我需要打开一个 .bi5
文件并阅读其内容以长话短说.问题:我有数以万计的 .bi5
文件,其中包含我需要解压缩和处理(读取、转储到 pandas)的时间序列数据.
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-series data that I need to decompress and process (read, dump into pandas).
我最终为 lzma
库安装了 Python 3(我通常使用 2.7),因为我遇到了使用 Python 2.7 的 lzma
后向端口编译的噩梦,所以我承认并使用 Python 3 运行,但没有成功.问题多得不胜枚举,长问题没人看!
I ended up installing Python 3 (I use 2.7 normally) specifically for the lzma
library, as I ran into compiling nightmares using the lzma
back-ports for Python 2.7, so I conceded and ran with Python 3, but with no success. The problems are too numerous to divulge, no one reads long questions!
我已经包含了其中一个 .bi5
文件,如果有人能够设法将其放入 Pandas Dataframe 并向我展示他们是如何做到的,那将是理想的.
I have included one of the .bi5
files, if someone could manage to get it into a Pandas Dataframe and show me how they did it, that would be ideal.
ps这个文件只有几kb,它会在一秒钟内下载.首先十分感谢.
ps the fie is only a few kb, it will download in a second. Thanks very much in advance.
(文件)http://www.filedropper.com/13hticks
下面的代码应该可以解决问题.首先,它打开一个文件并在 lzma 中对其进行解码,然后使用 struct 解压二进制数据.
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
最重要的是要知道正确的格式.我四处搜索并尝试猜测,'>3i2f'
(或 >3I2f
)效果很好.(这是大端 3 个整数 2 个浮点数.您的建议:'i4f'
不会产生合理的浮点数 - 无论是大端还是小端.)对于 struct
和格式语法请参阅 文档.
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
<小时>
更新
将 bi5_to_df
的输出与 https://github.com/ninety47/进行比较杜高斯贝,我从那里编译并运行 test_read_bi5
.输出的第一行是:
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
和 bi5_to_df
在同一个输入文件上给出:
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
所以一切似乎都很好(ninety47 的代码重新排列了列).
So everything seems to be fine (ninety47's code reorders columns).
另外,使用 '>3I2f'
而不是 '>3i2f'
可能更准确(即 unsigned int
而不是int
).
Also, it's probably more accurate to use '>3I2f'
instead of '>3i2f'
(i.e. unsigned int
instead of int
).
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