Is there a neat way to check for the inclusion in an interval of all values in an array using pandas (or another python tool)?

Question

pandas.Interval can be used to define if a value falls within an interval in a neat way, e.g.:

In [1]: import numpy as np

In [2]: import pandas as pd

In [3]: iv = pd.Interval(0, 5.5)

In [4]: 4.37 in iv
Out[4]: True

Is it possible to check inclusion for all elements of an array instead of a single value? The result would be the same as in:

In [5]: arr = np.array(((1,8),(-4,3.5)))

In [6]: arr
Out[6]:
array([[ 1. ,  8. ],
       [-4. ,  3.5]])

In [7]: (arr > iv.left) & (arr <= iv.right)
Out[7]:
array([[ True, False],
       [False,  True]])

But using a simpler syntax which is cool about pd.Interval. Something like the below which does not work:

In [8]: arr in iv
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-15-a118a68ee023> in <module>()
----> 1 arr in iv

pandas/_libs/interval.pyx in pandas._libs.interval.Interval.__contains__()

ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

Answer 1

Check with vectorize

def youf(x,iv):
    return x in iv

vfunc = np.vectorize(youf)

iv = pd.Interval(0, 5.5)

vfunc(arr, iv)
Out[27]: 
array([[ True, False],
       [False,  True]])