numpy where with array of tuples

Question

Why can't I find the location of a tuple in an array? Afterall, the bottom expression prints True

foo = numpy.array([(5, 30), (5,), 5])
bar = numpy.where(foo==foo[0])
print(bar)

Prints (array([], dtype=int64),)

print((5,30)==foo[0])

Prints True

foo is a dtype=object array (different size tuples). Many of the usual numeric array operations, including comparison, are not implemented for this dtype. foo should be a list. — hpaulj
– hpaulj, Commented Nov 9, 2015 at 18:58
foo = numpy.array([(5, 30), (5,), 5]) ----> NOWADAYS ValueError: setting an array element with a sequence. The requested array has an inhomogeneous shape after 1 dimensions. The detected shape was (3,) + inhomogeneous part. — pippo1980
– pippo1980, Commented Feb 13, 2024 at 23:10

Leb · Accepted Answer · 2015-11-09 13:23:32Z

2

It's because:

import numpy

foo = numpy.array([(5, 30), (5,), 5])
bar = numpy.where(foo==foo[0])
print(foo==foo[0])

False

That's why you're getting an empty array. A list comprehension alternative is [v for v in foo if v == foo[0]] will result in [(5, 30)]

answered Nov 9, 2015 at 13:23

Leb

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