topic Re: Read float (single precision) from binary in Python Questions

Read float (single precision) from binary

AlfredBaldenweck — Thu, 09 Nov 2023 15:46:04 GMT

Hi all,

I'm trying to retrieve some float values from a blob field.

My workflow is get the blob field as text (hex) and parse it.

I can parse it no problem, but the issue is that the floats are returning what they're actually stored as, e.g. 1.2 is actually 1.2000000476837158203125, etc.
I can't round because each float has a different number of significant digits.

Weirdly enough, doubles parsed analogously come exactly as they should.

Does anyone have any ideas of how to get the correct values?

'''Float (single precision)''' a = binVal[6:8] + binVal[4:6] + binVal[2:4] + binVal[0:2] print(a) # 3F99999A domCd = struct.unpack('!f', bytes.fromhex(a))[0] print(domCd) # 1.2000000476837158 '''Double (double precision)''' a = (binVal[14:16] + binVal[12:14] + binVal[10:12] + binVal[8:10] + binVal[6:8] + binVal[4:6] + binVal[2:4] + binVal[0:2]) print(a) # 3FF3333333333333 domCd = struct.unpack('!d', bytes.fromhex(a))[0] print(domCd) # 1.2

Re: Read float (single precision) from binary

VinceAngelo — Thu, 09 Nov 2023 16:25:07 GMT

It's not possible to represent 1.2000000000 in floating point representation. That's just how IEEE-754 works.

There's lots of web sites that show this, but this is the one I used.

- V

Re: Read float (single precision) from binary

AlfredBaldenweck — Thu, 09 Nov 2023 16:29:06 GMT

Yes, I understand that.

However, the GUI that can read this field natively (read: arcmap) can pull the correct values, so how can I?

Also, as I understand it, neither can doubles, and yet...