We Keep Coding

Lost files while tried to move files using python shutil.move

I had 120 files in my source folder which I need to move to a new directory (destination). The destination is made in the function I wrote, based on the string in the filename. For example, here is the function I used.
path ='/path/to/source'
dropbox='/path/to/dropbox'
files = = [os.path.join(path,i).split('/')[-1] for i in os.listdir(path) if i.startswith("SSE")]
sam_lis =list()
for sam in files:
sam_list =sam.split('_')[5]
sam_lis.append(sam_list)
sam_lis =pd.unique(sam_lis).tolist()
# Using the above list
ID = sam_lis
def filemover(ID,files,dropbox):
"""
Function to move files from the common place to the destination folder
"""
for samples in ID:
for fs in files:
if samples in fs:
desination = dropbox + "/"+ samples + "/raw/"
if not os.path.isdir(desination):
os.makedirs(desination)
for rawfiles in fnmatch.filter(files, pat="*"):
if samples in rawfiles:
shutil.move(os.path.join(path,rawfiles),
os.path.join(desination,rawfiles))
In the function, I am creating the destination folders, based on the ID's derived from the files list. When I tried to run this for the first time it threw me FILE NOT exists error.
However, later when I checked the source all files starting with SSE were missing. In the beginning, the files were there. I want some insights here;
Whether or not os.shutil.move moves the files to somewhere like a temp folder instead of destination folder?
whether or not the os.shutil.move deletes the files from the source in any circumstance?
Is there any way I can test my script to find the potential reasons for missing files?
Any help or suggestions are much appreciated?

It is late but people don't understand the op's question. If you move a file into a non-existing folder, the file seems to become a compressed binary and get lost forever. It has happened to me twice, once in git bash and the other time using shutil.move in Python. I remember the python happens when your shutil.move destination points to a folder instead of to a copy of the full file path.
For example, if you run the code below, a similar situation to what the op described will happen:
src_folder = r'C:/Users/name'
dst_folder = r'C:/Users/name/data_images'
file_names = glob.glob(r'C:/Users/name/*.jpg')
for file in file_names:
file_name = os.path.basename(file)
shutil.move(os.path.join(src_folder, file_name), dst_folder)
Note that dst_folder in the else block is just a folder. It should be dst_folder + file_name. This will cause what the Op described in his question. I find something similar on the link here with a more detailed explanation of what went wrong: File moving mistake with Python

shutil.move does not delete your files, if for any reason your files failed to move to a given location, check the directory where your code is stored, for a '+' folder your files are most likely stored there.

FileNotFound Error: [Errno 2] No such file or directory b when iterating through list of files in Python on Windows?

I am getting a FileNotFound error when iterating through a list of files in Python on Windows.
The specific error I get looks like:
FileNotFoundError: File b'fileName.csv' does not exist
In my code, I first ask for input on where the file is located and generate a list using os (though I also tried glob):
directory = input('In what directory are your files located?')
fileList = [s for s in os.listdir(directory) if s.endswith('.csv')]
When I print the list, it does not contain byte b before any strings, as expected (but I still checked). My code seems to break at this step, which generates the error:
for file in fileList:
pd.read_csv(file) # breaks at this line
I have tried everything I could find on Stack Overflow to solve this problem. That includes:
Putting an r or b or rb right before the path string
Using both the relative and absolute file paths
Trying different variations of path separators (/, \, \\, etc.)
I've been dealing with Windows-related issues lately (since I normally work in Mac or Linux) so that was my first suspicion. I'd love another set of eyes to help me figure out where the snag is.

A2A.
Although the list was generated correctly because the full directory path was used, the working directory when the file was being run was .. You can verify this by running os.getcwd(). When later iterating through the list of file names, the program could not find those file names in the . directory, as it shouldn't. That list is just a list of file names; there was no directory tied to it so it used the current directory.
The easiest fix here is to change the directory the file is running through after the input. So,
directory = input('In what directory are your files located?')
os.chdir(directory) # Points os to given directory to avoid byte issue
If you need to access multiple directories, you could either do this right before you switch to each directory or save the full file path into your list instead.

Python - [Errno 2] No such file or directory,

I am trying to make a minor modification to a python script made by my predecessor and I have bumped into a problem. I have studied programming, but coding is not my profession.
The python script processes SQL queries and writes them to an excel file, there is a folder where all the queries are kept in .txt format. The script creates a list of the queries found in the folder and goes through them one by one in a for cycle.
My problem is if I want to rename or add a query in the folder, I get a "[Errno 2] No such file or directory" error. The script uses relative path so I am puzzled why does it keep making errors for non-existing files.
queries_pathNIC = "./queriesNIC/"
def queriesDirer():
global filelist
l = 0
filelist = []
for file in os.listdir(queries_pathNIC):
if file.endswith(".txt"):
l+=1
filelist.append(file)
return(l)
Where the problem arises in the main function:
for round in range(0,queriesDirer()):
print ("\nQuery :",filelist[round])
file_query = open(queries_pathNIC+filelist[round],'r'); # problem on this line
file_query = str(file_query.read())
Contents of queriesNIC folder
00_1_Hardware_WelcomeNew.txt
00_2_Software_WelcomeNew.txt
00_3_Software_WelcomeNew_AUTORENEW.txt
The scripts runs without a problem, but if I change the first query name to
"00_1_Hardware_WelcomeNew_sth.txt" or anything different, I get the following error message:
FileNotFoundError: [Errno 2] No such file or directory: './queriesNIC/00_1_Hardware_WelcomeNew.txt'
I have also tried adding new text files to the folder (example: "00_1_Hardware_Other.txt") and the script simply skips processing the ones I added altogether and only goes with the original files.
I am using Python 3.4.
Does anyone have any suggestions what might be the problem?
Thank you

The following approach would be an improvement. The glob module can produce a list of files ending with .txt quite easily without needing to create a list.
import glob, os
queries_pathNIC = "./queriesNIC/"
def queriesDirer(directory):
return glob.glob(os.path.join(directory, "*.txt"))
for file_name in queriesDirer(queries_pathNIC):
print ("Query :", file_name)
with open(file_name, 'r') as f_query:
file_query = f_query.read()
From the sample you have given, it is not clear if you need further access to the round variable or the file list.

taking data from files which are in folder

How do I get the data from multiple txt files that placed in a specific folder. I started with this could not fix. It gives an error like 'No such file or directory: '.idea' (??)
(Let's say I have an A folder and in that, there are x.txt, y.txt, z.txt and so on. I am trying to get and print the information from all the files x,y,z)
def find_get(folder):
for file in os.listdir(folder):
f = open(file, 'r')
for data in open(file, 'r'):
print data
find_get('filex')
Thanks.

If you just want to print each line:
import glob
import os
def find_get(path):
for f in glob.glob(os.path.join(path,"*.txt")):
with open(os.path.join(path, f)) as data:
for line in data:
print(line)
glob will find only your .txt files in the specified path.
Your error comes from not joining the path to the filename, unless the file was in the same directory you were running the code from python would not be able to find the file without the full path. Another issue is you seem to have a directory .idea which would also give you an error when trying to open it as a file. This also presumes you actually have permissions to read the files in the directory.
If your files were larger I would avoid reading all into memory and/or storing the full content.

First of all make sure you add the folder name to the file name, so you can find the file relative to where the script is executed.
To do so you want to use os.path.join, which as it's name suggests - joins paths. So, using a generator:
def find_get(folder):
for filename in os.listdir(folder):
relative_file_path = os.path.join(folder, filename)
with open(relative_file_path) as f:
# read() gives the entire data from the file
yield f.read()
# this consumes the generator to a list
files_data = list(find_get('filex'))
See what we got in the list that consumed the generator:
print files_data
It may be more convenient to produce tuples which can be used to construct a dict:
def find_get(folder):
for filename in os.listdir(folder):
relative_file_path = os.path.join(folder, filename)
with open(relative_file_path) as f:
# read() gives the entire data from the file
yield (relative_file_path, f.read(), )
# this consumes the generator to a list
files_data = dict(find_get('filex'))
You will now have a mapping from the file's name to it's content.
Also, take a look at the answer by #Padraic Cunningham . He brought up the glob module which is suitable in this case.

The error you're facing is simple: listdir returns filenames, not full pathnames. To turn them into pathnames you can access from your current working directory, you have to join them to the directory path:
for filename in os.listdir(directory):
pathname = os.path.join(directory, filename)
with open(pathname) as f:
# do stuff
So, in your case, there's a file named .idea in the folder directory, but you're trying to open a file named .idea in the current working directory, and there is no such file.
There are at least four other potential problems with your code that you also need to think about and possibly fix after this one:
You don't handle errors. There are many very common reasons you may not be able to open and read a file--it may be a directory, you may not have read access, it may be exclusively locked, it may have been moved since your listdir, etc. And those aren't logic errors in your code or user errors in specifying the wrong directory, they're part of the normal flow of events, so your code should handle them, not just die. Which means you need a try statement.
You don't do anything with the files but print out every line. Basically, this is like running cat folder/* from the shell. Is that what you want? If not, you have to figure out what you want and write the corresponding code.
You open the same file twice in a row, without closing in between. At best this is wasteful, at worst it will mean your code doesn't run on any system where opens are exclusive by default. (Are there such systems? Unless you know the answer to that is "no", you should assume there are.)
You don't close your files. Sure, the garbage collector will get to them eventually--and if you're using CPython and know how it works, you can even prove the maximum number of open file handles that your code can accumulate is fixed and pretty small. But why rely on that? Just use a with statement, or call close.
However, none of those problems are related to your current error. So, while you have to fix them too, don't expect fixing one of them to make the first problem go away.

Full variant:
import os
def find_get(path):
files = {}
for file in os.listdir(path):
if os.path.isfile(os.path.join(path,file)):
with open(os.path.join(path,file), "r") as data:
files[file] = data.read()
return files
print(find_get("filex"))
Output:
{'1.txt': 'dsad', '2.txt': 'fsdfs'}
After the you could generate one file from that content, etc.
Key-thing:
os.listdir return a list of files without full path, so you need to concatenate initial path with fount item to operate.
there could be ideally used dicts :)
os.listdir return files and folders, so you need to check if list item is really file

You should check if the file is actually file and not a folder, since you can't open folders for reading. Also, you can't just open a relative path file, since it is under a folder, so you should get the correct path with os.path.join. Check below:
import os
def find_get(folder):
for file in os.listdir(folder):
if not os.path.isfile(file):
continue # skip other directories
f = open(os.path.join(folder, file), 'r')
for line in f:
print line

simple python file search and record to csv

import os, csv
f=open("C:\\tempa\\file.csv", 'wb') #write to an existing blank csv file
w=csv.writer(f)
for path, dirs, files, in os.walk("C:\\tempa"):
for filename in files:
w.writerow([filename])
running win7 64bit latest python, using anaconda spyder, pyscripter issue persists regardless of the ide.
I have some media in folders in tempa jpg, pdf and mov... and I wanted to get a file list of all of them, and the code works but it stops without any issue at row 113, nothing special with the file it stops on, no weird characters.
I could have 3 blocks of code one for each folder to work around this weird bug. but it shouldnt have an issue.. the folders are all in the root folder without going too deep in sub folders:
C:\
-tempa
-jpg
-pdf
-mov
I have heard there are issues with os.walk but I didn't expext anything weird like this.
Maybe I need an f=close?

You were examining the file before it was fully closed. (f won't be closed until, at least, it is no longer referenced by any in-scope variable name.) If you examine a file before it is closed, you may not see the final, partial, data buffer.
Use the file object's context manager to ensure that the file is flushed and closed in all cases:
import os, csv
with open("C:\\tempa\\file.csv", 'wb') as f: #write to an existing blank csv file
w=csv.writer(f)
for path, dirs, files, in os.walk("C:\\tempa"):
for filename in files:
w.writerow([filename])
# Now no need for f.close()