How to open a binary file as raw string in python?

Question

I have a macro written in imageJ script. I need to rewrite this macro to python3.4. I have opened a binary file in reading mode:

b_f = open("image.bin", "rb")
OverScan = 0
sizeY = 480
reg = OverScan + 10

Then I came into problems when trying to find a way how to open b_f as a raw string. In imageJ script it looks like that: s=File.openAsRawString(b_f,2*192*(1+sizeY)*reg); File.openAsRawString(path, count) - Opens a file and returns up to the first count bytes as a string.

Is there some easy way in python how to open a binary file as raw string? I am totally new to python. Thanks for your help in advance.

Not yet, I will try thanks a lot, but how can I specify there that it should be read as raw string? — Ivana
– Ivana, Commented Nov 5, 2015 at 15:10

Eric Dill · Accepted Answer · 2015-11-05 15:05:42Z

2

After you have opened a file, that file reference has a read() method on it which takes the number of bytes that you want to read in.

with open("image.bin", "rb") as b_f:
    OverScan = 0
    sizeY = 480
    reg = OverScan + 10
    binary_data = b_f.read(2*192*(1+sizeY)*reg)

binary_data will now be of type bytes and hold the number of bytes that you have asked for

answered Nov 5, 2015 at 15:05

Eric Dill

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Ivana Over a year ago

Thank you very much for your answer!

Gabriel Grant · Accepted Answer · 2015-11-05 15:14:37Z

1

If you wish to read the entire file into memory, simply calling the file's read() method (with no arguments) will do.

Eg:

s = open("image.bin", "rb").read()

If you only wish to read up to a specific number of bytes (as in @Eric Dill's example), that can be passed as a parameter to the read method:

s = open("image.bin", "rb").read(SOME_NUMBER_OF_BYTES)

edited Nov 5, 2015 at 15:14

answered Nov 5, 2015 at 15:07

Gabriel Grant

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