I wrote a python function that writes a integer to a file in binary form. But are there any ways that I can reduce the number of steps, is there a more efficient way to write a integer to a file?

def IntWrite(FileName, Integer):
    #convert to hexadecimal
    data = str('{0:0x}'.format(Integer))
    #pad number if odd length
    if(len(data) % 2 != 0):
    #split into chucks of two
    info = [data[i:i+2] for i in range(0, len(data), 2)]
    #convert each chuck to a byte
    data2 = "".join([chr(int(a, 16)) for a in info])
    #write to file
    file(FileName, "wb").write(data2)

#convert back to integer
def IntRead(FileName):
    RawData = file(FileName, "rb").read()
    HexData = "".join(['{0:0x}'.format(ord(b)) for b in RawData])
    return int(HexData, 16)

IntWrite("int_data.txt", 100000)
print IntRead("int_data.txt")
  • 2
    \$\begingroup\$ This is more of a SO question. See struct.pack, struct.unpack, in this case with a format character of i or I, unless you need more than 32 bits. \$\endgroup\$ Nov 13 '13 at 3:02
  • \$\begingroup\$ … but it's also acceptable as a Code Review question. \$\endgroup\$ Nov 13 '13 at 5:58
  • \$\begingroup\$ Sure, especially if there is a need to handle arbitrarily large integers, which the original code seems to offer. Otherwise the best code is that which you don't have to write. :) \$\endgroup\$ Nov 13 '13 at 13:16

Python comes with a built-in pickle module for serialization:

>>> import pickle
>>> with open('data', 'wb') as f: pickle.dump(9 ** 33, f)
>>> with open('data', 'rb') as f: print(pickle.load(f))

(P.S. I wouldn't use a filename ending with .txt to store binary data.)

  • 1
    \$\begingroup\$ Note that pickle's binary format is incompatible with the original question. \$\endgroup\$ Nov 13 '13 at 15:51
  • \$\begingroup\$ Yes, that's right: I'm assuming here that the OP is starting from scratch. \$\endgroup\$ Nov 13 '13 at 15:54

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