Find multiple elements in string in Python

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.

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

asked Nov 23 '18 at 8:29

M4I3X

286

@Ev.Kounis my bad

– SilverSlash
Nov 23 '18 at 8:40

add a comment |

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.

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

asked Nov 23 '18 at 8:29

M4I3X

286

@Ev.Kounis my bad

– SilverSlash
Nov 23 '18 at 8:40

add a comment |

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.

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

asked Nov 23 '18 at 8:29

M4I3X

286

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.

python

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

asked Nov 23 '18 at 8:29

M4I3X

286

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

asked Nov 23 '18 at 8:29

M4I3X

286

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

edited Dec 11 '18 at 9:56

Cœur

19.1k9114155

asked Nov 23 '18 at 8:29

M4I3X

286

asked Nov 23 '18 at 8:29

M4I3X

286

asked Nov 23 '18 at 8:29

M4I3X

286

@Ev.Kounis my bad

– SilverSlash
Nov 23 '18 at 8:40

add a comment |

@Ev.Kounis my bad

– SilverSlash
Nov 23 '18 at 8:40

@Ev.Kounis my bad

– SilverSlash
Nov 23 '18 at 8:40

add a comment |

4 Answers
4

active

oldest

votes

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 *.

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

1

This is the way to do it. +1

– Ev. Kounis
Nov 23 '18 at 8:44

Thanks, im using this now.

– M4I3X
Nov 23 '18 at 9:06

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

– M4I3X
Nov 23 '18 at 9:15

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__)

– Samuel GIFFARD
Nov 23 '18 at 9:22

I solved it, i was reading lines from a file but forgot to assign a variable to it^^

– M4I3X
Nov 23 '18 at 9:32

add a comment |

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']

edited Nov 23 '18 at 8:48

answered Nov 23 '18 at 8:46

Raniz

8,52112356

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.

– Samuel GIFFARD
Nov 23 '18 at 8:48

That was intentional though

– Raniz
Nov 23 '18 at 8:49

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

– Robin Nemeth
Nov 23 '18 at 9:24

add a comment |

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

answered Nov 23 '18 at 8:41

Sandesh34

254113

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

– Andrey Suglobov
Nov 23 '18 at 8:57

add a comment |

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']

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

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 *.

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

1

This is the way to do it. +1

– Ev. Kounis
Nov 23 '18 at 8:44

Thanks, im using this now.

– M4I3X
Nov 23 '18 at 9:06

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

– M4I3X
Nov 23 '18 at 9:15

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__)

– Samuel GIFFARD
Nov 23 '18 at 9:22

I solved it, i was reading lines from a file but forgot to assign a variable to it^^

– M4I3X
Nov 23 '18 at 9:32

add a comment |

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 *.

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

1

This is the way to do it. +1

– Ev. Kounis
Nov 23 '18 at 8:44

Thanks, im using this now.

– M4I3X
Nov 23 '18 at 9:06

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

– M4I3X
Nov 23 '18 at 9:15

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__)

– Samuel GIFFARD
Nov 23 '18 at 9:22

I solved it, i was reading lines from a file but forgot to assign a variable to it^^

– M4I3X
Nov 23 '18 at 9:32

add a comment |

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 *.

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

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 *.

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

answered Nov 23 '18 at 8:42

Samuel GIFFARD

390112

1

This is the way to do it. +1

– Ev. Kounis
Nov 23 '18 at 8:44

Thanks, im using this now.

– M4I3X
Nov 23 '18 at 9:06

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

– M4I3X
Nov 23 '18 at 9:15

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__)

– Samuel GIFFARD
Nov 23 '18 at 9:22

I solved it, i was reading lines from a file but forgot to assign a variable to it^^

– M4I3X
Nov 23 '18 at 9:32

add a comment |

1

This is the way to do it. +1

– Ev. Kounis
Nov 23 '18 at 8:44

Thanks, im using this now.

– M4I3X
Nov 23 '18 at 9:06

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

– M4I3X
Nov 23 '18 at 9:15

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__)

– Samuel GIFFARD
Nov 23 '18 at 9:22

I solved it, i was reading lines from a file but forgot to assign a variable to it^^

– M4I3X
Nov 23 '18 at 9:32

This is the way to do it. +1

– Ev. Kounis
Nov 23 '18 at 8:44

Thanks, im using this now.

– M4I3X
Nov 23 '18 at 9:06

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

– M4I3X
Nov 23 '18 at 9:15

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__)

– Samuel GIFFARD
Nov 23 '18 at 9:22

I solved it, i was reading lines from a file but forgot to assign a variable to it^^

– M4I3X
Nov 23 '18 at 9:32

add a comment |

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']

edited Nov 23 '18 at 8:48

answered Nov 23 '18 at 8:46

Raniz

8,52112356

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.

– Samuel GIFFARD
Nov 23 '18 at 8:48

That was intentional though

– Raniz
Nov 23 '18 at 8:49

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

– Robin Nemeth
Nov 23 '18 at 9:24

add a comment |

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']

edited Nov 23 '18 at 8:48

answered Nov 23 '18 at 8:46

Raniz

8,52112356

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.

– Samuel GIFFARD
Nov 23 '18 at 8:48

That was intentional though

– Raniz
Nov 23 '18 at 8:49

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

– Robin Nemeth
Nov 23 '18 at 9:24

add a comment |

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']

edited Nov 23 '18 at 8:48

answered Nov 23 '18 at 8:46

Raniz

8,52112356

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']

edited Nov 23 '18 at 8:48

answered Nov 23 '18 at 8:46

Raniz

8,52112356

edited Nov 23 '18 at 8:48

answered Nov 23 '18 at 8:46

Raniz

8,52112356

answered Nov 23 '18 at 8:46

Raniz

8,52112356

answered Nov 23 '18 at 8:46

Raniz

8,52112356

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.

– Samuel GIFFARD
Nov 23 '18 at 8:48

That was intentional though

– Raniz
Nov 23 '18 at 8:49

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

– Robin Nemeth
Nov 23 '18 at 9:24

add a comment |

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.

– Samuel GIFFARD
Nov 23 '18 at 8:48

That was intentional though

– Raniz
Nov 23 '18 at 8:49

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

– Robin Nemeth
Nov 23 '18 at 9:24

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.

– Samuel GIFFARD
Nov 23 '18 at 8:48

That was intentional though

– Raniz
Nov 23 '18 at 8:49

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

– Robin Nemeth
Nov 23 '18 at 9:24

add a comment |

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

answered Nov 23 '18 at 8:41

Sandesh34

254113

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

– Andrey Suglobov
Nov 23 '18 at 8:57

add a comment |

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

answered Nov 23 '18 at 8:41

Sandesh34

254113

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

– Andrey Suglobov
Nov 23 '18 at 8:57

add a comment |

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

answered Nov 23 '18 at 8:41

Sandesh34

254113

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

answered Nov 23 '18 at 8:41

Sandesh34

254113

answered Nov 23 '18 at 8:41

Sandesh34

254113

answered Nov 23 '18 at 8:41

Sandesh34

254113

answered Nov 23 '18 at 8:41

Sandesh34

254113

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

– Andrey Suglobov
Nov 23 '18 at 8:57

add a comment |

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

– Andrey Suglobov
Nov 23 '18 at 8:57

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

– Andrey Suglobov
Nov 23 '18 at 8:57

add a comment |

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']

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

add a comment |

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']

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

add a comment |

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']

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

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']

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

answered Nov 23 '18 at 8:47

Andrey Suglobov

1649

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