Decompress and read Dukascopy .bi5 tick files

Question

I need to open a .bi5 file and read the contents to cut a long story short. The problem: I have tens of thousands of .bi5 files containing time-ser

Answer 1

import requests
import struct
from lzma import LZMADecompressor, FORMAT_AUTO

# for download compressed EURUSD 2020/06/15/10h_ticks.bi5 file
res = requests.get("https://www.dukascopy.com/datafeed/EURUSD/2020/06/15/10h_ticks.bi5", stream=True)
print(res.headers.get('content-type'))

rawdata = res.content

decomp = LZMADecompressor(FORMAT_AUTO, None, None)
decompresseddata = decomp.decompress(rawdata)

firstrow = struct.unpack('!IIIff', decompresseddata[0: 20])
print("firstrow:", firstrow)
# firstrow: (436, 114271, 114268, 0.9399999976158142, 0.75)
# time = 2020/06/15/10h + (1 month) + 436 milisecond

secondrow = struct.unpack('!IIIff', decompresseddata[20: 40])
print("secondrow:", secondrow)
# secondrow: (537, 114271, 114267, 4.309999942779541, 2.25)

# time = 2020/06/15/10h + (1 month) + 537 milisecond
# ask = 114271 / 100000 = 1.14271
# bid = 114267 / 100000 = 1.14267
# askvolume = 4.31
# bidvolume = 2.25

# note that 00 -> is january
# "https://www.dukascopy.com/datafeed/EURUSD/2020/00/15/10h_ticks.bi5" for january
# "https://www.dukascopy.com/datafeed/EURUSD/2020/01/15/10h_ticks.bi5" for february

#  iterating
print(len(decompresseddata), int(len(decompresseddata) / 20))
for i in range(0, int(len(decompresseddata) / 20)):
    print(struct.unpack('!IIIff', decompresseddata[i * 20: (i + 1) * 20]))

Answer 2

The code below should do the trick. First, it opens a file and decodes it in lzma and then uses struct to unpack the binary data.

import lzma
import struct
import pandas as pd


def bi5_to_df(filename, fmt):
    chunk_size = struct.calcsize(fmt)
    data = []
    with lzma.open(filename) as f:
        while True:
            chunk = f.read(chunk_size)
            if chunk:
                data.append(struct.unpack(fmt, chunk))
            else:
                break
    df = pd.DataFrame(data)
    return df

The most important thing is to know the right format. I googled around and tried to guess and '>3i2f' (or >3I2f) works quite good. (It's big endian 3 ints 2 floats. What you suggest: 'i4f' doesn't produce sensible floats - regardless whether big or little endian.) For struct and format syntax see the docs.

df = bi5_to_df('13h_ticks.bi5', '>3i2f')
df.head()
Out[177]: 
      0       1       2     3     4
0   210  110218  110216  1.87  1.12
1   362  110219  110216  1.00  5.85
2   875  110220  110217  1.00  1.12
3  1408  110220  110218  1.50  1.00
4  1884  110221  110219  3.94  1.00

---

Update

To compare the output of bi5_to_df with https://github.com/ninety47/dukascopy, I compiled and run test_read_bi5 from there. The first lines of the output are:

time, bid, bid_vol, ask, ask_vol
2012-Dec-03 01:00:03.581000, 131.945, 1.5, 131.966, 1.5
2012-Dec-03 01:00:05.142000, 131.943, 1.5, 131.964, 1.5
2012-Dec-03 01:00:05.202000, 131.943, 1.5, 131.964, 2.25
2012-Dec-03 01:00:05.321000, 131.944, 1.5, 131.964, 1.5
2012-Dec-03 01:00:05.441000, 131.944, 1.5, 131.964, 1.5

And bi5_to_df on the same input file gives:

bi5_to_df('01h_ticks.bi5', '>3I2f').head()
Out[295]: 
      0       1       2     3    4
0  3581  131966  131945  1.50  1.5
1  5142  131964  131943  1.50  1.5
2  5202  131964  131943  2.25  1.5
3  5321  131964  131944  1.50  1.5
4  5441  131964  131944  1.50  1.5

So everything seems to be fine (ninety47's code reorders columns).

Also, it's probably more accurate to use '>3I2f' instead of '>3i2f' (i.e. unsigned int instead of int).

Answer 3

Did you try using numpy as to parse the data before transfer it to pandas. Maybe is a long way solution, but I will allow you to manipulate and clean the data before you made the analysis in Panda, also the integration between them are pretty straight forward,