I'm trying to filter a date retrieved from a .csv file, but no combination I try seems to work. The date comes in as "2011-10-01 19:25:01" or "year-month-date hour:min:sec".
I want just the year, month and date but I get can't seem to get ride of the time from the string:
date = bug[2] # Column in which the date is located
date = date.replace('\"','') #getting rid of the quotations
mdate = date.replace(':','')
re.split('$[\d]+',mdate) # trying to get rid of the trailing set of number (from the time)
Thanks in advance for the advice.
If your source is a string, you'd probably better use strptime
import datetime
string = "2011-10-01 19:25:01"
dt = datetime.datetime.strptime(string, "%Y-%m-%d %H:%M:%S")
After that, use
dt.year
dt.month
dt.day
to access the data you want.
Use datetime to parse your input as a datetime object, then output it in whatever format you like: http://docs.python.org/library/datetime.html
I think you're confusing the circumflex for start of line and dollar for end of line. Try ^[\d-]+.
If the format is always "YYYY-MM-DD HH:mm:ss", then try this:
date = date[1:11]
In a prompt:
>>> date = '"2012-01-12 15:13:20"'
>>> date[1:11]
'2012-01-12'
>>>
No need for regex
>>> date = '"2011-10-01 19:25:01"'
>>> date.strip('"').split()[0]
'2011-10-01'
One problem with your code is that in your last regular expression, $ matches the end of the string, so that regular expression will never match anything. You could do this much more simply by splitting by spaces and only taking the first result. After removing the quotation marks, the line
date.split()
will return ["2011-10-01", "19:25:01"], so the first element of that list is what you need.
Related
I'm attempting to read data from multiple text files and move the data into a two-dimensional array. The data needs to remain in a specific order.
Could regex assist with this?
If you have any insight on how to improve this section of the code please let me know.
the datetime module provides (most) everything date-related
from datetime import datetime
date = "Sat 30-Mar-1996 7:40 PM"
fmt = "%a %d-%b-%Y %I:%M %p"
a = datetime.strptime(date, fmt)
print(a.year)
>>> 1996
You can parse the date-time string very easily by splitting its components and using iterable unpacking, e.g.,
def parse_date(d):
day_of_week, date, hhmm, ampm = d.split()
day_of_month, month, year = date.split('-')
hour, minute = hhmm.split(':')
return (year, month, day_of_month,
hour if ampm=='AM' or str(int(hour)+12), minute,
day_of_week)
and later, in the body of the loop
year, m, dom, h, m, dow = parse_date(fields[-1].strip())
or, if you are interested only in year
year, *_ = parse_date(fields[-1].strip())
You're probably looking for regular expressions, which are a very powerful way to analyze and extract data from strings. For an intro into them, I'd check out this site or the python docs, but in your case I think you probably want something like '| ([a-zA-Z]*) ([0-9]*)-([a-zA-Z]*)-([0-9]*) ([0-9:]* [a-zA-Z]*) |' would work. A more specific description of the format the time would be in is necessary for a 100% correct regex [short for regular expressions].
To use regex in python, you want the re library. First, create the pattern matcher with matcher = re.compile(your_regex_string_here). Then, find the match with result = matcher.match(file_contents). (You could also just do result = re.match(regex_string,file_contents).) Whatever your regex, anything surrounded by parentheses is known as a "capturing group", which can be extracted from the result with result.group(); result.group(0) will return full match, and result.group(n) will return the contents of the nth capturing group - that is, the nth set of parentheses. In the above example, result.group(4) would return the year, though you could get any of the day of the week, day, month, year, and time by using groups 1-5.
The DateTime module as mentioned in another answer is also a great tool.
My file name looks like as follow:
show_data_paris_12112019.xlsx
I want to extract the date only and I have tried this script:
date = os.path.basename(xls)
pattern = r'(?<=_)_*(?=\.xlsx)'
re.search(pattern, date).group(0)
event_date = re.search(pattern, date).group(0)
event_date_obj = datetime.strptime (event_date, '%Y%m%d')
but It gives me errors. How can I fix this?
Thank you.
It looks to me like the regex you're using is also at fault, and so it fails when trying to group(0) from the empty return.
Assuming all your dates are stored as digits the following regex i've made seems to work quite well.
(?!.+_)\d+(?=\.xlsx)
The next issue is when formatting the date it experiences an issue with the way you're formatting the date, to me it looks like 12112019 would be the 12/11/2019 obviously this could also be the 11/12/2019 but the basic is that we change the way strftime formats the date.
So for the date / month / year format we would use
# %d%m%Y
event_date_obj = datetime.strptime(event_date, '%d%m%Y')
And we would simply swap %d and %m for the month / date / year format. So your complete code would look something like this:
date = os.path.basename(xls)
pattern = "(?!.+_)\d+(?=\.xlsx)"
event_date = re.search(pattern, date).group(0)
event_date_obj = datetime.strptime (event_date, '%d%m%Y')
For further information on how to use strftime see https://strftime.org/.
_* matches a sequence of zero or more underscore characters.
(?<=_) means that it has to be preceded by an underscore.
(?=\.xlsx) means that it has to be followed by .xlsx.
So this will match the ___ in foo____.xlsx. But it doesn't match your filename, because the data is between the underscore and .xlsx.
You should match \d+ rather than _* between the lookbehind and lookahead.
pattern = r'(?<=_)\d+(?=\.xlsx)'
And if the data is always 8 digits, use \d{8} to be more specific.
I have several thousand files which feature datetime in their file name.
Sadly the devider between the datetime blocks are not always the same.
Example:
Data_trul-100A1-Berlin_2019-01-31_150480.dat
Data_tral-2000B2-Frankf-2018_02_27-190200.dat
Data_bash-300003_Hambrg_2017-04-12_210500.dat
I managed to find the datetime part in the string with a regular expression
import re
strings = ['Data_trul-100A1-Berlin_2019-01-31_150430.dat',
'Data_tral-2000B2-Frankf-2018_02_27-190200.dat',
'Data_bash-300003_Hambrg_2017-04-12_210500.dat']
for part_string in strings:
match = re.search('\d{4}[-_]\d{2}[-_]\d{2}[-_]\d{6}', part_string)
print(match.group())
However, now I am stuck to convert the group to datetime
from datetime import datetime
date = datetime.strptime(match.group(), "%Y-%m-%d_%H%M%S")
because I need to specify dashes or underscores.
I came up with the following solution to just replace it, but that feels like cheating.
for part_string in strings:
part_string = part_string.replace('-',"_")
match = re.search('\d{4}_\d{2}_\d{2}_\d{6}', part_string)
date = datetime.strptime(match.group(), "%Y_%m_%d_%H%M%S")
print(date)
Is there a more elegant way? Using regex to find the divider and pass it on to strptime?
You could change your regular expression to find 4 separate elements
match = re.search('(\d{4})[-_](\d{2})[-_](\d{2})[-_](\d{6})', part_string)
Then combine them into one standard string format
fixedstring = "{}_{}_{}_{}".format(match.groups())
date = datetime.strptime(match.group(), "%Y_%m_%d_%H%M%S")
Of course at this point you could just split the HHMMSS part of the time into their own elements and build the datetime object directly,
m = re.search('(\d{4})[-_](\d{2})[-_](\d{2})[-_](\d{2})(\d{2})(\d{2})', part_string)
date = datetime.datetime(year=m.group(0),
month=m.group(1),
day=m.group(2),
hour=m.group(3),
minute=m.group(4),
second=m.group(5))
i am trying to do string manipulation based on format. str.replace(old,new) alllows changing by specific string pattern. is it possible to find and replace by format? for example,
i want to find all datetime like value in a long string and replace it with another format
assuming % is wildcard for number and datetime is %%/%%/%%T%%:%%
str.replace(%%/%%/%%T%%:%%, 'dummy value')
EDIT:
sorry i should have been more clearer. re.sub seems like I can use that, but how do it substitute it with a date converted value. in this case, e.g.
YY/MM/DDTHH:MM to (YY/MM/DD HH:MM)+8 hours
The easiest way to do this is probably using a combination of regular expression syntax, applying re.sub and using the fact that the repl parameter can be a function that takes a match and returns a string to replace it, and datetime's syntax for strptime and strftime:
>>> from datetime import datetime
>>> import re
>>> def replacer(match):
return datetime.strptime(
match.group(), # matched text
'%y/%m/%dT%H:%M', # source format in datetime syntax
).strftime('%d %B %Y at %H.%M') # destination format in datetime syntax
>>> re.sub(
r'\d{2}/\d{2}/\d{2}T\d{2}:\d{2}', # source format in regex syntax
replacer, # function to process match
'The date and time was 12/12/12T12:12 exactly.', # string to process
)
'The date and time was 12 December 2012 at 12.12 exactly.'
The only downside of this is that you need to define the source format in both datetime and re syntax, which isn't very DRY; if they don't match, you'll get nowhere.
I have a string
string = "Friday07:48 AM"
How do I get rid of "Friday"? I could simply use a replace() function but this string could also be any other day of the week. So it could look like:
string = "Sunday07:48 AM"
How do I only get "07:48 AM"?
We can utilize the fact that every day of the week in English ends in the substring 'day' to locate that within your string, and then go from three indices farther from where 'day' starts until the end of your string.
date_str = "Friday07:48 AM"
new_str = date_str[date_str.index('day')+3:]
new_str # '07:48 AM'
As an aside, never name a string 'string' or 'str', because those are special words in Python.
Well for everything except Saturday and Wednesday you could just grab it by:
day = string[0:6]
but in the all cases IF the time stamp in the front always holds the same index you could:
day = string[:-8]
Try it in the shell
How about using regular expressions?
import re
re.search("([a-z]+)(.*)",string,flags=re.I).group(2)