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Python: Keep line formatting (breaklines) while using re.substitute for template
How can I do a line break (line continuation) in Python?Emulate a do-while loop in Python?How to get line count cheaply in Python?Why do people write the #!/usr/bin/env python shebang on the first line of a Python script?What is the standard Python docstring format?String formatting: % vs. .formatHow can I print literal curly-brace characters in python string and also use .format on it?Why is reading lines from stdin much slower in C++ than Python?Pythonic way to create a long multi-line string“Large data” work flows using pandas
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I'm taking information about points from multiple files and create new files for each point based on template. The problematic file looks like this
...
Point number:"Number"
info
info
...
Point number:"Number"
info
info
info
info
...
etc
Number of line with information can vary.
I'm uing re.findall() to search for points like this
point_info = re.findall(r"(?ms)^P.*?$.*?^;$", file.read())
But when i try to put it in my template, i'm getting
['Point number:"Number" n info n info n info n info n ']
I use .substitute to insert info into template. Template looks like this
Template
...
...
...
$point_info
...
...
While input code
d = 'point_info ': point_info
result = src.substitute(d)
output_file.write(result) #opened beforehand
How can i get my output to look exactly like in main file, keeping in mind that amount of lines can differ?
(template has many other $variables, just not mentioned here)
python templates string-formatting
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
add a comment |
I'm taking information about points from multiple files and create new files for each point based on template. The problematic file looks like this
...
Point number:"Number"
info
info
...
Point number:"Number"
info
info
info
info
...
etc
Number of line with information can vary.
I'm uing re.findall() to search for points like this
point_info = re.findall(r"(?ms)^P.*?$.*?^;$", file.read())
But when i try to put it in my template, i'm getting
['Point number:"Number" n info n info n info n info n ']
I use .substitute to insert info into template. Template looks like this
Template
...
...
...
$point_info
...
...
While input code
d = 'point_info ': point_info
result = src.substitute(d)
output_file.write(result) #opened beforehand
How can i get my output to look exactly like in main file, keeping in mind that amount of lines can differ?
(template has many other $variables, just not mentioned here)
python templates string-formatting
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
add a comment |
I'm taking information about points from multiple files and create new files for each point based on template. The problematic file looks like this
...
Point number:"Number"
info
info
...
Point number:"Number"
info
info
info
info
...
etc
Number of line with information can vary.
I'm uing re.findall() to search for points like this
point_info = re.findall(r"(?ms)^P.*?$.*?^;$", file.read())
But when i try to put it in my template, i'm getting
['Point number:"Number" n info n info n info n info n ']
I use .substitute to insert info into template. Template looks like this
Template
...
...
...
$point_info
...
...
While input code
d = 'point_info ': point_info
result = src.substitute(d)
output_file.write(result) #opened beforehand
How can i get my output to look exactly like in main file, keeping in mind that amount of lines can differ?
(template has many other $variables, just not mentioned here)
python templates string-formatting
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
I'm taking information about points from multiple files and create new files for each point based on template. The problematic file looks like this
...
Point number:"Number"
info
info
...
Point number:"Number"
info
info
info
info
...
etc
Number of line with information can vary.
I'm uing re.findall() to search for points like this
point_info = re.findall(r"(?ms)^P.*?$.*?^;$", file.read())
But when i try to put it in my template, i'm getting
['Point number:"Number" n info n info n info n info n ']
I use .substitute to insert info into template. Template looks like this
Template
...
...
...
$point_info
...
...
While input code
d = 'point_info ': point_info
result = src.substitute(d)
output_file.write(result) #opened beforehand
How can i get my output to look exactly like in main file, keeping in mind that amount of lines can differ?
(template has many other $variables, just not mentioned here)
python templates string-formatting
python templates string-formatting
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
asked Mar 27 at 1:24
LegithLegith
61 bronze badge
61 bronze badge
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I managed to solve this by using .join() function. point_info =''.join(point_info)
gave me expected result
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I managed to solve this by using .join() function. point_info =''.join(point_info)
gave me expected result
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
add a comment |
I managed to solve this by using .join() function. point_info =''.join(point_info)
gave me expected result
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
add a comment |
I managed to solve this by using .join() function. point_info =''.join(point_info)
gave me expected result
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
I managed to solve this by using .join() function. point_info =''.join(point_info)
gave me expected result
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
answered Mar 27 at 16:29
LegithLegith
61 bronze badge
61 bronze badge
add a comment |
add a comment |
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