parallel=True)
def doit(A, Q, n):
for i in prange(len(Q)):
Q[i] = np.sum(A[i:i+n] <= A[i+n])
A = np.random.random(1000*1000)
n = 5000
Q = np.zeros(len(A)-n)
doit(A, Q, n)
回答5:
You can use the np.quantile instead of sum(A[i:i+n] <= A[i+n]) * 1.0 / n. That may be as good as it gets. Not sure if there really is a better approach for your question.
来源：https://stackoverflow.com/questions/53130434/rolling-comparison-between-a-value-and-a-past-window-with-percentile-quantile