numpy array equivalent of pandas.shift() function?

user_8684461 0
Viewed 1204

I have an array 

    [False False False ...  True  True  True]

I want to check if the previous value == current value. In pandas, I can use something like...

np.where(df[col name].shift(1).eq(df[col name]), True, False)

I tried using scipy shift but the output isn't correct so maybe I am using it wrong? 

np.where(shift(long_gt_price, 1) == (long_gt_price),"-", "Different")

Just to show you what I mean when I say it produces the incorrect output:The left column is the shift(1) and the right column is the unshifted column so the left column should equal the square diagonal up to it at least thats my understanding / what I want the False / True at 5 down on the left and 4 on the right therefore doesnt make any sense to me.

enter image description here

pandas shift scipy python-3.x numpy
3 Answers

Why not use slicing

arr[1:] == arr[:-1]

Result wouls be slightly shorter array but there is no need to handle border cases.

user_49894510

This seems to be what you want:

shift_by = 1
arr = np.array([False, False, False, True, True, True]).tolist() ## array -> list
shift_arr = [np.nan]*shift_by + arr[:-shift_by]
np.equal(arr,shift_arr)

For purely numpy:

shift_by = 1
arr = np.array([False, False, False, True, True, True])
np.concatenate([np.array([False]*shift_by),np.equal(arr[shift_by:],arr[:-shift_by])])
user_130700320

A simple function that shifts 1d-arrays, in a similar way to pandas:

def arr_shift(arr: np.ndarray, shift: int) -> np.ndarray:
    if shift == 0:
        return arr
    nas = np.empty(abs(shift))
    nas[:] = np.nan
    if shift > 0:
        res = arr[:-shift]
        return np.concatenate((nas,res))
    res = arr[-shift:]
    return np.concatenate((res,nas))

this is suposed to work with numerical arrays, as the shifted value is replaced by np.NAN. It is trivial to select another "null" value by just filling the nas array with whatever you want.

user_52585370
Related