如何将每个第 n 项保留在列表中并使其余项为零

Question

I am trying to model and fit to noisy data over a long time series and I want to see what happens to my fit if I remove a substantial amount of my data.

I have a long time-series of data and I am only interested in every nth item. However I still want to plot this list over time but with every other unwanted element removed.

For example, for n=4, the list

a = [1,2,3,4,5,6,7,8,9,10...]

Should become

a_new = [1,0,0,0,5,0,0,0,9,0...]

I don't mind if the position of the nth item is at the start or end of the sequence, my series is effectively arbitrary and so long that it won't matter what I delete. For example 'a_new' could also be:

a_new = [0,0,0,4,0,0,0,8,0,0...]

Ideally the solution wouldn't depend on the length of the list, but I can have that length as a variable.

Edit 1:

I actually wanted empty elements, not zero's, (if that's possible?) so:

a_new = [1,,,,5,,,,9...] 

Edit 2:

I needed to remove the corresponding elements from my time series too so that when everything is plotted, each data element has the same index as the time series element.

Thanks!

Answer 1

Use a list comprehension with a ternary conditional that takes the mod of each element on the number n :

>>> a = [1,2,3,4,5,6,7,8,9,10]
>>> n = 4
>>> [i if i % n == 0 else 0 for i in a]
[0, 0, 0, 4, 0, 0, 0, 8, 0, 0]

---

In case the data does not proceed incrementally, which is most likely, use enumerate so the mod is taken on the index and not on the element:

>>> [v if i % n == 0 else 0 for i, v in enumerate(a)]
[1, 0, 0, 0, 5, 0, 0, 0, 9, 0]

The starting point can also be easily changed when using enumerate :

>>> [v if i % n == 0 else 0 for i, v in enumerate(a, 1)] # start indexing from 1
[0, 0, 0, 4, 0, 0, 0, 8, 0, 0]

---

If you intend to remove your unwanted data rather than replace them, then a filter using if (instead of the ternary operator) in the list comprehension can handle this:

>>> [v for i, v in enumerate(a, 1) if i % n == 0]
[4, 8]

Answer 2

[0 if i%4 else num for i, num in enumerate(a)]

Answer 3

Here's a working example to filter functions given a certain step K:

def filter_f(data, K=4):
    if K <= 0:
        return data

    N = len(data)
    f_filter = [0 if i % K else 1 for i in range(N)]
    return [a * b for a, b in zip(data, f_filter)]

f_input = range(10)

for K in range(10):
    print("Original function: {0}".format(f_input))
    print("Filtered function (step={0}): {1}".format(
        K, filter_f(f_input, K)))
    print("-" * 80)

Output:

Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=0): [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=1): [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=2): [0, 0, 2, 0, 4, 0, 6, 0, 8, 0]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=3): [0, 0, 0, 3, 0, 0, 6, 0, 0, 9]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=4): [0, 0, 0, 0, 4, 0, 0, 0, 8, 0]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=5): [0, 0, 0, 0, 0, 5, 0, 0, 0, 0]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=6): [0, 0, 0, 0, 0, 0, 6, 0, 0, 0]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=7): [0, 0, 0, 0, 0, 0, 0, 7, 0, 0]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=8): [0, 0, 0, 0, 0, 0, 0, 0, 8, 0]
--------------------------------------------------------------------------------
Original function: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Filtered function (step=9): [0, 0, 0, 0, 0, 0, 0, 0, 0, 9]
--------------------------------------------------------------------------------

Answer 4

Alternatively, irrespective of the programming language, you can use function:

$f(i)=(1-(-1)^{\floor((i-1)/k)+\floor(i/k)})/2$

This function produces 1 every k-th element. For k=4, this generates

f(i)=[0,0,0,1,0,0,0,1,0,0,0,1] for i=[1,2,3,4,5,6,7,8,9,10,11,12]

The function that you want would be then i*f(i) .