# Remove consecutive duplicates in a NumPy array Code Answer

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I would like to remove duplicates which follow each other, but not duplicates along the whole array. Also, I want to keep the ordering unchanged.

So if the input is `[0 0 1 3 2 2 3 3]` the output should be `[0 1 3 2 3]`

I found a way using `itertools.groupby()` but I am looking for a faster NumPy solution.

```a[np.insert(np.diff(a).astype(np.bool), 0, True)]
Out[99]: array([0, 1, 3, 2, 3])
```

The general idea is to use `diff` to find the difference between two consecutive elements in the array. Then we only index those which give `non-zero` differences elements. But since the length of `diff` is shorter by 1. So before indexing, we need to `insert` the `True` to the beginning of the diff array.

Explanation:

```In [100]: a
Out[100]: array([0, 0, 1, 3, 2, 2, 3, 3])

In [101]: diff = np.diff(a).astype(np.bool)

In [102]: diff
Out[102]: array([False,  True,  True,  True, False,  True, False], dtype=bool)

In [103]: idx = np.insert(diff, 0, True)

In [104]: idx
Out[104]: array([ True, False,  True,  True,  True, False,  True, False], dtype=bool)

In [105]: a[idx]
Out[105]: array([0, 1, 3, 2, 3])
```