Python Extract Elements from Array

I looks like you are looking for numpy.where, here is a simple example to get you started:

In [18]: import numpy as np

In [19]: a = np.array([[0,3],[1,2],[2,3],[3,2],[4,5],[5,1]])

In [20]: a[np.where((a[:, 0] > 1) & (a[:, 0] < 5))[0]]
Out[20]: 
array([[2, 3],
       [3, 2],
       [4, 5]])

In [21]: a[np.where(~((a[:, 0] > 1) & (a[:, 0] < 5)))[0]]
Out[21]: 
array([[0, 3],
       [1, 2],
       [5, 1]])

The first element in row can be your index, and second your value. numpy.where checks whether condition is true or false, and returns a binary array (actually tuple of arrays), once we have binary array, we can index the original array based on that.

If you want, you can use masks

mask = np.ones(len(X), dtype=bool)
mask[idx] = False
train = X[mask]
test = X[idx]

# you can also use this for test
test = X[np.logical_not(mask)]

When building train, you need to iterate through all of your source data.

Using enumerate should make things easy:

>>> data = list(range(8000))
>>> train, test = [], []
>>> for i, value in enumerate(data):
...     if 1995 <= i <= 1999 or 3995 <= i <= 3999 or 5995 <= i <= 5999 or 7995 <= i <= 7999:
...         test.append(value)
...     else:
...         train.append(value)
...
>>> test
[1995, 1996, 1997, 1998, 1999, 3995, 3996, 3997, 3998, 3999, 5995, 5996, 5997, 5998, 5999, 7995, 7996, 7997, 7998, 7999]
>>> len(train)
7980

This is one possibility, assuming array is the name of the list containing 8000 elements:

idx = {1995, 1996, 1997, 1998, 1999, 3995, 3996, 3997, 3998, 3999, 5995, 5996, 5997, 5998, 5999, 7995, 7996, 7997, 7998, 7999}

test = [array[x] for x in idx]

train = [x for i, x in enumerate(array) if i not in idx]