Parse complex text files in python

Question

I have a txt file in the below format

*15*EN*5562*CB*56198898~ZA*QS*70*EA~ZA*QA*3246*EA~ZA*QP*16020*EA~ZA*QT*12*EA
*16*EN*9966*CB*60526588~ZA*QS*148*EA~ZA*QA*4261*EA~ZA*QP*18000*EA~ZA*QT*6*EA
*17*EN*9973*CB*61540958~ZA*QS*133*EA~ZA*QA*3475*EA~ZA*QP*15600*EA~ZA*QT*12*EA
*18*EN*9980*CB*61540385~ZA*QS*24*EA~ZA*QA*2157*EA~ZA*QP*10062*EA

how do I convert this to csv file in the expected format?

line EN   CB            QS    QA    QP    QT 
15  5562  56198898~ZA   70   3246  16020   12
16  9966  60526588~ZA   148  4261  18000    6
17  9973  61540958~ZA   133  3475  15600   12
18  9980  61540385~ZA    24  2157  10062   NAN

To get the output you want, you need to do some custom work. Typically one row in the csv is reserved for headers, in your csv its mixed in with the data so you need to parse that out :/ — DannyMoshe
– DannyMoshe, Commented Oct 22, 2019 at 17:48

GiovaniSalazar · Accepted Answer · 2019-10-22 17:41:42Z

1

like this

import pandas as pd

df = pd.read_csv('myfile.csv',sep="*",engine='python')

print(df)

answered Oct 22, 2019 at 17:41

GiovaniSalazar

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Eric Wilson · Accepted Answer · 2019-10-22 17:40:15Z

0

You have tagged this question with pandas but the csv module should handle this just fine.

Here is a slightly modified example from the documentation:

>>> import csv
>>> with open('eggs.csv', newline='') as csvfile:
...     spamreader = csv.reader(csvfile, delimiter='*')
...     for row in spamreader:
...         print(', '.join(row))
Spam, Spam, Spam, Spam, Spam, Baked Beans
Spam, Lovely Spam, Wonderful Spam

answered Oct 22, 2019 at 17:40

Eric Wilson

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