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python string find and look up on string until found space from json [closed]

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Below is my data.
'{"Data": {"a":5647953897,"b":"323299901059958183671030","c":1605858513465}}{"Data": {"a":5647953897,"b":"323299901059958183671030","c":1605858513465}}'
My output should be as below
[{"a":5647953897,"b":"323299901059958183671030","c":1605858513465},{"a":5647953897,"b":"323299901059958183671030","c":1605858513465}]

This should solve your case.
from json import JSONDecoder, JSONDecodeError
import re
NOT_WHITESPACE = re.compile(r'[^\s]')
data = '''{"Data": {"a":5647953897,"b":"323299901059958183671030","c":1605858513465}}{"Data": {"a":5647953897,"b":"323299901059958183671030","c":1605858513465}}'''
def decode_stacked(document, pos=0, decoder=JSONDecoder()):
while True:
match = NOT_WHITESPACE.search(document, pos)
if not match:
return
pos = match.start()
try:
obj, pos = decoder.raw_decode(document, pos)
except JSONDecodeError:
raise
yield obj
for obj in decode_stacked(data):
print(obj)

Create email with the firsts letters of the name [closed]

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Is there any magic python way to get the firsts letters of each name and join with '#company.com' to create a email address?
Here is how I did:
name = ['Elon Reeve Musk']
full_name = [word[0].lower() for word in name[0].split(' ')]
firts_letters = "".join(full_name)
username = '%s#company.com' %(firts_letters)
The result is erm#company.com

Here's some "python magic"
name = ['Elon Reeve Musk']
f"{''.join(filter(str.isupper, name[0]))}#company.com".lower()
>>> erm#company.com
Whether this is better than your method is debatable. Most of the time a few lines of easy to read code is better than a one line hack.
My recommendation would be
name = ['Elon Reeve Musk']
initials = ''.join(word[0] for word in name[0].split())
f'{initials.lower()}#company.com'
>>> erm#company.com

Very simple python cycle string [closed]

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I want to write in a file which changes the name with the counter.
for j in range(30):
fileout.write("bla bla")
and fileout should be
file001 if j=1
file002 if j=2
...
...
I cannot format it in the right way.
Anyone can help me?
Thank you

You can try to use the .zfill( < number of digits > ) option. It will complet your string with 0.
For example, this code should work :
radical ='file'
numberOfDigits = 3
for j in range( 30 ) :
fileout = open ( radical + '%s' %(str(j).zfill(numberOfDigits)) + '.txt' , 'w' )
fileout.write('bla bla')
fileout.close()

Not sure if below means what you needed ?
Code below will create file000 - file029 with 'bla bla' inside
for i in range(30):
with open("file%03d" % i, 'w') as f:
f.write('bla bla')

Create correct loop output with python function [closed]

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Could someone help me with following function:
def nagios_chart():
alpha = [chr(item).upper() for item in range(ord('b'), ord('l')+1)]
for idx, column in enumerate(alpha):
print worksheet.write_column('column + 1', nagios_data[idx])
if __name__ == '__main__':
nagios_chart()
I need output like this:
worksheet.write_column('B1', nagios_data[0])
worksheet.write_column('C1', nagios_data[1])
worksheet.write_column('D1', nagios_data[2])
worksheet.write_column('E1', nagios_data[3])
worksheet.write_column('F1', nagios_data[4])
worksheet.write_column('G1', nagios_data[5])
worksheet.write_column('H1', nagios_data[6])
worksheet.write_column('I1', nagios_data[7])
worksheet.write_column('J1', nagios_data[8])
worksheet.write_column('K1', nagios_data[9])
worksheet.write_column('L1', nagios_data[10])

To use the column as a variable, remove it from the quotes so it isn't treated as a string literal. Then you can concatenate it with '1'.
def nagios_chart():
alpha = [chr(item).upper() for item in range(ord('b'), ord('l')+1)]
for idx, column in enumerate(alpha):
print worksheet.write_column(column + '1', nagios_data[idx])

How to format this line according to PEP 8? [closed]

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I'm trying to adhere to PEP 8, with a 78 character limit on the length of my lines.
I have the following statement:
startTime = time.strptime(request.GET.get('st', (dt.datetime.now() - dt.timedelta(days=1)).strftime("%d/%b/%Y:%H:%M:%S")), "%d/%b/%Y:%H:%M:%S")
How should I format it so that it adheres to PEP8 (where should I break it into new lines?)

Avoid writing such overly convoluted code in the first place:
if 'st' in request.GET:
startTime = time.strptime(request.GET['st'], "%d/%b/%Y:%H:%M:%S")
else:
startTime = (dt.datetime.now() - dt.timedelta(days=1)).timetuple()
Why generate a default to be parsed out to a timetuple again instead of just going there straight?

startTime = time.strptime(
request.GET.get(
'st',
(
dt.datetime.now() - dt.timedelta(days=1)
).strftime("%d/%b/%Y:%H:%M:%S")
),
"%d/%b/%Y:%H:%M:%S"
)
This one solution, but you use more variables for this. For example:
time_format = "%d/%b/%Y:%H:%M:%S"
yesterday_date = dt.datetime.now() - dt.timedelta(days=1
This would make the code more readable

Split it up into different variables:
a_date = (dt.datetime.now() - dt.timedelta(days=1)).strftime("%d/%b/%Y:%H:%M:%S")
req = request.GET.get('st', a_date)
startTime = time.strptime(req, "%d/%b/%Y:%H:%M:%S")