Following-up from this question years ago, is there a canonical "shift" function in numpy? I don't see anything from the documentation.
Here's a simple version of what I'm looking for:
def shift(xs, n):
if n >= 0:
return np.r_[np.full(n, np.nan), xs[:-n]]
else:
return np.r_[xs[-n:], np.full(-n, np.nan)]
Using this is like:
In [76]: xs
Out[76]: array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9.])
In [77]: shift(xs, 3)
Out[77]: array([ nan, nan, nan, 0., 1., 2., 3., 4., 5., 6.])
In [78]: shift(xs, -3)
Out[78]: array([ 3., 4., 5., 6., 7., 8., 9., nan, nan, nan])
This question came from my attempt to write a fast rolling_product yesterday. I needed a way to "shift" a cumulative product and all I could think of was to replicate the logic in np.roll()
.
So np.concatenate()
is much faster than np.r_[]
. This version of the function performs a lot better:
def shift(xs, n):
if n >= 0:
return np.concatenate((np.full(n, np.nan), xs[:-n]))
else:
return np.concatenate((xs[-n:], np.full(-n, np.nan)))
An even faster version simply pre-allocates the array:
def shift(xs, n):
e = np.empty_like(xs)
if n >= 0:
e[:n] = np.nan
e[n:] = xs[:-n]
else:
e[n:] = np.nan
e[:n] = xs[-n:]
return e
There is no single function that does what you want. Your definition of shift is slightly different than what most people are doing. The ways to shift an array are more commonly looped:
>>>xs=np.array([1,2,3,4,5])
>>>shift(xs,3)
array([3,4,5,1,2])
However, you can do what you want with two functions.
Consider a=np.array([ 0., 1., 2., 3., 4., 5., 6., 7., 8., 9.])
:
def shift2(arr,num):
arr=np.roll(arr,num)
if num<0:
np.put(arr,range(len(arr)+num,len(arr)),np.nan)
elif num > 0:
np.put(arr,range(num),np.nan)
return arr
>>>shift2(a,3)
[ nan nan nan 0. 1. 2. 3. 4. 5. 6.]
>>>shift2(a,-3)
[ 3. 4. 5. 6. 7. 8. 9. nan nan nan]
After running cProfile on your given function and the above code you provided, I found that the code you provided makes 42 function calls while shift2
made 14 calls when arr is positive and 16 when it is negative. I will be experimenting with timing to see how each performs with real data.
For those who want to just copy and paste the fastest implementation of shift, there is a benchmark and conclusion(see the end). In addition, I introduce fill_value parameter and fix some bugs.
import numpy as np
import timeit
# enhanced from IronManMark20 version
def shift1(arr, num, fill_value=np.nan):
arr = np.roll(arr,num)
if num < 0:
arr[num:] = fill_value
elif num > 0:
arr[:num] = fill_value
return arr
# use np.roll and np.put by IronManMark20
def shift2(arr,num):
arr=np.roll(arr,num)
if num<0:
np.put(arr,range(len(arr)+num,len(arr)),np.nan)
elif num > 0:
np.put(arr,range(num),np.nan)
return arr
# use np.pad and slice by me.
def shift3(arr, num, fill_value=np.nan):
l = len(arr)
if num < 0:
arr = np.pad(arr, (0, abs(num)), mode='constant', constant_values=(fill_value,))[:-num]
elif num > 0:
arr = np.pad(arr, (num, 0), mode='constant', constant_values=(fill_value,))[:-num]
return arr
# use np.concatenate and np.full by chrisaycock
def shift4(arr, num, fill_value=np.nan):
if num >= 0:
return np.concatenate((np.full(num, fill_value), arr[:-num]))
else:
return np.concatenate((arr[-num:], np.full(-num, fill_value)))
# preallocate empty array and assign slice by chrisaycock
def shift5(arr, num, fill_value=np.nan):
result = np.empty_like(arr)
if num > 0:
result[:num] = fill_value
result[num:] = arr[:-num]
elif num < 0:
result[num:] = fill_value
result[:num] = arr[-num:]
else:
result[:] = arr
return result
arr = np.arange(2000).astype(float)
def benchmark_shift1():
shift1(arr, 3)
def benchmark_shift2():
shift2(arr, 3)
def benchmark_shift3():
shift3(arr, 3)
def benchmark_shift4():
shift4(arr, 3)
def benchmark_shift5():
shift5(arr, 3)
benchmark_set = ['benchmark_shift1', 'benchmark_shift2', 'benchmark_shift3', 'benchmark_shift4', 'benchmark_shift5']
for x in benchmark_set:
number = 10000
t = timeit.timeit('%s()' % x, 'from __main__ import %s' % x, number=number)
print '%s time: %f' % (x, t)
benchmark result:
benchmark_shift1 time: 0.265238
benchmark_shift2 time: 0.285175
benchmark_shift3 time: 0.473890
benchmark_shift4 time: 0.099049
benchmark_shift5 time: 0.052836
shift5 is winner! It's OP's third solution.
One way to do it without spilt the code into cases
with array:
def shift(arr, dx, default_value):
result = np.empty_like(arr)
get_neg_or_none = lambda s: s if s < 0 else None
get_pos_or_none = lambda s: s if s > 0 else None
result[get_neg_or_none(dx): get_pos_or_none(dx)] = default_value
result[get_pos_or_none(dx): get_neg_or_none(dx)] = arr[get_pos_or_none(-dx): get_neg_or_none(-dx)]
return result
with matrix it can be done like this:
def shift(image, dx, dy, default_value):
res = np.full_like(image, default_value)
get_neg_or_none = lambda s: s if s < 0 else None
get_pos_or_none = lambda s : s if s > 0 else None
res[get_pos_or_none(-dy): get_neg_or_none(-dy), get_pos_or_none(-dx): get_neg_or_none(-dx)] = \
image[get_pos_or_none(dy): get_neg_or_none(dy), get_pos_or_none(dx): get_neg_or_none(dx)]
return res
If you want a one-liner from numpy and aren't too concerned about performance, try:
np.sum(np.diag(the_array,1),0)[:-1]
Explanation: np.diag(the_array,1)
creates a matrix with your array one-off the diagonal, np.sum(...,0)
sums the matrix column-wise, and ...[:-1]
takes the elements that would correspond to the size of the original array. Playing around with the 1
and :-1
as parameters can give you shifts in different directions.
You can convert ndarray
to Series
or DataFrame
with pandas
first, then you can use shift
method as you want.
Example:
In [1]: from pandas import Series
In [2]: data = np.arange(10)
In [3]: data
Out[3]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
In [4]: data = Series(data)
In [5]: data
Out[5]:
0 0
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
dtype: int64
In [6]: data = data.shift(3)
In [7]: data
Out[7]:
0 NaN
1 NaN
2 NaN
3 0.0
4 1.0
5 2.0
6 3.0
7 4.0
8 5.0
9 6.0
dtype: float64
In [8]: data = data.values
In [9]: data
Out[9]: array([ nan, nan, nan, 0., 1., 2., 3., 4., 5., 6.])
You can also do this with Pandas:
Using a 2356-long array:
import numpy as np
xs = np.array([...])
Using scipy:
from scipy.ndimage.interpolation import shift
%timeit shift(xs, 1, cval=np.nan)
# 956 µs ± 77.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
Using Pandas:
import pandas as pd
%timeit pd.Series(xs).shift(1).values
# 377 µs ± 9.42 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In this example, using Pandas was about ~8 times faster than Scipy
scipy.ndimage.interpolation.shift
) is the slowest solution listed in this page.shift4_numba
(defined below) if you want good all-aroundershift4_numba
import numba
@numba.njit
def shift4_numba(arr, num, fill_value=np.nan):
if num >= 0:
return np.concatenate((np.full(num, fill_value), arr[:-num]))
else:
return np.concatenate((arr[-num:], np.full(-num, fill_value)))
shift5_numba
import numba
@numba.njit
def shift5_numba(arr, num, fill_value=np.nan):
result = np.empty_like(arr)
if num > 0:
result[:num] = fill_value
result[num:] = arr[:-num]
elif num < 0:
result[num:] = fill_value
result[:num] = arr[-num:]
else:
result[:] = arr
return result
shift5
shift5_numba
, just remove the @numba.njit decorator.shift_scipy
: scipy.ndimage.interpolation.shift
(scipy 1.4.1) - The option from accepted answer, which is clearly the slowest alternative.shift1
: np.roll
and out[:num] xnp.nan
by IronManMark20 & gzcshift2
: np.roll
and np.put
by IronManMark20shift3
: np.pad
and slice
by gzcshift4
: np.concatenate
and np.full
by chrisaycockshift5
: using two times result[slice] = x
by chrisaycockshift#_numba
: @numba.njit decorated versions of the previous.The shift2
and shift3
contained functions that were not supported by the current numba (0.50.1).
Source: Stackoverflow.com