I've got 2 folders, each with a different CSV file inside (both have the same format):
I've written some python code to search within the "C:/Users/Documents" directory for CSV files which begin with the word "File"
import glob, os
inputfile = []
for root, dirs, files in os.walk("C:/Users/Documents/"):
for datafile in files:
if datafile.startswith("File") and datafile.endswith(".csv"):
inputfile.append([os.path.join(root, datafile)])
print(inputfile)
That almost worked as it returns:
[['C:/Users/Documents/Test A\\File 1.csv'], ['C:/Users/Documents/Test B\\File 2.csv']]
Is there any way I can get it to return this instead (no sub list and shows / instead of \):
['C:/Users/Documents/Test A/File 1.csv', 'C:/Users/Documents/Test B/File 2.csv']
The idea is so I can then read both CSV files at once later, but I believe I need to get the list in the format above first.
okay, I will paste an option here.
I made use of os.path.abspath to get the the path before join.
Have a look and see if it works.
import os
filelist = []
for folder, subfolders, files in os.walk("C:/Users/Documents/"):
for datafile in files:
if datafile.startswith("File") and datafile.endswith(".csv"):
filePath = os.path.abspath(os.path.join(folder, datafile))
filelist.append(filePath)
filelist
Result:
['C:/Users/Documents/Test A/File 1.csv','C:/Users/Documents/Test B/File 2.csv']
Related
I am trying to write a program in python that loops through data from various csv files within a folder. Right now I just want to see that the program can identify the files in a folder but I am unable to have my code print the file names in my folder. This is what I have so far, and I'm not sure what my problem is. Could it be the periods in the folder names in the file path?
import glob
path = "Users/Sarah/Documents/College/Lab/SEM EDS/1.28.20 CZTS hexane/*.csv"
for fname in glob.glob(path):
print fname
No error messages are popping up but nothing will print. Does anyone know what I'm doing wrong?
Are you on a Linux-base system ? If you're not, switch the / for \\.
Is the directory you're giving the full path, from the root folder ? You might need to
specify a FULL path (drive included).
If that still fails, silly but check there actually are files in there, as your code otherwise seems fine.
This code below worked for me, and listed csv files appropriately (see the C:\\ part, could be what you're missing).
import glob
path = "C:\\Users\\xhattam\\Downloads\\TEST_FOLDER\\*.csv"
for fname in glob.glob(path):
print(fname)
The following code gets a list of files in a folder and if they have csv in them it will print the file name.
import os
path = r"C:\temp"
filesfolders = os.listdir(path)
for file in filesfolders:
if ".csv" in file:
print (file)
Note the indentation in my code. You need to be careful not to mix tabs and spaces as theses are not the same to python.
Alternatively you could use os
import os
files_list = os.listdir(path)
out_list = []
for item in files_list:
if item[-4:] == ".csv":
out_list.append(item)
print(out_list)
Are you sure you are using the correct path?
Try moving the python script in the folder when the CSV files are, and then change it to this:
import glob
path = "./*.csv"
for fname in glob.glob(path):
print fname
I am combining two questions here because they are related to each other.
Question 1: I am trying to use glob to open all the files in a folder but it is giving me "Syntax Error". I am using Python 3.xx. Has the syntax changed for Python 3.xx?
Error Message:
File "multiple_files.py", line 29
files = glob.glob(/src/xyz/rte/folder/)
SyntaxError: invalid syntax
Code:
import csv
import os
import glob
from pandas import DataFrame, read_csv
#extracting
files = glob.glob(/src/xyz/rte/folder/)
for fle in files:
with open (fle) as f:
print("output" + fle)
f_read.close()
Question 2: I want to read input files, append "output" to the names and print out the names of the files. How can I do that?
Example: Input file name would be - xyz.csv and the code should print output_xyz.csv .
Your help is appreciated.
Your first problem is that strings, including pathnames, need to be in quotes. This:
files = glob.glob(/src/xyz/rte/folder/)
… is trying to divide a bunch of variables together, but the leftmost and rightmost divisions are missing operands, so you've confused the parser. What you want is this:
files = glob.glob('/src/xyz/rte/folder/')
Your next problem is that this glob pattern doesn't have any globs in it, so the only thing it's going to match is the directory itself.
That's perfectly legal, but kind of useless.
And then you try to open each match as a text file. Which you can't do with a directory, hence the IsADirectoryError.
The answer here is less obvious, because it's not clear what you want.
Maybe you just wanted all of the files in that directory? In that case, you don't want glob.glob, you want listdir (or maybe scandir): os.listdir('/src/xyz/rte/folder/').
Maybe you wanted all of the files in that directory or any of its subdirectories? In that case, you could do it with rglob, but os.walk is probably clearer.
Maybe you did want all the files in that directory that match some pattern, so glob.glob is right—but in that case, you need to specify what that pattern is. For example, if you wanted all .csv files, that would be glob.glob('/src/xyz/rte/folder/*.csv').
Finally, you say "I want to read input files, append "output" to the names and print out the names of the files". Why do you want to read the files if you're not doing anything with the contents? You can do that, of course, but it seems pretty wasteful. If you just want to print out the filenames with output appended, that's easy:
for filename in os.listdir('/src/xyz/rte/folder/'):
print('output'+filename)
This works in http://pyfiddle.io:
Doku: https://docs.python.org/3/library/glob.html
import csv
import os
import glob
# create some files
for n in ["a","b","c","d"]:
with open('{}.txt'.format(n),"w") as f:
f.write(n)
print("\nFiles before")
# get all files
files = glob.glob("./*.*")
for fle in files:
print(fle) # print file
path,fileName = os.path.split(fle) # split name from path
# open file for read and second one for write with modified name
with open (fle) as f,open('{}{}output_{}'.format(path,os.sep, fileName),"w") as w:
content = f.read() # read all
w.write(content.upper()) # write all modified
# check files afterwards
print("\nFiles after")
files = glob.glob("./*.*") # pattern for all files
for fle in files:
print(fle)
Output:
Files before
./d.txt
./main.py
./c.txt
./b.txt
./a.txt
Files after
./d.txt
./output_c.txt
./output_d.txt
./main.py
./output_main.py
./c.txt
./b.txt
./output_b.txt
./a.txt
./output_a.txt
I am on windows and would use os.walk (Doku) instead.
for d,subdirs,files in os.walk("./"): # deconstruct returned aktDir, all subdirs, files
print("AktDir:", d)
print("Subdirs:", subdirs)
print("Files:", files)
Output:
AktDir: ./
Subdirs: []
Files: ['d.txt', 'output_c.txt', 'output_d.txt', 'main.py', 'output_main.py',
'c.txt', 'b.txt', 'output_b.txt', 'a.txt', 'output_a.txt']
It also recurses into subdirs.
I am new to python. I have successful written a script to search for something within a file using :
open(r"C:\file.txt) and re.search function and all works fine.
Is there a way to do the search function with all files within a folder? Because currently, I have to manually change the file name of my script by open(r"C:\file.txt),open(r"C:\file1.txt),open(r"C:\file2.txt)`, etc.
Thanks.
You can use os.walk to check all the files, as the following:
import os
for root, _, files in os.walk(path):
for filename in files:
with open(os.path.join(root, filename), 'r') as f:
#your code goes here
Explanation:
os.walk returns tuple of (root path, dir names, file names) in the folder, so you can iterate through filenames and open each file by using os.path.join(root, filename) which basically joins the root path with the file name so you can open the file.
Since you're a beginner, I'll give you a simple solution and walk through it.
Import the os module, and use the os.listdir function to create a list of everything in the directory. Then, iterate through the files using a for loop.
Example:
# Importing the os module
import os
# Give the directory you wish to iterate through
my_dir = <your directory - i.e. "C:\Users\bleh\Desktop\files">
# Using os.listdir to create a list of all of the files in dir
dir_list = os.listdir(my_dir)
# Use the for loop to iterate through the list you just created, and open the files
for f in dir_list:
# Whatever you want to do to all of the files
If you need help on the concepts, refer to the following:
for looops in p3: http://www.python-course.eu/python3_for_loop.php
os function Library (this has some cool stuff in it): https://docs.python.org/2/library/os.html
Good luck!
You can use the os.listdir(path) function:
import os
path = '/Users/ricardomartinez/repos/Salary-API'
# List for all files in a given PATH
file_list = os.listdir(path)
# If you want to filter by file type
file_list = [file for file in os.listdir(path) if os.path.splitext(file)[1] == '.py']
# Both cases yo can iterate over the list and apply the operations
# that you have
for file in file_list:
print(file)
#Operations that you want to do over files
as the title would imply I am looking to create a script that will allow me to print a list of file names in a directory to a CSV file.
I have a folder on my desktop that contains approx 150 pdf's. I'd like to be able to have the file names printed to a csv.
I am brand new to Python and may be jumping out of the frying pan and into the fire with this project.
Can anyone offer some insight to get me started?
First off you will want to start by grabbing all of the files in the directory, then simply by writing them to a file.
from os import listdir
from os.path import isfile, join
import csv
onlyfiles = [f for f in listdir("./") if isfile(join("./", f))]
with open('file_name.csv', 'w') as print_to:
writer = csv.writer(print_to)
writer.writerow(onlyfiles)
Please Note
"./" on line 5 is the directory you want to grab the files from.
Please replace 'file_name.csv' with the name of the file you want to right too.
The following will create a csv file with all *.pdf files:
from glob import glob
with open('/tmp/filelist.csv', 'w') as fout:
# write the csv header -- optional
fout.write("filename\n")
# write each filename with a newline characer
fout.writelines(['%s\n' % fn for fn in glob('/path/to/*.pdf')])
glob() is a nice shortcut to using listdir because it supports wildcards.
import os
csvpath = "csvfile.csv"
dirpath = "."
f = open("csvpath, "wb")
f.write(",".join(os.listdir(dirpath)))
f.close()
This may be improved to present filenames in way that you need, like for getting them back, or something. For instance, this most probably won't include unicode filenames in UTF-8 form but make some mess out of the encoding, but it is easy to fix all that.
If you have very big dir, with many files, you may have to wait some time for os.listdir() to get them all. This also can be fixed by using some other methods instead of os.listdir().
To differentiate between files and subdirectories see Michael's answer.
Also, using os.path.isfile() or os.path.isdir() you can recursively get all subdirectories if you wish.
Like this:
def getall (path):
files = []
for x in os.listdir(path):
x = os.path.join(path, x)
if os.path.isdir(x): files += getall(x)
else: files.append(x)
return files
I have a large numbers of fasta files (these are just text files) in different subfolders. What I need is a way to search through the directories for files that have the same name and concatenate these into a file with the name of the input files. I can't do this manually as I have 10000+ genes that I need to do this for.
So far I have the following Python code that looks through one of the directories and then uses those file names to search through the other directories. This returns a list that has the full path for each file.
import os
from os.path import join, abspath
path = '/directoryforfilelist/' #Directory for source list
listing = os.listdir(path)
for x in listing:
for root, dirs, files in os.walk('/rootdirectorytosearch/'):
if x in files:
pathlist = abspath(join(root,x))
Where I am stuck is how to concatenate the files it returns that have the same name. The results from this script look like this.
/directory1/file1.fasta
/directory2/file1.fasta
/directory3/file1.fasta
/directory1/file2.fasta
/directory2/file2.fasta
/directory3/file2.fasta
In this case I would need the end result to be two files named file1.fasta and file2.fasta that contain the text from each of the same named files.
Any leads on where to go from here would be appreciated. While I did this part in Python anyway that gets the job done is fine with me. This is being run on a Mac if that matters.
Not tested, but here's roughly what I'd do:
from itertools import groupby
import os
def conc_by_name(names):
for tail, group in groupby(names, key=os.path.split):
with open(tail, 'w') as out:
for name in group:
with open(name) as f:
out.writelines(f)
This will create the files (file1.fasta and file2.fasta in your example) in the current folder.
For each file of your list, allocate the target file in append mode, read each line of your source file and write it to the target file.
Assuming that the target folder is empty to start with, and is not in /rootdirectorytosearch.