This question already has answers here:
String formatting: % vs. .format vs. f-string literal
(16 answers)
Closed 4 years ago.
I am using a Markov chain. When the chain arrives at a particular state, two files (a .png and an .mp3) need to open.
s is the current state of the chain, an integer from 1-59.
I can't seem to find how to open the file with the same number as 's'.
I'm sure it has something to do with %str formatting, but I can't seem to implement it.
img = Image.open('/.../.../s.png')
img.show()
You should use the following line in your code:
img = Image.open('/.../.../{0}.png'.format(s))
You can format a string using a variable like this
>>> s = 10
>>> '/path/to/file/{}.png'.format(s)
'/path/to/file/10.png'
Related
This question already has answers here:
Convert a Unicode string to a string in Python (containing extra symbols)
(12 answers)
Closed 3 years ago.
I need to convert word
name = 'Łódź'
to ASCII characters
output: 'Lodz'
I can't import any library like unicodedata.
I need to do it in clear python.
I've tried to encode than decode and nothing worked.
Well, a simple method would be to map and replace. This also does not require any special imports.
name = 'Łódź'
name=name.replace('Ł','L')
name=name.replace('ó','o')
name=name.replace('ź','z')
print(name)
This question already has answers here:
Regex match one of two words
(2 answers)
Closed 3 years ago.
I want to make a list of several PNG in a folder based on multiple references. So in the list I want the PNG that have the string "7029113" OR "7031503" in their name. This is what I got so far, I only need to know how to do OR with regex, and probably my wildcards are wrong too I'm not sure.
render_path = "C:/BatchRender/Renaming"
os.chdir(render_path)
list_files = glob.glob("*.png")
r = re.compile(".*7029113.*" OR ".*7031503.*")
list_40 = list(filter(r.match, list_files))
This is one way of doing it.
r = re.compile('.*(7029113|7031503).*')
This question already has answers here:
Extract file name from path, no matter what the os/path format
(22 answers)
Extract file name with a regular expression
(3 answers)
Get Filename Without Extension in Python
(6 answers)
Extracting extension from filename in Python
(33 answers)
Regex: Get Filename Without Extension in One Shot?
(10 answers)
Closed 3 years ago.
I am new in the regular expression and trying to extract the file name from a string which is basically a file path.
string = "input_new/survey/argentina-attributes.csv"
string_required = argentina-attributes
i know i can do this by below code.
string.split('/')[2].split('.')[0]
but I am looking to do this by using regular expression so if in future the formate of the path changes(input_new/survey/path/path/argentina-attributes.csv) should not affect the output.
i know kind of similar question asked before but I am looking for a pattern which will work for my use case.
Try this,
>>> import re
>>> string = "input_new/survey/argentina-attributes.csv"
Output:
>>> re.findall(r'[^\/]+(?=\.)',string) # or re.findall(r'([^\/]+)\.',string)
['argentina-attributes']
Referred from here
Try this:
string = "input_new/survey/argentina-attributes.csv"
new_string = string.split('/')[-1].split('.')[0]
print(new_string)
This question already has answers here:
How to use a variable inside a regular expression?
(12 answers)
Closed 5 years ago.
I want to eventually be able to increment an integer within my Regex, but the braces are preventing me from doing so.
So far I have:
start = 6
m = re.search(r"(.{{n},}).*?\1".format(n=start), s)
return m.group(1)
However, I get `ValueError: Single '}' encountered in format string
I am using Python 2.7.
What about using a different method of string formatting:
m = re.search(r"({.%s,}).*?\1" % start, s)
This question already has an answer here:
Python 3.4.1 script syntax error, arcpy & [duplicate]
(1 answer)
Closed 7 years ago.
I am very new to Python and am getting this small problem. I am using Python 3.3
There is a variable I declared in my code
file_name = "resource\email\ham\6.txt"
However, when I look for the variable, it appends additional numbers
>>file_name
'resource\\email\\ham\x06.txt'
Is there a reason why it behaves as so? If not, how do I remove those additional characters? Also, why are they there?
Use r raw string:
file_name = r"resource\email\ham\6.txt"
Double \:
file_name = "resource\\email\\ham\\6.txt"
Or /:
file_name = "resource/email/ham/6.txt"
\ has a special meaning in python, it is used to escape characters.