How can I automatically extract part of a string that has a .csv extension. The following example shows the complex string that I am trying to extract 2010_USACE_VA_minmax.csv from. A simple slice won't work in my case, instead I need some sort of pattern matching.
sample = "1001 15707 May 08 23:01 2010_USACE_VA_metadata.xml\r\n-rw-rw-r-- 1 311 1001 1784 May 08 23:01 2010_USACE_VA_minmax.csv\r\ndrwxrwxr-x 2 311 2013"
Intended output
2010_USACE_VA_minmax.csv
If you know these are white-space separated and the names themselves do not contain any white space themselves, and you're trying to find a token that ends with .csv, you could also do
>>> tokens = sample.split()
>>> matches = [ i for i in tokens if i.endswith('.csv') ]
>>> matches
['2010_USACE_VA_minmax.csv']
The same behaviour is achievable with the regular expression \S+\.csv(?!\S), which is not quite so readable:
>>> import re
>>> re.findall(r'\S+\.csv(?!\S)', sample)
['2010_USACE_VA_minmax.csv']
Here \S+ means at least 1 consecutive non-whitespace characters, \. is the literal . character, and (?!\S) means that the .csv cannot be succeeded by a non-whitespace character (negative zero-width lookahead assertion).
However, it looks like you're parsing the output of the ls *nix command - yet another way would be to find matching files with the glob module:
>>> from glob import glob
>>> glob('*.csv')
['2010_USACE_VA_minmax.csv']
ls then the proper way to do it is not to parse it at all, but to use glob.ls *.csv then no parsing needed at allls by itself!This regex extracted the csv file. There might be a more robust regex, I'm not perfect at it. But this works:
FYI: I used this to test: Pythex
The circle brackets are important as they are your capture group to extract what you are looking for.
(\s\w+\.csv)
If you want to handle spaces in the filename, I believe this should work:
(\s[\w,\s-]+\.csv)
Here is infrmation on regex in Python: https://docs.python.org/3/library/re.html
If there were no spaces in the path:
print(sample[:sample.find(".csv")+4].rsplit(None, 1)[1])
2010_USACE_VA_minmax.csv
The output also looks like it comes from a unix command so might be an idea to use a linux tool to parse it, if it is a unix command the format is most probably consistent so you can split the lines to get the filenames:
sample = "1001 15707 May 08 23:01 2010_USACE_VA_metadata.xml\r\n-rw-rw-r-- 1 311 1001 1784 May 08 23:01 2010_USACE_VA_minmax.csv\r\ndrwxrwxr-x 2 311 2013"
for line in sample.splitlines():
f = line.rsplit(None, 1)[1]
print(f)
2010_USACE_VA_metadata.xml
2010_USACE_VA_minmax.csv
2013
I presume 2013 comes from you having truncated some of the output.
If you are using subprocess to run the command and you didn't need any of the other data, ls can take a wildcard:
from subprocess import check_output
f = check_output(["ls","*.csv"])
Or to get the permissions etc.. as per you own command:
data = check_output(["ls","-l","*.csv"])
That will give you just the .csv files and their permissions so you just need to iterate over the output again with splitlines and every file at the end will be a csv file.
import re
mobj = re.search(r'\s\d{4}_[^ ]*csv',sample)
print(mobj.group())
2010_USACE_VA_minmax.csv
