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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

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])

Improving a "Four Fours" puzzle solver [closed]

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import itertools
import math
import time
from time import time
from math import factorial
from math import sqrt
def pretty(string):
string=string.replace("(4)","4")
string=string.replace("factorial4","factorial(4)")
string=string.replace("sqrt4","sqrt(4)")
return string
def test(n):
start=time()
fails=0
for i in range(0,n+1):
if(fours(i))!=None:
print(fours(i))
else:
print("Failed: "+str(i))
fails+=1
return("\nFailed "+str(fails)+" out of "+str(n)+"\n\nTotal time: "+str(time()-start)[:4]+"\nAverage time: "+str((time()-start)/n)[:4])
def fours(goal):
operators = ['-','/','+','*','sqrt','^','factorial',"."]
brackets = ["{0}{1}{0}{2}{0}{3}{0}",
"({0}{1}{0}){2}{0}{3}{0}",
"({0}{1}{0}{2}{0}){3}{0}",
"({0}{1}({0}{2}{0})){3}{0}",
"(({0}{1}{0}){2}{0}){3}{0}",
"{0}{1}({0}{2}({0}{3}{0}))",
"{0}{1}(({0}{2}{0}){3}{0})",
"({0}{1}{0}){2}({0}{3}{0})"]
for combination in itertools.product(operators, repeat=3):
for bracket in brackets:
try:
formula = bracket.format("(4)", *combination).replace(".(4","(.4")
except ValueError:
pass
try:
if eval(formula)==goal:
return(pretty((formula + " = " + str(int(eval(formula))))))
except:
pass
print(test(20))
Here is the code for a solver for the "Four Fours"
puzzle.http://en.wikipedia.org/wiki/Four_fours
It basically work, but the problem is that it can only use the + - / * operators, because it can't have two operators in a row(Solutions like this (4-4)/4+factorial(4) aren't allowed because of this "+factorial" part).What I could do(but would be to slow) would be to make pairings like this.
['-', '/', '+', '*', '-sqrt', '-^', '-factorial', '-.', '/sqrt', '/^', '/factorial', '/.', '+sqrt', '+^', '+factorial', '+.', 'sqrt', '^', 'factorial', '.']
This is far too many operators.
What I would like to do, would be to try something like this
formula = bracket.format(["(4)","(4*)","(4-)","(4/)"], *combination).replace(".(4","(.4")
Except this syntax is invalid.
How can I do this?
Or if you have better ideas(which I'm sure some of you do) I am open to suggestions.

You could try converting it into pre or post-fix. You can allow it to use ['!','sqrt','.'] anytime. '^' would only be used if at least one four is still available to be pushed on the stack. And ['+','-','/','*'] would only be pushed if there are more operators than numbers (or something like that).

Return outside function error - help needed [closed]

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def getPressAve(odbname):
odb=openOdb(odbname)
lastFrame=odb.steps['Step-1'].frames[-1]
pressure=lastFrame.fieldOutputs['CPRESS']
press=[[0,0]] # sets the first element to [0,0]
for n in pressure.values:
gridPt=part1.nodes.getFromLabel(n.nodeLabel)
coord=assemb.getCoordinates(gridPt)
press.append([n.nodeLabel,n.data,coord])
press=avePress=press[1:] # removes the first element
press.sort(Comp_X)
print ('pressure extracted')
index=0
while index<len(press):
sum=0
tally=0
if index!=0:
sum=sum+press[index-1][1]
tally=tally+1
if index!=1:
sum=sum+press[index-2][1]
tally=tally+1
if index!=2:
sum=sum+press[index][1]
tally=tally+1
if index<len(press)-1:
sum=sum+press[index+1][1]
tally=tally+1
if index<len(press)-2:
sum=sum+press[index+2][1]
tally=tally+1
average=sum/tally
avePress[index][1]=average
index=index+1
odb.close()
print ('pressure averaged')
return avePress

In Python, indentation matters. As is, you're defining a function called getPressAve which does only this:
odb=openOdb(odbname)
After you've defined your function, you go on to do
lastFrame=odb.steps['Step-1'].frames[-1]
and such outside of the function. That's not what you want. The solution is to indent everything after that odb=openOdb(odbname) line to that level, so those lines are interpreted as being part of the body of the function.

You forgot to properly indent your code:
def getPressAve(odbname):
odb=openOdb(odbname)
...
print ('pressure averaged')
return avePress
As yours has not been, the return keyword is featured outside of a function, and hence the error: SyntaxError: 'return' outside function.