Python, open a file when the name is not fully known - python

I have a list of files with names such as these:
20140911_085234.csv
20140912_040056.csv
What is known is the first part which is the date (the second is a random number). How can I open the correct file if I know the date?
Update: There is one file per day.

As #isedev says, you could use the fnmatch method to find all the files with the "date" pattern. The code could be like this:
from fnmatch import fnmatch
import os
folder_path = '/home/Desktop/project'
all_files = os.listdir(folder_path)
content_file = 'Hello World'
_date = '20140911'
_pattern = _date + '*'
for file_name in all_files:
if fnmatch(file_name, _pattern):
with open(os.path.join(folder_path, file_name), 'wb') as f:
f.write(content_file)
I hope it helps you!

Using glob :
import time
import glob
import os
def open_file_by_date(date):
path = "/path/to/file"
files = glob.glob1(path, date + "_*.csv")
for file in files:
with open(os.path.join(path, file), 'wb') as f:
#do your stuff with file
if __name__ == "__main__":
today = time.strftime("%Y%m%d")
open_file_by_date(today)

Related

Python Validate Filename is of specified format or not with date part

I am trying to validate the filenames which are having datepart in their name. What i am trying to do is i want to check a filename with specified format where datpart is different for different file. If the filename doesn't match it should give us the message file not found.
Filename e.g =abc_sales_2020-09-01_exp.csv, abc_sales_2020-09-02_exp.csv,abc_sales_2020-09-03_exp.csv. Only the datepart changes rest remains the same.
from datetime import date
def get_filename_datetime():
return "ak_sales_" + str(date.today()) + "_abc"+".csv"
name = get_filename_datetime() print("NAME", name) path = "aks/" +
name print("PATH", path);
with open(path, "r") as f:
f.read()
You can use regex to filter specific files and then open them with pandas:
import re
import glob
import pandas as pd
for file in glob.glob('*.csv'):
r = re.findall('\d\d\d\d-\d\d-\d\d', file)
if r:
df = pd.read_csv(file)

Python compare filename with folder name

I just start python and i have to compare filename with folder name to launch the good sh script. (i'm using airflow)
import glob
import os
import shutil
from os import path
odsPath = '/apps/data/02_ODS/'
receiptPath = '/apps/data/80_DATA/01_Receipt/'
for files in os.listdir(receiptPath):
if(files.startswith('MEM_ZMII') or files.startswith('FMS') and files.endswith('.csv')):
parsedFiles = files.split('_')
pattern = '_'.join(parsedFiles[0:2])
fileName = '_'.join(parsedFiles[2:5])
fileName = fileName.split('-')[0].lower()
# print('appCode: ', pattern)
# print('fileName: ', fileName)
for odsFolder in os.listdir(odsPath):
if(odsFolder == fileName):
print('it exist: ', str(fileName))
else:
print('it\'s not')
I got 3 files in receiptPath , it only matching for 1 file, but not the others. Can someone help me?
Thank a lot!
Ok, your problem is that you overwrite your variable fileName, so at the end of the first for loop, it only keeps the last value, which is material_makt. The solution consists in saving all the filenames in a list fileNames_list, and then you can check if (odsFolder in fileNames_list) :
import glob
import os
import shutil
from os import path
odsPath = '/apps/data/02_ODS/'
receiptPath = '/apps/data/80_DATA/01_Receipt/'
fileNames_list = []
for files in os.listdir(receiptPath):
if(files.startswith('MEM_ZMII') or files.startswith('FMS') and files.endswith('.csv')):
parsedFiles = files.split('_')
pattern = '_'.join(parsedFiles[0:2])
fileName = '_'.join(parsedFiles[2:5])
fileName = fileName.split('-')[0].lower()
fileNames_list.append(fileName)
for odsFolder in os.listdir(odsPath):
if (odsFolder in fileNames_list):
print('it exist:', str(odsFolder))
else:
print('it\'s not')
Output :
it exist: zcormm_familymc
it exist: kpi_obj_data
it exist: material_makt

Change filename for multiple files

I want to change filename for all my files in a folder. They all end with a date and time like "filename 2019-05-20 1357" and I want the date first for all files. How can I do that simplest way?
#!/usr/bin/python3
import shutil, os, re
r = re.compile(r"^(.*) (\d{4}-\d{2}-\d{2} \d{4})$")
for f in os.listdir():
m = r.match(f)
if m:
shutil.move(f, "{} {}".format(m.group(2), m.group(1)))
Quick and roughly tested version
Here is my Implementation:
from datetime import datetime
import os
path = '/Users/name/desktop/directory'
for _, file in enumerate(os.listdir(path)):
os.rename(os.path.join(path, file), os.path.join(path, str(datetime.now().strftime("%d-%m-%Y %H%M"))+str(file)))
Output Format:
20-05-2019 1749filename.ext
import os
import re
import shutil
dir_path = '' # give the dir name
comp = re.compile(r'\d{4}-\d{2}-\d{2}')
for file in os.listdir(dir_path):
if '.' in file:
index = [i for i, v in enumerate(file,0) if v=='.'][-1]
name = file[:index]
ext = file[index+1:]
else:
ext=''
name = file
data = comp.findall(name)
if len(data)!=0:
date= comp.findall(name)[0]
rest_name = ' '.join(comp.split(name)).strip()
new_name = '{} {}{}'.format(date,rest_name,'.'+ext)
print('changing {} to {}'.format(name, new_name))
shutil.move(os.path.join(dir_path,name), os.path.join(dir_path, new_name))
else:
print('file {} is not change'.format(name))

I have a ".txt "file which consists of various filenames and I want to search each filename in a folder where these files are actually kept

Suppose I have a text file aiq_hits.txt.
Each line in this file corresponds a filename
ant1.aiq
ant2.aiq
ant3.aiq
ant4.aiq
I want to match each line of my textfile (ant1.aiq,ant2.aiq and so on) with filenames which are present at some specific place(R:\Sample) and extract matching files into some other place (R:\sample\wsa).
I have an idea that I need to use functions like os.walk() and fnmatch.fnmatch(), shutil.copy() but I am not able to implement them
My code:
import os
import shutil
import fnmatch
with open("aiq_hits.txt","r") as in_file:
for line in in_file:
I am stuck here
import os
import shutil
sourceDir = "R:\\Sample"
targetDir = "R:\\Sample\\wsa"
existingFiles = set(f for f in os.listdir(sourceDir) if os.path.isfile(os.path.join(sourceDir, f)))
infilepath = "aiq_hits.txt"
with open(infilepath) as infile:
for line in infile:
fname = line.strip()
if fname not in existingFiles: continue
shutil.move(os.path.join(sourceDir, fname), os.path.join(targetDir, fname))
I hope this will suffice:
import os
def match_files(url,file_read, dest):
f = open(file_read, 'rb')
file_list = os.listdir(url)
print(file_list)
saved_path = os.getcwd()
print("Current working directory is " + saved_path)
os.chdir(url)
match = []
for file_name in f:
file_name = file_name.strip()
if file_name in file_list:
match.append(file_name)
os.rename(os.path.join(url, file_name), os.path.join(dest, file_name))
os.chdir(saved_path)
print match
here, url is source directory or folder from which u want to match files, file_read is the name of file (with path) in which list of file names is given, dest is the destination folder.
this code moves the matching files from url to dest, i.e. these files won't remin in url after running the code.
Alternatively you could use the glob module which allows you to enter in a expression for the file name\extension which will then return a list that you can loop over.
I'd use this module if the source directory can have files with the same extension that you want to exclude from being looped over
Also I'm assuming that the file name list is not large and so storing it in a list wont be an issue
eg (I haven't tested the below )
from glob import glob
import os
import shutil
src = 'R:\\Sample'
dst = "R:\\Sample\\wsa"
in_file_list = "aiq_hits.txt"
list_Of_files = glob(os.path.join(src, 'ant*.aiq'))
data = []
with open(in_file_list) as reader:
data += reader.readlines()
for row in list_Of_files:
file_path, file_name = os.path.split(row)
if file_name in data:
shutil.copy2(row, os.path.join(dst, file_name))
# or if you want to move the file
# shutil.move(row, os.path.join(dst, file_name))

Get rows from all .txt files in directory using python

I have some txt files in a directory and I need to get the last 15 lines from all of them. How could I do it using python?
I chose this code:
from os import listdir
from os.path import isfile, join
dir_path= './'
files = [ f for f in listdir(dir_path) if isfile(join(dir_path,f)) ]
out = []
for file in files:
filedata = open(join(dir_path, file), "r").readlines()[-15:]
out.append(filedata)
f = open(r'./fin.txt','w')
f.writelines(out)
f.close()
but I get the error "TypeError: writelines() argument must be a sequence of strings". I think it's because of Russian letters in the lines.
import os
from collections import deque
for filename in os.listdir('/some/path'):
# might want to put a check it's actually a file here...
# (join it to a root path, or anything else....)
# and sanity check it's text of a usable kind
with open(filename) as fin:
last_15 = deque(fin, 15)
deque will automatically discard the oldest entry and peak the max size to be 15, so it's an efficient way of keeping just the "last" 'n' items.
Try this:
from os import listdir
from os.path import isfile
for filepath in listdir("/path/to/folder")
if isfile(filepath): # if need
last_five_lines = open(filepath).readlines()[-15:]
# or, one line:
x = [open(f).readlines()[-15:] for f in listdir("/path/to/folder") if isfile(f)]
Updated:
lastlines = []
for file in files:
lastlines += open(join(dir_path, file), "r").readlines()[-15:]
with open('./fin.txt', 'w') as f:
f.writelines(lastlines)
from os import listdir
from os.path import isfile, join
dir_path= '/usr/lib/something'
files = [ f for f in listdir(dir_path) if isfile(join(dir_path,f)) ]
for file in files:
filedata = open(join(dir_path, file), "r").readlines()[-15:]
#do something with the filedata
Hope this helps:
import os
current_dir = os.getcwd()
dir_objects = os.listdir(current_dir)
dict_of_last_15 = {}
for file in dir_objects:
file_obj = open(file, 'rb')
content = file_obj.readlines()
last_15_lines = content[-15:]
dict_of_last_15[file] = last_15_lines
print "#############: %s" % file
print dict_of_last_15[file]
file_to_check.close()

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