I am trying to use pandas.read_csv to read files that contain the date in their names. I used the below code to do the job. The problem is that the files name is not consistent as the number of date change the pattern. I was wondering if there is a way to let the code read the file with parts of the name is the date in front of the file name?
for x in range(0,10):
dat = 20170401+x
dat2 = dat+15
file_name='JS_ALL_V.'+str(dat)+'_'+str(dat2)+'.csvp.gzip'
df = pd.read_csv(file_name,compression='gzip',delimiter='|')
You can use glob library to read file names in unix style
Below is its hello world:
import glob
for name in glob.glob('dir/*'):
print name
An alternative of using glob.glob() (since it seems not working) is os.listdir() as explained in this question in order to have a list containing all the elements (or just the files) in your path.
I'm fairly new to python. I'd like to save the text that is printed by at this script as a variable. (The variable is meant to be written to a file later, if that matters.) How can I do that?
import fnmatch
import os
for file in os.listdir("/Users/x/y"):
if fnmatch.fnmatch(file, '*.txt'):
print(file)
you can store it in variable like this:
import fnmatch
import os
for file in os.listdir("/Users/x/y"):
if fnmatch.fnmatch(file, '*.txt'):
print(file)
my_var = file
# do your stuff
or you can store it in list for later use:
import fnmatch
import os
my_match = []
for file in os.listdir("/Users/x/y"):
if fnmatch.fnmatch(file, '*.txt'):
print(file)
my_match.append(file) # append insert the value at end of list
# do stuff with my_match list
You can store it in a list:
import fnmatch
import os
matches = []
for file in os.listdir("/Users/x/y"):
if fnmatch.fnmatch(file, '*.txt'):
matches.append(file)
Both answers already provided are correct, but Python provides a nice alternative. Since iterating through an array and appending to a list is such a common pattern, the list comprehension was created as a one-stop shop for the process.
import fnmatch
import os
matches = [filename for filename in os.listdir("/Users/x/y") if fnmatch.fnmatch(filename, "*.txt")]
While NSU's answer and the others are all perfectly good, there may be a simpler way to get what you want.
Just as fnmatch tests whether a certain file matches a shell-style wildcard, glob lists all files matching a shell-style wildcard. In fact:
This is done by using the os.listdir() and fnmatch.fnmatch() functions in concert…
So, you can do this:
import glob
matches = glob.glob("/Users/x/y/*.txt")
But notice that in this case, you're going to get full pathnames like '/Users/x/y/spam.txt' rather than just 'spam.txt', which may not be what you want. Often, it's easier to keep the full pathnames around and os.path.basename them when you want to display them, than to keep just the base names around and os.path.join them when you want to open them… but "often" isn't "always".
Also notice that I had to manually paste the "/Users/x/y/" and "*.txt" together into a single string, the way you would at the command line. That's fine here, but if, say, the first one came from a variable, rather than hardcoded into the source, you'd have to use os.path.join(basepath, "*.txt"), which isn't quite as nice.
By the way, if you're using Python 3.4 or later, you can get the same thing out of the higher-level pathlib library:
import pathlib
matches = list(pathlib.Path("/Users/x/y/").glob("*.txt"))
Maybe defining an utility function is the right path to follow...
def list_ext_in_dir(e,d):
"""e=extension, d= directory => list of matching filenames.
If the directory d cannot be listed returns None."""
from fnmatch import fnmatch
from os import listdir
try:
dirlist = os.listdir(d)
except OSError:
return None
return [fname for fname in dirlist if fnmatch(fname,e)]
I have put the dirlist inside a try except clause to catch the
possibility that we cannot list the directory (non-existent, read
permission, etc). The treatment of errors is a bit simplistic, but...
the list of matching filenames is built using a so called list comprehension, that is something that you should investigate as soon as possible if you're going to use python for your programs.
To close my post, an usage example
l_txtfiles = list_ext_in_dir('*.txt','/Users/x/y;)
Say that I have this string "D:\Users\Zache\Downloads\example.obj" and I want to copy another file to the same directory as example.obj. How do I do this in a way that´s not hardcoded?
"example" can also be something else (user input). I'm using filedialog2 to get the big string.
This is for an exporter with a basic GUI.
os.path.dirname() gives you the directory portion of a given filename:
>>> import os.path
>>> os.path.dirname(r"D:\Users\Zache\Downloads\example.obj")
'D:\\Users\\Zache\\Downloads'
You can solve it with str.split but this should be solved with os.path.split
I am trying to execute f = open('filename') in python.
However, I dont know the full name of the file. All I know is that it starts with 's12' and ends with '.ka',I know the folder where it's located, and I know it is the only file in that folder that starts and ends with "s12" and ".ka". Is there a way to do this?
Glob is your friend:
from glob import glob
filename = glob('s12*.ka')[0]
Careful though, glob returns a list of all files matching this pattern so you might want to assert that you get the file you actually want somehow.
I am trying to write a small python script to rename a bunch of filenames by searching and replacing. For example:
Original filename:
MyMusic.Songname.Artist-mp3.iTunes.mp3
Intendet Result:
Songname.Artist.mp3
what i've got so far is:
#!/usr/bin/env python
from os import rename, listdir
mustgo = "MyMusic."
filenames = listdir('.')
for fname in fnames:
if fname.startswith(mustgo):
rename(fname, fname.replace(mustgo, '', 1))
(got it from this site as far as i can remember)
Anyway, this will only get rid of the String at the beginning, but not of those in the filename.
Also I would like to maybe use a seperate file (eg badwords.txt) containing all the strings that should be searched for and replaced, so that i can update them without having to edit the whole code.
Content of badwords.txt
MyMusic.
-mp3
-MP3
.iTunes
.itunes
I have been searching for quite some time now but havent found anything. Would appreciate any help!
Thank you!
import fnmatch
import re
import os
with open('badwords.txt','r') as f:
pat='|'.join(fnmatch.translate(badword)[:-1] for badword in
f.read().splitlines())
for fname in os.listdir('.'):
new_fname=re.sub(pat,'',fname)
if fname != new_fname:
print('{o} --> {n}'.format(o=fname,n=new_fname))
os.rename(fname, new_fname)
# MyMusic.Songname.Artist-mp3.iTunes.mp3 --> Songname.Artist.mp3
Note that it is possible for some files to be overwritten (and thus
lost) if two names get reduced to the same shortened name after
badwords have been removed. A set of new fnames could be kept and
checked before calling os.rename to prevent losing data through
name collisions.
fnmatch.translate takes shell-style patterns and returns the
equivalent regular expression. It is used above to convert badwords
(e.g. '.iTunes') into regular expressions (e.g. r'\.iTunes').
Your badwords list seems to indicate you want to ignore case. You
could ignore case by adding '(?i)' to the beginning of pat:
with open('badwords.txt','r') as f:
pat='(?i)'+'|'.join(fnmatch.translate(badword)[:-1] for badword in
f.read().splitlines())