This question already has answers here:
Extract file name from path, no matter what the os/path format
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Extract file name with a regular expression
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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)
Related
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:
How to get only the last part of a path in Python?
(10 answers)
Closed 4 years ago.
I want to get the substring from a path from the end to a certain character, take for example the following path:
my_path = "/home/Desktop/file.txt"
My intention is to do something like:
my_path.substring(end,"/")
So I can get the name of the file that is located between the end of the string and the character "/", in this case "file.txt"
The easiest approach, IMHO, would be to split the string:
filename = my_path.split('/')[-1]
use the os.path.basename for this
In [1]: import os
In [2]: os.path.basename('/home/Desktop/file.txt')
Out[2]: 'file.txt'
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'
This question already has answers here:
What's the u prefix in a Python string?
(5 answers)
Closed 6 years ago.
I am trying to parse the 'Meghan' part from the line:
link = http://python-data.dr-chuck.net/known_by_Meghan.html
...with the following regex:
print re.findall('by_(\S+).html$',link)
I am getting the output:
[u'Meghan']
Why I am getting the 'u'?
It means unicode. Depending on what you'll do with it, you can ignore it for the most part, of you can convert it to ascii by doing .encode('ascii')
This question already has answers here:
What is the difference between re.search and re.match?
(9 answers)
Closed 7 years ago.
I am trying to match string with mypattern, somehow I do not get correct result. Can you please point where am I wrong?
import re
mypattern = '_U_[R|S]_data.csv'
string = 'X003_U_R_data.csv'
re.match(mypattern, string)
I like to compile the regex statement first. Then I do whatever kind of matching/searching I would like.
mypattern = re.compile(ur'_U_[R|S]_data.csv')
Then
re.search(mypattern, string)
Here's a great website for regex creation- https://regex101.com/#python