How to compare numpy strings in Python 3

Question

The following demonstrates the problem:

import io
import numpy as np

a = np.loadtxt(io.StringIO("val1 val2\nval3 val4"), \
               dtype=np.dtype([("col1", "S10"), ("col2", "S10")]))
print("looks weired: %s"%(a["col1"][0]))
assert(a["col1"][0] == "val1")

I don't understand how I should compare the strings. On my system (numpy 1.6.2, python 3.2.2) the output looks like this:

>>> 
looks weired: b'val1'
Traceback (most recent call last):
  File "D:/..../bug_sample.py", line 7, in <module>
    assert(a["col1"][0] == "val1")
AssertionError

Kos · Accepted Answer · 2013-11-13 15:03:47Z

4

This is not numpy-related:

>>> b"asd" == "asd"
False

In Python 3 bytes objects don't compare equal to strings. So either:

compare against b"val1" instead of "val1" so that the types match,
decode the bytes object into a string (like .decode('utf-8') and compare with "val1".

answered Nov 13, 2013 at 15:03

Kos

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3 Comments

user2988134 Over a year ago

Thanks Kos, but why is the value returned as a byte array if it is declared "S10"? I would expect it to be a np.str. The behavoir of numpy is not intuitive.

Kos Over a year ago

That depends on your intuition. :-) Maybe they just didn't want the typestrings to change meaning in Python 3. See also docs.scipy.org/doc/numpy/reference/…

user2988134 Over a year ago

Kos, I am new to py3k so maybe I oversee something fundamental... but the result of this simply looks wrong: a = np.loadtxt(io.StringIO("val1 val2\nval3 val4"), dtype=np.dtype([("col1", np.str, 4), ("col2", "S4")])) print(a)

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