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Python split regex matches into two variables

I am trying to split a re.findall match into two variables - one for the date and one for the time - but i can't find a way to split the list up somehow. Any help appreciated!
txt = "created_at': datetime.datetime(2023, 1, 17, 11, 38, 26, tzinfo"
x = re.findall("datetime.datetime\((.+?)\, (.+?)\, (.+?)\, (.+?)\, (.+?)\, (.+?)\, tzinfo", txt)
print(x)
print(x[0:4])
This is the results I get
[('2023', '1', '17', '11', '38', '26')]
[('2023', '1', '17', '11', '38', '26')]
It seems like the re.findall doesn't create a list with each find but just puts it all into one entry. The first 3 numbers are the date, the last 3 the time. I really don't know how to approach this without being able to grab each item individually.

You can use a list comprehension to extract the required fields for each match.
res = [o[:4] for o in x]

removing multiple pipes from a list

so i have some data i have been trying to clean up, its a list and it looks like this
a = [\nlondon\n\n18\n\n20\n\n30\n\n\n\n\njapan\n\n6\n\n80\n\n2\n\n\n\n\nSpain]
i have tried to clean it up by doing this
a.replace("\n", "|")
the output turn out like this :
[london||18||20||30||||japan||6||80||2|||Spain]
if i do this:
a.replace("\n","")
i get this:
[london,"", "", 18,"","",20"","",30,"","","",""japan,"",""6,"","",80,"","",2"","","","",Spain]
can anyone explain why i am having multiple pipes, spaces and whats the best way to clean the data.

Assuming that your input is:
s = '\nlondon\n\n18\n\n20\n\n30\n\n\n\n\njapan\n\n6\n\n80\n\n2\n\n\n\n\nSpain'
The issue is that there are multiple '\n' in-between data, therefore just replacing each '\n' with another character (say '|') will give you as many of the new characters as there were '\n'.
The simplest approach is to use str.split() to get the non-blank data:
l = list(s.split())
print(l)
# ['london', '18', '20', '30', 'japan', '6', '80', '2', 'Spain']
or, combine it with str.join(), if you want to have it separated by '|':
t = '|'.join(s.split())
print(t)
# london|18|20|30|japan|6|80|2|Spain

I tried it and got this:
a = ['\nlondon\n\n18\n\n20\n\n30\n\n\n\n\njapan\n\n6\n\n80\n\n2\n\n\n\n\nSpain']
print(a[0].replace("\n", ""))
Output:
london182030japan6802Spain
Could you please clarify the exact input and the expected output? it does not seem correct yet and I have taken some liberties.
If your input was a string you can use split():
a = '\nlondon\n\n18\n\n20\n\n30\n\n\n\n\njapan\n\n6\n\n80\n\n2\n\n\n\n\nSpain'
print(a.split())
Output:
['london', '18', '20', '30', 'japan', '6', '80', '2', 'Spain']

python regex for incomplete decimals numbers

I have a string of numbers which may have incomplete decimal reprisentation
for example
a = '1. 1,00,000.00 1 .99 1,000,000.999'
desired output
['1','1,00,000.00','1','.99','1,000,000.999']
so far i have tried the following 2
re.findall(r'[-+]?(\d+(?:[.,]\d+)*)',a)
which gives
['1', '1,00,000.00', '1', '99', '1,000,000.999']
which makes .99 to 99 which is not desired
while
re.findall(r'[-+]?(\d*(?:[.,]\d+)*)',a)
gives
['1', '', '', '1,00,000.00', '', '', '1', '', '.99', '', '1,000,000.999', '']
which gives undesirable empty string results as well
this is for finding currency values in a string so the commas separators don't have a set pattern or mat not be present at all

My suggestion is to use the regex below:
I've implemented a snippet in python.
import re
a = '1. 1,00,000.00 1 .99 1,000,000.999'
result = re.split('/\.?\d\.?\,?/', a)
print result
Output:
['1', '1,00,000.00', '1', '.99', '1,000,000.999']

You can use re.split:
import re
a = '1. 1,00,000.00 1 .99 1,000,000.999'
d = re.split('(?<=\d)\.\s+|(?<=\d)\s+', a)
Output:
['1', '1,00,000.00', '1', '.99', '1,000,000.999']

This regex will give you your desired output:
([0-9]+(?=\.))|([0-9,]+\.[0-9]+)|([0-9]+)|(\.[0-9]+)
You can test it here: https://regex101.com/r/VfQIJC/6

extract substring between multiple words in a pandas dataframe

I have a pandas data frame where I need to extract sub-string from each row of a column based on the following conditions
We have start_list ('one','once I','he') and end_list ('fine','one','well').
The sub-string should be preceded by any of the elements of the start_list.
The sub-string may be succeeded by any of the elements of the end_list.
When any of the elements of the start_list is available then the succeeding sub string should be extracted with/without the presence of the elements of the end_list.
Example Problem:
df = pd.DataFrame({'a' : ['one was fine today', 'we had to drive', ' ','I
think once I was fine eating ham ', 'he studies really
well
and is polite ', 'one had to live well and prosper',
'43948785943one by onej89044809', '827364hjdfvbfv',
'&^%$&*+++===========one kfnv dkfjn uuoiu fine', 'they
is one who makes me crazy'],
'b' : ['11', '22', '33', '44', '55', '66', '77', '', '88',
'99']})
Expected Result:
df = pd.DataFrame({'a' : ['was', '','','was ','studies really','had to live',
'by','','kfnv dkfjn uuoiu','who makes me crazy'],
'b' : ['11', '22', '33', '44', '55', '66', '77', '',
'88','99']})

I think this should work for you. This solution requires Pandas of course and also the built-in library functools.
Function: remove_preceders
This function takes as input a collection of words start_list and str string. It looks to see if any of the items in start_list are in string, and if so returns only the piece of string that occurs after said items. Otherwise, it returns the original string.
def remove_preceders(start_list, string):
for word in start_list:
if word in string:
string = string[string.find(word) + len(word):]
return string
Function: remove_succeders
This function is very similar to the first, except it returns only the piece of string that occurs before the items in end_list.
def remove_succeeders(end_list, string):
for word in end_list:
if word in string:
string = string[:string.find(word)]
return string
Function: to_apply
How do you actually run the above functions? The apply method allows you to run complex functions on a DataFrame or Series, but it will then look for as input either a full row or single value, respectively (based on whether you're running on a DF or S).
This function takes as input a function to run & a collection of words to check, and we can use it to run the above two functions:
def to_apply(func, words_to_check):
return functools.partial(func, words_to_check)
How to Run
df['no_preceders'] = df.a.apply(
to_apply(remove_preceders,
('one', 'once I', 'he'))
)
df['no_succeders'] = df.a.apply(
to_apply(remove_succeeders,
('fine', 'one', 'well'))
)
df['substring'] = df.no_preceders.apply(
to_apply(remove_succeeders,
('fine', 'one', 'well'))
)
And then there's one final step to remove the items from the substring column that were not affected by the filtering:
def final_cleanup(row):
if len(row['a']) == len(row['substring']):
return ''
else:
return row['substring']
df['substring'] = df.apply(final_cleanup, axis=1)
Results
Hope this works.

Why is requests is adding a part of a list to repeated query URIs?

I am trying to place a variable into a POST request using the Requests library. Here is my code:
import requests
message = "if i can\'t let it go out of my mind"
split_message = message.split()
initial_request = requests.get('http://ws.spotify.com/search/1/track?q='.join(split_message[:3]))
print initial_request.content
The outcome I am getting is this (which is an error, as it is messing up the URI):
No connection adapters were found for 'ifhttp://ws.spotify.com/search/1/track?q=ihttp://ws.spotify.com/search/1/track?q=can't'
I would like the request URI to look like this:
"http://ws.spotify.com/search/1/track?q=if i can\'t"
What am I missing here? Is there a better way to pass a variable to the request URI? I tried using a dictionary as the payload, but I can't place a variable into a dictionary.

String.join is very counter intuitive, the string is something you join with not join to.
Examine the following code:
>>> x = range(5)
>>> x
[0, 1, 2, 3, 4]
>>> x = [str(c) for c in x]
>>> x
['0', '1', '2', '3', '4']
>>> "-".join(x)
'0-1-2-3-4'
Here the "joiner" (-) is inserted between every element in our array, which we've converted to strings using list comprehension.
Here is your code:
'http://ws.spotify.com/search/1/track?q='.join(split_message[:3])
Your joining string is 'http://ws.spotify.com/search/1/track?q=' and so this is inserted between every element in the array split_message[:3].
So lets examine whats going on:
>>> message = "if i can\'t let it go out of my mind"
>>> split_message = message.split()
>>> split_message
['if', 'i', "can't", 'let', 'it', 'go', 'out', 'of', 'my', 'mind']
>>> split_message[:3]
['if', 'i', "can't"]
Here the array is 3 items long, which explains why the output string is:
ifhttp://ws.spotify.com/search/1/track?q=ihttp://ws.spotify.com/search/1/track?q=can't
But adding in some line breaks:
if
http://ws.spotify.com/search/1/track?q=
i
http://ws.spotify.com/search/1/track?q=
can't'
Notice the joining string is inserted twice, which explains why at first glance it just looks like its mucking up the URI.
What you want instead is:
request = 'http://ws.spotify.com/search/1/track?q='+"%20".join(split_message[:3])
Notice, in the above we join using %20: like so: "%20".join(split_message[:3]) and add this the request prefix. Which gives the URL below, with the spaces correctly encoded:
"http://ws.spotify.com/search/1/track?q=if%20i%20can't"

initial_request = requests.get('http://ws.spotify.com/search/1/track?q=' +
' '.join(split_message[:3]))
s.join(seq) sticks together the elements of seq with the string s stuck between them. For example, 'a'.join('bcd') == 'bacad'. Note that your desired string probably isn't what you really should be sending; you should probably use + instead of spaces. The requests module can handle url parameter encoding for you:
params = {'q': ' '.join(split_message[:3])}
initial_request = requests.get('http://ws.spotify.com/search/1/track',
params=params)