Python Lists – Return list of every nth element before known substring in a list

I have an array like so:

['6.00', '1.85', 'X', '2.25', '1.26', 'X', '6.50', '1.80', 'X', '6.50', '1.95', 'X', '34.00', '5.50']

Does anyone know how I could return an array of every 3rd element before ‘X’?:

['6.00', '2.25', '6.50', '6.50']

I have been removing every 2nd and 3rd element by the following code, but it breaks when there are more elements past the last ‘X’ in the list or when the ‘X’ is not evenly spaced:

data = ['6.00', '1.85', 'X', '2.25', '1.26', 'X', '6.50', '1.80', 'X', '6.50', '1.95', 'X', '34.00', '5.50']
x = np.array(data)
x = np.delete(x, np.arange(1, x.size, 3))
x = np.delete(x, np.arange(1, x.size, 2))

>Solution :

By using offset of positions (found by np.nonzero) where x == 'X':

x[np.nonzero(x == 'X')[0] - 2]

array(['6.00', '2.25', '6.50', '6.50'], dtype='<U5')

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