# Shift elements in a numpy array

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### Question :

Shift elements in a numpy array

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 : xs
Out: array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.])

In : shift(xs, 3)
Out: array([ nan,  nan,  nan,   0.,   1.,   2.,   3.,   4.,   5.,   6.])

In : shift(xs, -3)
Out: 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
``````

Not numpy but scipy provides exactly the shift functionality you want,

``````import numpy as np
from scipy.ndimage.interpolation import shift

xs = np.array([ 0.,  1.,  2.,  3.,  4.,  5.,  6.,  7.,  8.,  9.])

shift(xs, 3, cval=np.NaN)
``````

where default is to bring in a constant value from outside the array with value `cval`, set here to `nan`. This gives the desired output,

``````array([ nan, nan, nan, 0., 1., 2., 3., 4., 5., 6.])
``````

and the negative shift works similarly,

``````shift(xs, -3, cval=np.NaN)
``````

Provides output

``````array([  3.,   4.,   5.,   6.,   7.,   8.,   9.,  nan,  nan,  nan])
``````

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.

### Benchmark

``````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
``````

### Conclusion

shift5 is winner! It’s OP’s third solution.

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.

You can convert `ndarray` to `Series` or `DataFrame` with `pandas` first, then you can use `shift` method as you want.

Example:

``````In : from pandas import Series

In : data = np.arange(10)

In : data
Out: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])

In : data = Series(data)

In : data
Out:
0    0
1    1
2    2
3    3
4    4
5    5
6    6
7    7
8    8
9    9
dtype: int64

In : data = data.shift(3)

In : data
Out:
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 : data = data.values

In : data
Out: array([ nan,  nan,  nan,   0.,   1.,   2.,   3.,   4.,   5.,   6.])
``````

# Benchmarks & introducing Numba

## 1. Summary

• The accepted answer (`scipy.ndimage.interpolation.shift`) is the slowest solution listed in this page.
• Numba (@numba.njit) gives some performance boost when array size smaller than ~25.000
• “Any method” equally good when array size large (>250.000).
• The fastest option really depends on
(2)  Amount of shift you need to do.
• Below is the picture of the timings of all different methods listed on this page (2020-07-11), using constant shift = 10. As one can see, with small array sizes some methods are use more than +2000% time than the best method. ## 2. Detailed benchmarks with the best options

• Choose `shift4_numba` (defined below) if you want good all-arounder ## 3. Code

### 3.1 `shift4_numba`

• Good all-arounder; max 20% wrt. to the best method with any array size
• Best method with medium array sizes: ~ 500 < N < 20.000.
• Caveat: Numba jit (just in time compiler) will give performance boost only if you are calling the decorated function more than once. The first call takes usually 3-4 times longer than the subsequent calls.
``````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)))
``````

### 3.2. `shift5_numba`

• Best option with small (N <= 300.. 1500) array sizes. Treshold depends on needed amount of shift.
• Good performance on any array size; max + 50% compared to the fastest solution.
• Caveat: Numba jit (just in time compiler) will give performance boost only if you are calling the decorated function more than once. The first call takes usually 3-4 times longer than the subsequent calls.
``````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
``````

### 3.3. `shift5`

• Best method with array sizes ~ 20.000 < N < 250.000
• Same as `shift5_numba`, just remove the @numba.njit decorator.

# 4 Appendix

## 4.1 Details about used methods

• `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 & gzc
• `shift2`: `np.roll` and `np.put` by IronManMark20
• `shift3`: `np.pad` and `slice` by gzc
• `shift4`: `np.concatenate` and `np.full` by chrisaycock
• `shift5`: using two times `result[slice] = x` by chrisaycock
• `shift#_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).

## 4.2.3 Raw timings, few best methods

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

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.

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
``````