How to cut a substring from a formated string in python?

Question

The string is formated as '[DATE] record flag: X1.X2.X3.X4.YEAR.NUM;', where DATA is a date string; there is one or two spaces between : and X1; X1, X2, X4 are composed of one or more chars; X3 consisits of zero or more chars; YEAR and NUM are 4 and 3 digits, respectively.

Here is an exmaple: s = '[2011-03-13] record flag: NW.SENSOR..MH1.2011.012;'

How do I cut 'NW' and 'SENSOR' from s?

heemayl · Accepted Answer · 2016-05-08 10:37:15Z

1

Using str.split:

>>> s = '[2011-03-13] record flag: NW.SENSOR..MH1.2011.012;'

>>> s.split(' ')[3].split('.')
['NW', 'SENSOR', '', 'MH1', '2011', '012;']

>>> out = s.split(' ')[3].split('.')

>>> out[0]
'NW'

>>> out[1]
'SENSOR'

Using re.search:

>>> s = '[2011-03-13] record flag: NW.SENSOR..MH1.2011.012;'

>>> out = re.search(r':\s+([^.]+)\.([^.]+)', s)

>>> out.group(1)
'NW'

>>> out.group(2)
'SENSOR'

answered May 8, 2016 at 10:37

heemayl

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2 Comments

wsdzbm Over a year ago

peferct, I like the re solution. but btw, why add r before the pattern since without it the expression also gives good results?

heemayl Over a year ago

@Lee No, r stands for raw. Its there so that the Python takes the string as raw and pass it as it is to Regex without any internal manipulation..

Roel Zinkstok · Accepted Answer · 2016-05-08 10:37:48Z

1

Use some splitting and stripping:

parts = s.split(":")[1].strip().split(".")
parts[0] # Should be NW
parts[1] # Should be SENSOR

answered May 8, 2016 at 10:37

Roel Zinkstok

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