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If given the string "John Doe;Lodging;123;050617", which is a line within a file, I need to remove everything before and including the first semicolon, and after and including the last one. How can I do that without removing my ability to later split the remaining substring?

Previous questions do not address removing the portions once separated while retaining the portion needed for further use.

Tayen Madsen
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  • What do you want the string to look like? That would help make this question more clear. What final results do you want? – Garry Polley Jul 26 '17 at 20:51
  • The same answers apply there. Just split and then join what you want. – cs95 Jul 26 '17 at 20:59
  • I need Lodging;123 (Which will change because these are inputs from a .txt file). I then need to split the remaining substring and add each half to a list of categories and values, respectively. – Tayen Madsen Jul 26 '17 at 21:02

8 Answers8

2

Stepwise for clarity:

string = "John Doe;Lodging;123;050617"
lst = string.split(';')
lst = lst[1:-1]
string = ';'.join(lst)
print(string)
>>> 'Lodging;123'

As one line:

';'.join('John Doe;Lodging;123;050617'.split(';')[1:-1])
>>> 'Lodging;123'
Jason Stein
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test = "John Doe;Lodging;123;050617"
';'.join(test.split(';')[1:-1])
Zoidberg
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2
s = "John Doe;Lodging;123;050617"
new_s = ';'.join(s.split(';')[1:-1])

This will make new_s = 'Lodging;123'

jacoblaw
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0

The find method will give you the index of the first instance of a character in a string

original_s="John Doe;Lodging;123;050617"
idx = original_s.find(';')
new_s = original_s[i+1:]
print(new_s)
mdornfe1
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0

Use str.partition and str.rpartition.

s = "John Doe;Lodging;123;050617"
s.partition(';')[-1].rpartition(';')[0] # Lodging;123

In terms of efficiency, this is probably the best solution. However, I find the split/join method more readable.

Jared Goguen
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There are a few methods you could try. The simplest one is to split the string and reassemble it:

data = 'John Doe;Lodging;123;050617'
';'.join(data.split(';')[1:-1])

You could also use find and rfind (like find, but from the end of the string backwards:

data[data.find(';')+1:data.rfind(';')]

There's also partition and rpartition:

first, sep, remainder = data.partition(';')
middle, sep, final = remainder.rpartition(';')
Simon Fraser
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Try: 

s = "John Doe;Lodging;123;050617"
print(';'.join(s.split(';')[1:-1]))
Md. Rezwanul Haque
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Try this: use str.find() function to locate the first occurance of ';' and then substring from that point to the end using [:]

aStr = "John Doe;Lodging;123;050617"
aStr = aStr[aStr.find(';')+1:]
print(aStr)
Lodging;123;050617
nanojohn
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