Thanks, Travis. I'm not sure how I got it into my head that the dtype argument took the name of the type rather than the type object. The latter is certainly obvious to me now when I go back and read the Data-Type Descriptors section in the NumPy book. dtype=float it shall be! Doug

-----Original Message----- From: scipy-user-bounces@scipy.net [mailto:scipy-user-bounces@scipy.net] On Behalf Of Travis Oliphant Sent: February 24, 2006 16:17 To: SciPy Users List Subject: Re: [SciPy-user] Confused about enhanced-scalar dtypes

LATORNELL, Doug wrote:

...

I built NumPy from SVN this morning and got a surprise concerning dtype arguments:

In [24]: import numpy

In [25]: numpy.__version__ Out[25]: '0.9.6.2165'

In [26]: a = numpy.fromfile("profiles-O0", dtype="float", sep=' ') -------------------------------------------------------------

---

...

- --- exceptions.TypeError Traceback (most recent call last)

/ocean/dlatorne/SoG/dlatornell06/profiles/<ipython console>

TypeError: data type not understood

I don't recommend using strings like "float". Just use float (which is a type-object).

I don't think "float" is actually an accepted string. There are a few strings that are recognized and more could be added, but I would just use dtype=float.

...

Based on the tip in the middle of pg 19 of my copy of the NumPy book, I expected both dtype="float" and dtype="float_" to work.

dtype=float and dtype=numpy.float_ are similar. Note the absence of ""

...

I know there has been a bunch of work lately on things related to dtype, but I don't recall seeing anything about breaking the equivalence of the enhanced-scalar types (like float_) and the corresponding standard Python types (float, in this case). Did I miss, or misunderstand something, get unluck with the timing of my checkout this morning, or is this a bug?

I'm surprised "float" worked before and then doesn't work now. But, there have been enough changes between 0.9.3 and 0.9.5 that it's possible. But, don't do it that way :-)

-Travis

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