Title sounds more complicated than the facts really are. Given the data

```
data = [
np.array(['x'], dtype='object'),
np.array(['y'], dtype='object'),
np.array(['z'], dtype='object'),
np.array(['x', 'z', 'y'], dtype='object'),
np.array(['y', 'x'], dtype='object'),
]
s = pd.Series(data)
```

I would like to retrieve to elements of `s`

where `s == np.array(['x'])`

. The obvious way

```
c = np.array(['x'])
s[s==c]
```

does not work, since there is a ValueError in the comparison, complaining that "’Lengths must match to compare’, (5,), (1,)". I also tried

```
s[s=='x']
```

which only works if the elements of `s`

have all exactly one element themselves.

Is there a way to retrieve all elements of `s`

, where `s == c`

, *without* converting the elements to string?

### >Solution :

Use a list comprehension with `numpy.array_equal`

:

```
c = np.array(['x'])
out = s[[np.array_equal(a, c) for a in s]]
```

Alternative with a `partial`

function if you need to do this repeatedly (for the shorter syntax):

```
from functools import partial
eq_c = partial(np.array_equal, c)
out = s[map(eq_c, s)]
```

Output:

```
0 [x]
dtype: object
```