Classifying an array

Question

For instance I have this list:

A = [(1, 2.5), (2, 5.0), (3, 7.5), (4, 10.0)]

and I have another list with the following entry:

B = [2.5, 7.5, 12.5]

What I want is a program in Python where if the 2nd element of each tuple in A, namely 2.5, 5.0, 7.5, and 10.0, can be found in B, it will create another variable which looks like this:

C = [(1, 2.5), (3, 7.5)]

where it removes the other tuples in A whose 2nd elements are not in B. I tried to code it this way:

A = [(1, 2.5), (2, 5.0), (3, 7.5), (4, 10.0)]
D = np.asarray(list(A))
E = []
if D[:, 1] in B:
    E = np.append(E,D)
else:
    pass

but I got an error of:

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

Is 7,5 in B a typo for 7.5?

Holt
– Holt

2014-07-15 12:12:23 +00:00
Commented Jul 15, 2014 at 12:12 — Holt
– Holt, Commented Jul 15, 2014 at 12:12
B is also never defined if D[:,1] in B:

heinst
– heinst

2014-07-15 12:14:23 +00:00
Commented Jul 15, 2014 at 12:14 — heinst
– heinst, Commented Jul 15, 2014 at 12:14

unutbu · Accepted Answer · 2014-07-15 12:15:52Z

2

You could use np.n1d to determine which values in A[:, 1] are also in B:

In [104]: A = np.array([(1,2.5), (2,5.0), (3,7.5), (4,10.0)])

In [105]: B = [2.5, 7.5, 12.5]

In [106]: mask = np.in1d(A[:, 1], B)

In [107]: mask
Out[107]: array([ True, False,  True, False], dtype=bool)

In [108]: A[mask]
Out[108]: 
array([[ 1. ,  2.5],
       [ 3. ,  7.5]])

answered Jul 15, 2014 at 12:15

unutbu

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Comments

Holt · Accepted Answer · 2014-07-15 12:13:55Z

2

Why not simply the following:

>>> A = [(1,2.5), (2,5.0), (3,7.5), (4,10.0)]
>>> B = [2.5, 7,5, 12.5]
>>> C = [a for a in A if a[1] in B]
>>> C
[(1, 2.5), (3, 7.5)]

answered Jul 15, 2014 at 12:13

Holt

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1 Comment

Karl Knechtel Over a year ago

Because iterating in native Python - whether with a comprehension or a loop - usually defeats the purpose of Numpy.

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