1

my problem is that I need to find multiple elements in one string.

For example I got one string that looks like this:

line = if ((var.equals("INPUT")) || (var.equals("OUTPUT"))

and then i got this code to find everything between ' (" ' and ' ") '

char1 = '("'
char2 = '")'


add = line[line.find(char1)+2 : line.find(char2)]
list.append(add)

The current result is just:

['INPUT']

but I need the result to look like this:

['INPUT','OUTPUT', ...]

after it got the first match it stopped searching for other matches, but I need to find everything in that string that matches this search.

I also need to append every single match to the list.

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1
  • @Ev.Kounis my bad Commented Nov 23, 2018 at 8:40

4 Answers 4

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6

The simplest:

>>> import re
>>> s = """line = if ((var.equals("INPUT")) || (var.equals("OUTPUT"))"""
>>> r = re.compile(r'\("(.*?)"\)')
>>> r.findall(s)
['INPUT', 'OUTPUT']

The trick is to use .*? which is a non-greedy *.

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5 Comments

This is the way to do it. +1
Thanks, im using this now.
I got one more thing, how can i get the result of findall into a variable? If i try result = r.findall(s) i get this error TypeError: expected string or bytes-like object
Your code should work. Looks more like an encoding issue. You're Python 2 or Python 3? Check the type of your input. With something like print(my_input.__class__)
I solved it, i was reading lines from a file but forgot to assign a variable to it^^
2

You should look into regular expressions because that's a perfect fit for what you're trying to achieve.

Let's examine a regular expression that does what you want:

import re
regex = re.compile(r'\("([^"]+)"\)')

It matches the string (" then captures anything that isn't a quotation mark and then matches ") at the end.

By using it with findall you will get all the captured groups:

In [1]: import re

In [2]: regex = re.compile(r'\("([^"]+)"\)')

In [3]: line = 'if ((var.equals("INPUT")) || (var.equals("OUTPUT"))'

In [4]: regex.findall(line)
Out[4]: ['INPUT', 'OUTPUT']
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3 Comments

NB: Will not work if there's a " in the string that he wants to find. Non-greedy star operator *? is the clean way to go there.
That was intentional though
Regular experssions are not generally a good fit for brackets matching. For this particular task it works, but it can break easily (or get messy) if doing something slightly more complicated. Just a heads up to the OP
0

If you don't want to use regex, this will help you.

line = 'if ((var.equals("INPUT")) || (var.equals("OUTPUT"))'
char1 = '("'
char2 = '")'


add = line[line.find(char1)+2 : line.find(char2)]
list.append(add)
line1=line[line.find(char2)+1:]
add = line1[line1.find(char1)+2 : line1.find(char2)]
list.append(add)
print(list)

just add those 3 lines in your code, and you're done

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1 Comment

how that method will help if will be more than two ("...") in line?
0

if I understand you correct, than something like that is help you:

line = 'line = if ((var.equals("INPUT")) || (var.equals("OUTPUT"))'
items = []
start = 0
end = 0
c = 0;
while c < len(line):
    if line[c] == '(' and line[c + 1] == '"':
        start = c + 2
    if line[c] == '"' and line[c + 1] == ')':
        end = c
    if start and end:
        items.append(line[start:end])
        start = end = None
    c += 1

print(items)    # ['INPUT', 'OUTPUT']
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