Given an array of separators:
columns = ["Name:", "ID:", "Date:", "Building:", "Room:", "Notes:"]
and a string where some columns were left blank (and there is random white space):
input = "Name: JohnID:123:45Date: 8/2/17Building:Room:Notes: i love notes"
How can I get this:
["John", "123:45", "8/2/17", "", "", "i love notes"]
I've tried simply removing the substrings to see where I can go from there but I'm still stuck
import re
input = re.sub(r'|'.join(map(re.escape, columns)), "", input)
use the list to generate a regular expression by inserting (.*) in between, then use strip to remove spaces:
import re
columns = ["Name:", "ID:", "Date:", "Building:", "Room:", "Notes:"]
s = "Name: JohnID:123:45Date: 8/2/17Building:Room:Notes: i love notes"
result = [x.strip() for x in re.match("".join(map("{}(.*)".format,columns)),s).groups()]
print(result)
yields:
['John', '123:45', '8/2/17', '', '', 'i love notes']
the strip part can be handled by the regular expression at the expense of a more complex regex, but simpler overall expression:
result = re.match("".join(map("{}\s*(.*)\s*".format,columns)),s).groups()
more complex: if field data contains regex special chars, we have to escape them (not the case here):
result = re.match("".join(["{}\s*(.*)\s*".format(re.escape(x)) for x in columns]),s).groups()
How about using re.split?
>>> import re
>>> columns = ["Name:", "ID:", "Date:", "Building:", "Room:", "Notes:"]
>>> i = "Name: JohnID:123:45Date: 8/2/17Building:Room:Notes: i love notes"
>>> re.split('|'.join(map(re.escape, columns)), i)
['', ' John', '123:45', ' 8/2/17', '', '', ' i love notes']
To get rid of the whitespace, split on whitespace too:
>>> re.split(r'\s*' + (r'\s*|\s*'.join(map(re.escape, columns))) + r'\s*', i.strip())
['', 'John', '123:45', '8/2/17', '', '', ' i love notes']
Related
I need some help with a regex I am writing. I have a list of words that I want to match and words that might come after them (words meaning [A-Za-z/\s]+) I.e no parenthesis,symbols, numbers.
words = ['qtr','hard','quarter'] # keywords that must exist
test=['id:12345 cli hard/qtr Mix',
'id:12345 cli qtr 90%',
'id:12345 cli hard (red)',
'id:12345 cli hard work','Hello world']
excepted output is
['hard/qtr Mix', 'qtr', 'hard', 'hard work', None]
What I have tried so far
re.search(r'((hard|qtr|quarter)(?:[[A-Za-z/\s]+]))',x,re.I)
The problem with the pattern you have i.e.'((hard|qtr|quarter)(?:[[A-Za-z/\s]+]))', you have \s inside squared brackets [] which means to match the characters individually i.e. either \ or s, instead, you can just use space character i.e.
You can join all the words in words list by | to create the pattern '((qtr|hard|quarter)([a-zA-Z/ ]*))', then search for the pattern in each of strings in the list, if the match is found, take the group 0 and append it to the resulting list, else, append None:
pattern = re.compile('(('+'|'.join(words)+')([a-zA-Z/ ]*))')
result = []
for x in test:
groups = pattern.search(x)
if groups:
result.append(groups.group(0))
else:
result.append(None)
OUTPUT:
result
['hard/qtr Mix', 'qtr ', 'hard ', 'hard work', None]
And since you are including the space characters, you may end up with some values that has space at the end, you can just strip off the white space characters later.
Idea extracted from the existing answer and made shorter :
>>> pattern = re.compile('(('+'|'.join(words)+')([a-zA-Z/ ]*))')
>>> [pattern.search(x).group(0) if pattern.search(x) else None for x in test])
['hard/qtr Mix', 'qtr ', 'hard ', 'hard work', None]
As mentioned in comment :
But it is quite inefficient, because it needs to search for same pattern twice, once for pattern.search(x).group(0) and the other one for if pattern.search(x), and list-comprehension is not the best way to go about in such scenarios.
We can try this to overcome that issue :
>>> [v.group(0) if v else None for v in (pattern.search(x) for x in test)]
['hard/qtr Mix', 'qtr ', 'hard ', 'hard work', None]
You can put all needed words in or expression and put your word definition after that
import re
words = ['qtr','hard','quarter']
regex = r"(" + "|".join(words) + ")[A-Za-z\/\s]+"
p = re.compile(regex)
test=['id:12345 cli hard/qtr Mix(qtr',
'id:12345 cli qtr 90%',
'id:12345 cli hard (red)',
'id:12345 cli hard work','Hello world']
for string in test:
result = p.search(string)
if result is not None:
print(p.search(string).group(0))
else:
print(result)
Output:
hard/qtr Mix
qtr
hard
hard work
None
I want to parse and extract key, values from a given sentence which follow the following format:
I want to get [samsung](brand) within [1 week](duration) to be happy.
I want to convert it into a split list like below:
['I want to get ', 'samsung:brand', ' within ', '1 week:duration', ' to be happy.']
I have tried to split it using [ or ) :
re.split('\[|\]|\(|\)',s)
which is giving output:
['I want to get ',
'samsung',
'',
'brand',
' within ',
'1 week',
'',
'duration',
' to be happy.']
and
re.split('\[||\]|\(|\)',s)
is giving below output :
['I want to get ',
'samsung](brand) within ',
'1 week](duration) to be happy.']
Any help is appreciated.
Note: This is similar to stackoverflow inline links as well where if we type : go to [this link](http://google.com) it parse it as link.
As first step we split the string, and in second step we modify the string:
s = 'I want to get [samsung](brand) within [1 week](duration) to be happy.'
import re
s = re.split('(\[[^]]*\]\([^)]*\))', s)
s = [re.sub('\[([^]]*)\]\(([^)]*)\)', r'\1:\2', i) for i in s]
print(s)
Prints:
['I want to get ', 'samsung:brand', ' within ', '1 week:duration', ' to be happy.']
You may use a two step approach: process the [...](...) first to format as needed and protect these using some rare/unused chars, and then split with that pattern.
Example:
s = "I want to get [samsung](brand) within [1 week](duration) to be happy.";
print(re.split(r'｟([^｟｠]+)｠', re.sub(r'\[([^][]*)]\(([^()]*)\)', r'｟\1:\2｠', s)))
See the Python demo
The \[([^\][]*)]\(([^()]*)\) pattern matches
\[ - a [ char
([^\][]*) - Group 1 ($1): any 0+ chars other than [ and ]
]\( - ]( substring
([^()]*) - Group 2 ($2): any 0+ chars other than ( and )
\) - a ) char.
The ｟([^｟｠]+)｠ pattern just matches any ｟...｠ substring but keeps what is in between as it is captured.
You could replace the ]( pattern first, then split on [) characters
re.replace('\)\[', ':').split('\[|\)',s)
One approach, using re.split with a lambda function:
sentence = "I want to get [samsung](brand) within [1 week](duration) to be happy."
parts = re.split(r'(?<=[\])])\s+|\s+(?=[\[(])', sentence)
processTerms = lambda x: re.sub('\[([^\]]+)\]\(([^)]+)\)', '\\1:\\2', x)
parts = list(map(processTerms, parts))
print(parts)
['I want to get', 'samsung:brand', 'within', '1 week:duration', 'to be happy.']
I have a string that I want to split into a list of certain types. For example, I want to split Starter Main Course Dessert to [Starter, Main Course, Dessert]
I cannot use split() because it will split up the Main Course type. How can I do the splitting? Is regex needed?
If you have a list of acceptable words, you could use a regex union :
import re
acceptable_words = ['Starter', 'Main Course', 'Dessert', 'Coffee', 'Aperitif']
pattern = re.compile("("+"|".join(acceptable_words)+")", re.IGNORECASE)
# "(Starter|Main Course|Dessert|Coffee|Aperitif)"
menu = "Starter Main Course NotInTheList dessert"
print pattern.findall(menu)
# ['Starter', 'Main Course', 'dessert']
If you just want to specify which special substrings should be matched, you could use :
acceptable_words = ['Main Course', '\w+']
I think it's more practical to specify 'special' two-words tokens only.
special_words = ['Main Course', 'Something Special']
sentence = 'Starter Main Course Dessert Something Special Date'
words = sentence.split(' ')
for i in range(len(words) - 1):
try:
idx = special_words.index(str(words[i]) + ' ' + words[i+1])
words[i] = special_words[idx]
words[i+1] = None
except ValueError:
pass
words = list(filter(lambda x: x is not None, words))
print(words)
I've been looking around here, but I didn't find anything that was close to my problem. I'm using Python3.
I want to split a string at every whitespace and at commas. Here is what I got now, but I am getting some weird output:
(Don't worry, the sentence is translated from German)
import re
sentence = "We eat, Granny"
split = re.split(r'(\s|\,)', sentence.strip())
print (split)
>>>['We', ' ', 'eat', ',', '', ' ', 'Granny']
What I actually want to have is:
>>>['We', ' ', 'eat', ',', ' ', 'Granny']
I'd go for findall instead of split and just match all the desired contents, like
import re
sentence = "We eat, Granny"
print(re.findall(r'\s|,|[^,\s]+', sentence))
This should work for you:
import re
sentence = "We eat, Granny"
split = list(filter(None, re.split(r'(\s|\,)', sentence.strip())))
print (split)
Alternate way:
import re
sentence = "We eat, Granny"
split = [a for a in re.split(r'(\s|\,)', sentence.strip()) if a]
Output:
['We', ' ', 'eat', ',', ' ', 'Granny']
Works with both python 2.7 and 3
I have a string that may contain random segments of quoted and unquoted texts. For example,
s = "\"java jobs in delhi\" it software \"pune\" hello".
I want to separate out the quoted and unquoted parts of this string in python.
So, basically I expect the output to be:
quoted_string = "\"java jobs in delhi\"" "\"pune\""
unquoted_string = "it software hello"
I believe using a regex is the best way to do it. But I am not very good with regex. Is there some regex expression that can help me with this?
Or is there a better solution available?
I dislike regex for something like this, why not just use a split like this?
s = "\"java jobs in delhi\" it software \"pune\" hello"
print s.split("\"")[0::2] # Unquoted
print s.split("\"")[1::2] # Quoted
If your quotes are as basic as in your example, you could just split; example:
for s in (
'"java jobs in delhi" it software "pune" hello',
'foo "bar"',
):
result = s.split('"')
print 'text between quotes: %s' % (result[1::2],)
print 'text outside quotes: %s' % (result[::2],)
Otherwise you could try:
import re
pattern = re.compile(
r'(?<!\\)(?:\\\\)*(?P<quote>["\'])(?P<value>.*?)(?<!\\)(?:\\\\)*(?P=quote)'
)
for s in data:
print pattern.findall(s)
I explain the regex (I use it in ihih):
(?<!\\)(?:\\\\)* # find backslash
(?P<quote>["\']) # any quote character (either " or ')
# which is *not* escaped (by a backslash)
(?P<value>.*?) # text between the quotes
(?<!\\)(?:\\\\)*(?P=quote) # end (matching) quote
Debuggex Demo
/
Regex101 Demo
Use a regex for that:
re.findall(r'"(.*?)"', s)
will return
['java jobs in delhi', 'pune']
You should use Python's shlex module, it's very nice:
>>> from shlex import shlex
>>> def get_quoted_unquoted(s):
... lexer = shlex(s)
... items = list(iter(lexer.get_token, ''))
... return ([i for i in items if i[0] in "\"'"],
[i for i in items if i[0] not in "\"'"])
...
>>> get_quoted_unquoted("\"java jobs in delhi\" it software \"pune\" hello")
(['"java jobs in delhi"', '"pune"'], ['it', 'software', 'hello'])
>>> get_quoted_unquoted("hello 'world' \"foo 'bar' baz\" hi")
(["'world'", '"foo \'bar\' baz"'], ['hello', 'hi'])
>>> get_quoted_unquoted("does 'nested \"quotes\" work' yes")
(['\'nested "quotes" work\''], ['does', 'yes'])
>>> get_quoted_unquoted("what's up with single quotes?")
([], ["what's", 'up', 'with', 'single', 'quotes', '?'])
>>> get_quoted_unquoted("what's up when there's two single quotes")
([], ["what's", 'up', 'when', "there's", 'two', 'single', 'quotes'])
I think this solution is as simple as any other solution (basically a oneliner, if you remove the function declaration and grouping) and it handles nested quotes well etc.