Insert calculated values between consecutive values in array

Question

Let's say I have a simple array, like this one:

import numpy as np
a = np.array([1,2,3])

Which returns me, obviously:

array([1, 2, 3])

I'm trying to add calculated values between consecutive values in this array. The calculation should return me n equally spaced values between it's bounds.

To express myself in numbers, let's say I want to add 1 value between each pair of consecutive values, so the function should return me a array like this:

array([1, 1.5, 2, 2.5, 3])

Another example, now with 2 values between each pair:

array([1, 1.33, 1.66, 2, 2.33, 2.66, 3])

I know the logic and I can create myself a function which will do the work, but I feel numpy has specific functions that would make my code so much cleaner!

sunnytown · Accepted Answer · 2021-09-01 12:00:07Z

1

If your array is

import numpy as np

n = 2
a = np.array([1,2,5])

new_size = a.size + (a.size - 1) * n

x = np.linspace(a.min(), a.max(), new_size)
xp = np.linspace(a.min(), a.max(), a.size)
fp = a
result = np.interp(x, xp, fp)

returns: array([1. , 1.33333333, 1.66666667, 2. , 2.66666667, 3.33333333, 4. ])

If your array is always evenly spaced, you can just use

new_size = a.size + (a.size - 1) * n
result = np.linspace(a.min(), a.max(), new_size)

answered Sep 1, 2021 at 12:00

sunnytown

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1 Comment

momba Over a year ago

Fantastic! You even predicted the problem I didn't know I had

Tzane · Accepted Answer · 2021-09-01 12:02:05Z

0

Using linspace should do the trick:


a = np.array([1,2,3])
n = 1

temps = []

for i in range(1, len(a)):
    temps.append(np.linspace(a[i-1], a[i], num=n+1, endpoint=False))

# Add last final ending point
temps.append(np.array([a[-1]]))
new_a = np.concatenate(temps)
print(new_a)

answered Sep 1, 2021 at 12:02

Tzane

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Comments

U13-Forward · Accepted Answer · 2021-09-01 12:08:27Z

0

Try with np.arange:

a = np.array([1,2,3])
n = 2
print(np.arange(a.min(), a.max(), 1 / (n + 1)))

Output:

[1.         1.33333333 1.66666667 2.         2.33333333 2.66666667]

answered Sep 1, 2021 at 12:08

U13-Forward

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