I have this list and would like for all of the timestamps to have the same format (... = more elements):
timestampList = [...
"8:36 - Appointment1",
"9:21 - Appointment2",
"10:01 - Appointment3",
"11:52 - Appointment4",
"12:18 - Appointment5" ...]
Is there an easy way to make sure all timestamps in the list have the same format(HH:MM)? Is there perhaps a module that makes this possible? I have tried to resolve the problem but couldn't find a way of doing it. I want the list to look like this:
timestampList = [...
"08:36 - Appointment1",
"09:21 - Appointment2",
"10:01 - Appointment3",
"11:52 - Appointment4",
"12:18 - Appointment5" ...]
You can use re.sub and a lookahead regex from the beginning of the line. If we see that the timestamp starts with \d:, then prepend a "0":
>>> import re
>>> [re.sub(r"^(?=\d:)", "0", x) for x in timestamps]
['08:36 - Appointment1', '09:21 - Appointment2', '10:01 - Appointment3', '11:52 - Appointment4', '12:18 - Appointment5']
Related
As I SSH into CM, run commands and start reading the CLI output, I get the following
back:
# * A lot more output above but been removed *
terminal_output = """
[24;1H [79b[1GCommand: disp sys cust<<[23;0H[0;7m [79b[1G[0m[24;0H [79b[1G[1;0H[0;7m [79b[1G[0m[2;0H [79b[1G[3;1H[0J7[1;1H[0;7mdisplay system-parameters customer-options [0m8[1;65H[0;7mPage 1 of 12[0m[2;33HOPTIONAL FEATURES[4;8HG3 Version: [4;20HV20 [4;50HSoftware Package: [4;68HEnterprise [5;10HLocation: [5;20H2[6;10HPlatform: [6;20H28 [5;51HSystem ID (SID): [5;68H9990093751 [6;51HModule ID (MID): [6;68H1 [8;60HUSED[9;29HPlatform Maximum Ports: [9;53H 81000[9;60H 436[10;35HMaximum Stations: [10;53H 135[10;60H 110[11;27HMaximum XMOBILE Stations: [11;53H 41000[11;60H 0[12;17HMaximum Off-PBX Telephones - EC500: [12;53H 135[12;60H 2[13;17HMaximum Off-PBX Telephones - OPS: [13;53H 135[13;60H 40[14;17HMaximum Off-PBX Telephones - PBFMC: [14;53H 135[14;60H 0[15;17HMaximum Off-PBX Telephones - PVFMC: [15;53H 135[15;60H 0[16;17HMaximum Off-PBX Telephones - SCCAN: [16;53H 0[16;60H 0[17;22HMaximum Survivable Processors: [17;53H 313[17;62H 1[22;9H(NOTE: You must logoff & login to effect the permission changes.)[2;50H[0m
"""
It's a lot of ANSI escape codes (I think?) which sort of makes the output not too readable but anyways, what I'm trying to get back is the following from the text above:
Maximum Stations: 135 110
I know from my understanding that a Regex would be required for this.
The Regexes that I tried using but did not work:
r'Maximum Stations:\s*(\d+)(\d+)'
r'Maximum Stations: \d+'
If anyone knows how to filter out these ANSI character codes so they don't appear in the final output that'd be great too.
Thank you.
you can try the following
"(Maximum Stations:)\s\[\d*;\d*H\s*(\d*)\[\d*;\d*H\s*(\d*)"gm
it produces three groups the first with the maximum stations text then two more each with the number you wanted to capture. You would have to combine the groups to get your final output.
I don't know if this will be generic enough for your application though.
I have string like so:
"Job 1233:name_uuid (table n_Cars_1234567$20220316) done. Records: 24, with errors: 0."
I'd like to retieve the datte from the table name, so far I use:
"\$[0-9]+"
but this yields $20220316. How do I get only the date, without $?
I'd also like to get the table name: n_Cars_12345678$20220316
So far I have this:
pattern_table_info = "\(([^\)]+)\)"
pattern_table_name = "(?<=table ).*"
table_info = re.search(pattern_table_info, message).group(1)
table = re.search(pattern_table_name, table_info).group(0)
However I'd like to have a more simpler solution, how can I improve this?
EDIT:
Actually the table name should be:
n_Cars_12345678
So everything before the "$" sign and after "table"...how can this part of the string be retrieved?
You can use a regex with two capturing groups:
table\s+([^()]*)\$([0-9]+)
See the regex demo. Details:
table - a word
\s+ - one or more whitespaces
([^()]*) - Group 1: zero or more chars other than ( and )
\$ - a $ char
([0-9]+) - Group 2: one or more digits.
See the Python demo:
import re
text = "Job 1233:name_uuid (table n_Cars_1234567$20220316) done. Records: 24, with errors: 0."
rx = r"table\s+([^()]*)\$([0-9]+)"
m = re.search(rx, text)
if m:
print(m.group(1))
print(m.group(2))
Output:
n_Cars_1234567
20220316
You can write a single pattern with 2 capture groups:
\(table (\w+\$(\d+))\)
The pattern matches:
\(table
( Capture group 1
\w+\$ match 1+ word characters and $
(\d+) Capture group 2, match 1+ digits
) Close group 1
\) Match )
See a Regex demo and a Python demo.
import re
s = "Job 1233:name_uuid (table n_Cars_1234567$20220316) done. Records: 24, with errors: 0."
m = re.search(r"\(table (\w+\$(\d+))\)", s)
if m:
print(m.group(1))
print(m.group(2))
Output
n_Cars_1234567$20220316
20220316
Thanks in advance for the help. I am new to kdb/q, coming from a Python and C++ background.
Just a simple syntax question: I have a string with fields and their corresponding values
pp_str: "field_1:abc field_2:xyz field_3:kdb"
I wrote an atomic (scalar) function to extract the value of a given field.
get_field_value: {[field; pp_str] pp_fields: " " vs pp_str; pid_field: pp_fields[where like[pp_fields; field,":*"]]; start_i: (pid_field[0] ss ":")[0] + 1; end_i: count pid_field[0]; indices: start_i + til (end_i - start_i); pid_field[0][indices]}
show get_field_value["field_1"; pp_str]
"abc"
show get_field_value["field_3"; pp_str]
"kdb"
Now how do I generalize this so that if I input a vector of fields, I get a vector of values? I want to input ("field_1"; "field_2"; "field_3") and output ("abc"; "xyz"; "kdb"). I tried multiple approaches (below) but I just don't understand kdb/q's syntax well enough to vectorize my function:
/ Attempt 1 - Fail
get_field_value[enlist ("field_1"; "field_2"); pp_str]
/ Attempt 2 - Fail
get_field_value[; pp_str] /. enlist ("field_1"; "field_3")
/ Attempt 3 - Fail
fields: ("field_1"; "field_2")
get_field_value[fields; pp_str]
To run your function for each you could project the pp_str variable and use each for the others
q)get_field_value[;pp_str]each("field_1";"field_3")
"abc"
"kdb"
Kdb actually has built-in functionality to handle this: https://code.kx.com/q/ref/file-text/#key-value-pairs
q){#[;x](!/)"S: "0:y}[`field_1;pp_str]
"abc"
q)
q){#[;x](!/)"S: "0:y}[`field_1`field_3;pp_str]
"abc"
"kdb"
I think this might be the syntax you're looking for.
q)get_field_value[; pp_str]each("field_1";"field_2")
"abc"
"xyz"
I want to rename all files in a directory by tkinter listbox.
Got stuck at this point:
files_list = os.listdir(root.foldername)
print(files_list)
gives me
['1.mp4', '10.mp4', '2.mp4', '3.mp4', '4.mp4', '5.mp4', '6.mp4', '7.mp4', '8.mp4', '9.mp4']
values = [listbox.get(idx) for idx in listbox.curselection()]<br>
And
inlist = (', '.join(values))<br>
print(inlist)
gives me
Lost - 1x01 - Pilot(1), Lost - 1x02 - Pilot(2), Lost - 1x03 - Tabula Rasa, Lost - 1x04 - Walkabout, Lost - 1x05 - White Rabbit, Lost - 1x06 - House Of The Rising Sun, Lost - 1x07 - The Moth, Lost - 1x08 - Confidence Man, Lost - 1x09 - Solitary, Lost - 1x10 - Raised By Another
Now I'm looking for a solution to use os.rename in order to rename the files 1.mp4 till 10.mp4.
Additionally Python for whatever reason does not come with a built-in way to have natural sorting, so it sorts 1.mp4 followed by 10.mp4.
Thank you very much in advance.
For natural sorting take a look at Sorting alphanumeric strings in Python.
Then loop through all files and rename them, eg.
for i in range(len(files_list)):
old_file_name = files_list[i]
new_file_name = values[i] + '.mp4'
os.rename(old_file_name, new_file_name)
For assistance in dealing with pathnames see os.path.
I have a list of strings in python like this
['AM_B0_D0.0_2016-04-01T010000.flac.h5',
'AM_B0_D3.7_2016-04-13T215000.flac.h5',
'AM_B0_D10.3_2017-03-17T110000.flac.h5',
'AM_B0_D0.7_2016-10-21T104000.flac.h5',
'AM_B0_D4.4_2016-08-05T151000.flac.h5',
'AM_B0_D0.0_2016-04-01T010000.flac.h5',
'AM_B0_D3.7_2016-04-13T215000.flac.h5',
'AM_B0_D10.3_2017-03-17T110000.flac.h5',
'AM_B0_D0.7_2016-10-21T104000.flac.h5',
'AM_B0_D4.4_2016-08-05T151000.flac.h5']
I want to parse only the date and time (for example, 2016-08-05 15:10:00 )from these strings.
So far I used a for loop like the one below but it's very time consuming, is there a better way to do this?
for files in glob.glob("AM_B0_*.flac.h5"):
if files[11]=='_':
year=files[12:16]
month=files[17:19]
day= files[20:22]
hour=files[23:25]
minute=files[25:27]
second=files[27:29]
tindex=pd.date_range(start= '%d-%02d-%02d %02d:%02d:%02d' %(int(year),int(month), int(day), int(hour), int(minute), int(second)), periods=60, freq='10S')
else:
year=files[11:15]
month=files[16:18]
day= files[19:21]
hour=files[22:24]
minute=files[24:26]
second=files[26:28]
tindex=pd.date_range(start= '%d-%02d-%02d %02d:%02d:%02d' %(int(year), int(month), int(day), int(hour), int(minute), int(second)), periods=60, freq='10S')
Try this (based on the 2nd last '-', no need of if-else case):
filesall = ['AM_B0_D0.0_2016-04-01T010000.flac.h5',
'AM_B0_D3.7_2016-04-13T215000.flac.h5',
'AM_B0_D10.3_2017-03-17T110000.flac.h5',
'AM_B0_D0.7_2016-10-21T104000.flac.h5',
'AM_B0_D4.4_2016-08-05T151000.flac.h5',
'AM_B0_D0.0_2016-04-01T010000.flac.h5',
'AM_B0_D3.7_2016-04-13T215000.flac.h5',
'AM_B0_D10.3_2017-03-17T110000.flac.h5',
'AM_B0_D0.7_2016-10-21T104000.flac.h5',
'AM_B0_D4.4_2016-08-05T151000.flac.h5']
def find_second_last(text, pattern):
return text.rfind(pattern, 0, text.rfind(pattern))
for files in filesall:
start = find_second_last(files,'-') - 4 # from yyyy- part
timepart = (files[start:start+17]).replace("T"," ")
#insert 2 ':'s
timepart = timepart[:13] + ':' + timepart[13:15] + ':' +timepart[15:]
# print(timepart)
tindex=pd.date_range(start= timepart, periods=60, freq='10S')
In Place of using file[11] as hard coded go for last or 2nd last index of _ then use your code then you don't have to write 2 times same code. Or use regex to parse the string.