I'm trying to parse a csv file with times in the form of 6:30pm or 7am, or midnight. I've googled around and read the docs for regular expressions in the python docs but haven't been able to implement them successfully.
My first try to match them was:
re.findall(r'^d{1,2}(:d{1,2})?$', string)
But this didn't work. I have the parenthesis and the question mark there because sometimes there isn't always anything more than the hour. Also, I haven't even begun to think about how to match the am and pm.
Any help is appreciated!
First of all, to match digits you need \d, not just d.
re.findall(r'^\d{1,2}(:\d{1,2})?$', string)
Second, as written, your regex will only match a string which is exactly a single time and nothing else, because ^ means "beginning of string" and $ means "end of string. You can omit those if you want to find all of the times throughout the string:
re.findall(r'\d{1,2}(:\d{1,2})?', string)
As far as the am/pm goes, you can just add another optional group:
re.findall(r'\d{1,2}(:\d{1,2})?(am|pm)?', string)
Of course, because everything is optional besides the first 1 or 2 digits, you're also going to match any one or two digit number. You could instead require either at least either am/pm or a colon and two more digits:
re.findall(r'\d{1,2}((am|pm)|(:\d{1,2})(am|pm)?)', string)
But, findall behaves slightly oddly: if you have matching groups in your pattern, it'll only return the groups rather than the full match. Thus, you can change them to non-matching groups:
re.findall(r'\d{1,2}(?:(?:am|pm)|(?::\d{1,2})(?:am|pm)?)', string)
If you are strictly looking for a regex solution. You can use:
re.findall(r'^\d{1,2}(:\d{1,2})?$', string)
But wait
that's not all. There is a better way to do it without regex ;). You can use python CSV parsing powers.
import csv
string = "November,Monday,6:30pm,1989"
csv_reader = csv.reader( [ string ] )
for row in csv_reader:
print row
Output
['November', 'Monday', '6:30pm', '1989']
import re
regex = r'(\d{1,2})([.:](\d{1,2}))?[ ]?(am|pm)?'
groups = re.findall(regex, value)
group1 will give hr
group3 will give min
group4 will give am/pm
Examples :
12pm
12.30pm
12:30pm
2.30 am
all these examples are working
Related
I am new to regexes.
I have the following string : \n(941)\n364\nShackle\n(941)\nRivet\n105\nTop
Out of this string, I want to extract Rivet and I already have (941) as a string in a variable.
My thought process was like this:
Find all the (941)s
filter the results by checking if the string after (941) is followed by \n, followed by a word, and ending with \n
I made a regex for the 2nd part: \n[\w\s\'\d\-\/\.]+$\n.
The problem I am facing is that because of the parenthesis in (941) the regex is taking 941 as a group. In the 3rd step the regex may be wrong, which I can fix later, but 1st I needed help in finding the 2nd (941) so then I can apply the 3rd step on that.
PS.
I know I can use python string methods like find and then loop over the searches, but I wanted to see if this can be done directly using regex only.
I have tried the following regex: (?:...), (941){1} and the make regex literal character \ like this \(941\) with no useful results. Maybe I am using them wrong.
Just wanted to know if it is possible to be done using regex. Though it might be useful for others too or a good share for future viewers.
Thanks!
Assuming:
You want to avoid matching only digits;
Want to match a substring made of word-characters (thus including possible digits);
Try to escape the variable and use it in the regular expression through f-string:
import re
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
var1 = '(941)'
var2 = re.escape(var1)
m = re.findall(fr'{var2}\n(?!\d+\n)(\w+)', s)[0]
print(m)
Prints:
Rivet
If you have text in a variable that should be matched exactly, use re.escape() to escape it when substituting into the regexp.
s = '\n(941)\n364\nShackle\n(941)\nRivet\n105\nTop'
num = '(941)'
re.findall(rf'(?<=\n{re.escape(num)}\n)[\w\s\'\d\-\/\.]+(?=\n)', s)
This puts (941)\n in a lookbehind, so it's not included in the match. This avoids a problem with the \n at the end of one match overlapping with the \n at the beginning of the next.
I have an input string consisting of a sequence of real numbers separated by a single space. It is also acceptable for the string to contain only one real number (no spaces). My goal is to check whether the string structure matches the following (in this order):
optional (0/1): minus (-)
1/more digits
optional (1+): a period and 1/more digits
optional (0+): a group consisting of a space and the first group (the first three bullet points)
It should describe the string completely. If not, it should print an error message and exit.
My current regular expression is ^(-?\d+(\.?\d)*)( \1)*$ which I thought would be okay, but even the first group doesn't match all the real numbers individually. And I need it to check the string from the beginning to the end, including the spaces.
My code for this function looks like this:
import re
def structure_check(string):
structure = r"^(-?\d+(\.?\d)*)( \1)*$"
if re.match(structure,string):
return("OK")
else:
print("Input error")
exit()
It should accept strings like: 15 35 -45 8 -2.3 4564.18 56 etc., but it doesn't correspond to changes in the input (doesn't match) at all. It shouldn't match if there is too many spaces, incorrectly placed . or -, or if there are other characters than digits, periods, dashes (-) and spaces.
I could also do this with just the first group while iterating over a list created by splitting the input string by space, but I would prefer to check it according to my main goal, since I wouldn't have to split the input in the validation function and also to save some more code lines by checking the input alltogether (eg. for excess spaces, or unsupported characters, which I'd have to otherwise check separately).
Sorry if I missed any answered questions, I couldn't find any appropriate for my problem in Python. If you know about any, feel free to link them, please. And thank you, I am a beginner and started learning regex for a project just about yesterday.
You can use:
^((?:[+-]?\d+(?:[.]\d+)?)(?:[ \t]|$))*$
Demo and explantation
I added + to the optional sign. If you only want to match with no sign or -, just remove that from the optional character class.
You could also use an unrolled version to prevent matching a space at the end.
^-?\d+(?:\.\d+)?(?: -?\d+(?:\.\d+)?)*$
Regex demo
The backreference \1 will match exactly what is matched in group 1 and for your pattern will match for example 123 123 123
If you want to repeat the group, you could recurse the first group using the PyPi regex module and (?1)
^(-?\d+(?:\.\d+)?)(?: (?1))*$
See a Python example
Problem is in your regexp, to be specific, in ( \1)* part.
This, described, means: space and string that was matched in group 1 zero or more times
Thus, your regexp will match for the following, for example:
15 15 15
-5.3 -5.3 -5.3 -5.3
And so on.
To fix the regexp, I would replace the group reference with the actual group, like so:
^(-?\d+(\.?\d)*)( -?\d+(\.?\d)*)*$
I would also point out that this regexp allows the numbers to have multiple decimal dots, (e.g. 1.2.3 passes) however I'm not sure if that's intended or not.
In JavaScript you can use the method .test of regex. The regex should work in python.
let ok = /^(([+\-]?\d+(\.\d+)?)( |$))+$/.test("15 35 -45 8 -2.3 4564.18 56");
console.log(ok);
Explanation: (.\d+)? You must make the whole group optional. The number can be followed by a space or the end of a string ( |$). The pattern is repeated throughout the string so I wrapped the entire expression in a group. Insert ^ at the beginning of the regex and $ at the end of the regex to force the regex to check the string completely.
I am a beginner in Python and in regular expressions and now I try to deal with one exercise, that sound like that:
How would you write a regex that matches a number with commas for
every three digits? It must match the following:
'42'
'1,234'
'6,368,745'
but not the following:
'12,34,567' (which has only two digits between the commas)
'1234' (which lacks commas)
I thought it would be easy, but I've already spent several hours and still don't have write answer. And even the answer, that was in book with this exercise, doesn't work at all (the pattern in the book is ^\d{1,3}(,\d{3})*$)
Thank you in advance!
The answer in your book seems correct for me. It works on the test cases you have given also.
(^\d{1,3}(,\d{3})*$)
The '^' symbol tells to search for integers at the start of the line. d{1,3} tells that there should be at least one integer but not more than 3 so ;
1234,123
will not work.
(,\d{3})*$
This expression tells that there should be one comma followed by three integers at the end of the line as many as there are.
Maybe the answer you are looking for is this:
(^\d+(,\d{3})*$)
Which matches a number with commas for every three digits without limiting the number being larger than 3 digits long before the comma.
You can go with this (which is a slightly improved version of what the book specifies):
^\d{1,3}(?:,\d{3})*$
Demo on Regex101
I got it to work by putting the stuff between the carrot and the dollar in parentheses like so: re.compile(r'^(\d{1,3}(,\d{3})*)$')
but I find this regex pretty useless, because you can't use it to find these numbers in a document because the string has to begin and end with the exact phrase.
#This program is to validate the regular expression for this scenerio.
#Any properly formattes number (w/Commas) will match.
#Parsing through a document for this regex is beyond my capability at this time.
print('Type a number with commas')
sentence = input()
import re
pattern = re.compile(r'\d{1,3}(,\d{3})*')
matches = pattern.match(sentence)
if matches.group(0) != sentence:
#Checks to see if the input value
#does NOT match the pattern.
print ('Does Not Match the Regular Expression!')
else:
print(matches.group(0)+ ' matches the pattern.')
#If the values match it will state verification.
The Simple answer is :
^\d{1,2}(,\d{3})*$
^\d{1,2} - should start with a number and matches 1 or 2 digits.
(,\d{3})*$ - once ',' is passed it requires 3 digits.
Works for all the scenarios in the book.
test your scenarios on https://pythex.org/
I also went down the rabbit hole trying to write a regex that is a solution to the question in the book. The question in the book does not assume that each line is such a number, that is, there might be multiple such numbers in the same line and there might some kind of quotation marks around the number (similar to the question text). On the other hand, the solution provided in the book makes those assumptions: (^\d{1,3}(,\d{3})*$)
I tried to use the question text as input and ended up with the following pattern, which is way too complicated:
r'''(
(?:(?<=\s)|(?<=[\'"])|(?<=^))
\d{1,3}
(?:,\d{3})*
(?:(?=\s)|(?=[\'"])|(?=$))
)'''
(?:(?<=\s)|(?<=[\'"])|(?<=^)) is a non-capturing group that allows
the number to start after \s characters, ', ", or the start of the text.
(?:,\d{3})* is a non-capturing group to avoid capturing, for example, 123 in 12,123.
(?:(?=\s)|(?=[\'"])|(?=$)) is a non-capturing group that allows
the number to end before \s characters, ', ", or the end of the text (no newline case).
Obviously you could extend the list of allowed characters around the number.
I'm trying to parse a text document with data in the following format: 24036 -977. I need to separate the numbers into separate values, and the way I've done that is with the following steps.
values = re.search("(.*?)\s(.*)")
x = values.group(1)
y = values.gropu(2)
This does the job, however I was curious about why using (.*?) in the second group causes the regex to fail? I tested it in the online regex tester(https://regex101.com/r/bM2nK1/1), and adding the ? in causes the second group to return nothing. Now as far as I know .*? means to take any value unlimited times, as few times as possible, and the .* is just the greedy version of that. What I'm confused about is why the non greedy version.*? takes that definition to mean capturing nothing?
Because it means to match the previous token, the *, as few times as possible, which is 0 times. If you would it to extend to the end of the string, add a $, which matches the end of string. If you would like it to match at least one, use + instead of *.
The reason the first group .*? matches 24036 is because you have the \s token after it, so the fewest amount of characters the .*? could match and be followed by a \s is 24036.
#iobender has pointed out the answer to your question.
But I think it's worth mentioning that if the numbers are separated by space, you can just use split:
>>> '24036 -977'.split()
['24036', '-977']
This is simpler, easier to understand and often faster than regex.
I need to extract the date in format of: dd Month yyyy (20 August 2013).
I tried the following regex:
\d{2} (January|February|March|April|May|June|July|August|September|October|November|December) \d{4}
It works with regex testers (chcked with several the text - Monday, 19 August 2013), but It seems that Python doesn't understand it. The output I get is:
>>>
['August']
>>>
Can somebody please understand me why is that happening ?
Thank you !
Did you use re.findall? By default, if there's at least one capture group in the pattern, re.findall will return only the captured parts of the expression.
You can avoid this by removing every capture group, causing re.findall to return the entire match:
\d{2} (?:January|February|...|December) \d{4}
or by making a single big capture group:
(\d{2} (?:January|February|...|December) \d{4})
or, possibly more conveniently, by making every component a capture group:
(\d{2}) (January|February|...|December) (\d{4})
This latter form is more useful if you will need to process the individual day/month/year components.
It looks like you are only getting the data from the capture group, try this:
(\d{2} (?:January|February|March|April|May|June|July|August|September|October|November|December) \d{4})
I put a capture group around the entire thing and made the month a non-capture group. Now whatever was giving you "August" should give you the entire thing.
I just looked at some python regex stuff here
>>> p = re.compile('(a(b)c)d')
>>> m = p.match('abcd')
>>> m.group(0)
'abcd'
>>> m.group(1)
'abc'
>>> m.group(2)
'b'
Seeing this, I'm guessing (since you didn't show how you were actually using this regex) that you were doing group(1) which will now work with the regex I supplied above.
It also looks like you could have used group(0) to get the whole thing (if I am correct in the assumption that this is what you were doing). This would work in your original regex as well as my modified version.