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Get name of created file

I'm creating files in python and I want to print the full path of the already file created. Example:
dt = datetime.datetime.now()
datetime = dt.strftime("%d-%m")
output = open("c:\path\to\file_"+datetime+".txt","w")
How can I print the final name of the created file?
I tried:
print (output)
but it gives me a rare string.

For real file objects such as that,
print(output.name)

open() method doesn't return anything relevant. You can try on below methods...
1.print(output.__file__)
2. Import os
os.path.abspath("filename.ext"). # it will add the cwd to file name or if iterating the real path
os.path.dirname("filename.ext")
os.path.dirname("filename.ext")
3. Files= os.listdir(os.getcwd()) # it will list the files in cwd(you can give any path)
for i in files:
print(os.path.abspath(i))

Replacing part of a file name in a directory using python

I am trying to rename a set of files in a directory using python. The files are currently labelled with a Pool number, AR number and S number (e.g. Pool1_AR001_S13__fw_paired.fastq.gz.) Each file refers to a specific plant sequence name. I would like to rename these files by removing the 'Pool_AR_S' and replacing it with the sequence name e.g. 'Lbienne_dor5_GS1', while leaving the suffix (e.g. fw_paired.fastq.gz, rv_unpaired.fastq.gz), I am trying to read the files into a dictionary, but I am stuck as to what to do next. I have a .txt file containing the necessary information in the following format:
Pool1_AR010_S17 - Lbienne_lla10_GS2
Pool1_AR011_S18 - Lbienne_lla10_GS3
Pool1_AR020_S19 - Lcampanulatum_borau4_T_GS1
The code I have so far is:
from optparse import OptionParser
import csv
import os
parser = OptionParser()
parser.add_option("-w", "--wanted", dest="w")
parser.add_option("-t","--trimmed", dest="t")
parser.add_option("-d", "--directory", dest="working_dir", default="./")
(options, args) = parser.parse_args()
wanted_file = options.w
trimmomatic_output = options.t
#Read the wanted file and create a dictionary of index vs species identity
with open(wanted_file, 'rb') as species_sequence:
species_list = list(csv.DictReader(species_sequence, delimiter='-'))
print species_list
#Rename the Trimmomatic Output files according to the dictionary
for trimmed_sequence in os.listdir(trimmomatic_output):
os.rename(os.path.join(trimmomatic_output, trimmed_sequence),
os.path.join(trimmomatic_output, trimmed_sequence.replace(species_list[0], species_list[1]))
Please can you help me to replace half of the . I'm very new to python and to stack overflow, so I am sorry if this question has been asked before or if I have asked this in the wrong place.

First job is to get rid of all those modules. They may be nice, but for a job like yours they are very unlikely to make things easier.
Create a .py file in the directory where those .gz files reside.
import os
files = os.listdir() #files is of list type
#'txt_file' is the path of your .txt file containing those conversions
dic=parse_txt(txt_file) #omitted the body of parse_txt() func.Should return a dictionary by parsing that .txt file
for f in files:
pre,suf=f.split('__') #"Pool1_AR001_S13__(1)fw_paired.fastq.gz"
#(1)=assuming prefix and suffix are divided by double underscore
pre = dic[pre]
os.rename(f,pre+'__'+suf)
If you need help with parse_txt() function, let me know.

Here is a solution that I tested with Python 2. Its fine if you use your own logic instead of the get_mappings function. Refer comments in code for explanation.
import os
def get_mappings():
mappings_dict = {}
with(open('wanted_file.txt', 'r')) as f:
for line in f:
# if you have Pool1_AR010_S17 - Lbienne_lla10_GS2
# it becomes a list i.e ['Pool1_AR010_S17 ', ' Lbienne_lla10_GS2']
#note that there may be spaces before/after the names as shown above
text = line.split('-')
#trim is used to remove spaces in the names
mappings_dict[text[0].strip()] = text[1].strip()
return mappings_dict
#PROGRAM EXECUTION STARTS FROM HERE
#assuming all files are in the current directory
# if not replace the dot(.) with the path of the directory where you have the files
files = os.listdir('.')
wanted_names_dict = get_mappings()
for filename in files:
try:
#prefix='Pool1_AR010_S17', suffix='fw_paired.fastq.gz'
prefix, suffix = filename.split('__')
new_filename = wanted_names_dict[prefix] + '__' + suffix
os.rename(filename, new_filename)
print 'renamed', filename, 'to', new_filename
except:
print 'No new name defined for file:' + filename

Python3 Rename files in a directory importing the new names from a txt file

I have a directory containing multiple files.
The name of the files follows this pattern 4digits.1.4digits.[barcode]
The barcode specifies each file and it is composed by 7 leters.
I have a txt file where in one column I have that barcode and in the other column the real name of the file.
What I would like to do is to right a pyhthon script that automatically renames each file according to the barcode to it s new name written in the txt file.
Is there anybody that could help me?
Thanks a lot!

I will give you the logic:
1. read the text file that contains barcode and name.http://www.pythonforbeginners.com/files/reading-and-writing-files-in-python.
for each line in txt file do as follows:
2. Assign the value in first(barcode) and second(name) column in two separate variables say 'B' and 'N'.
3. Now we have to find the filename which has the barcode 'B' in it. the link
Find a file in python will help you do that.(first answer, 3 rd example, for your case the name you are going to find will be like '*B')
4. The previous step will give you the filename that has B as a part. Now use the rename() function to rename the file to 'N'. this link will hep you.http://www.tutorialspoint.com/python/os_rename.htm
Suggestion: Instead of having a txt file with two columns. You can have a csv file, that would be easy to handle.

The following code will do the job for your specific use-case, though can make it more general purpose re-namer.
import os # os is a library that gives us the ability to make OS changes
def file_renamer(list_of_files, new_file_name_list):
for file_name in list_of_files:
for (new_filename, barcode_infile) in new_file_name_list:
# as per the mentioned filename pattern -> xxxx.1.xxxx.[barcode]
barcode_current = file_name[12:19] # extracting the barcode from current filename
if barcode_current == barcode_infile:
os.rename(file_name, new_filename) # renaming step
print 'Successfully renamed %s to %s ' % (file_name, new_filename)
if __name__ == "__main__":
path = os.getcwd() # preassuming that you'll be executing the script while in the files directory
file_dir = os.path.abspath(path)
newname_file = raw_input('enter file with new names - or the complete path: ')
path_newname_file = os.path.join(file_dir, newname_file)
new_file_name_list = []
with open(path_newname_file) as file:
for line in file:
x = line.strip().split(',')
new_file_name_list.append(x)
list_of_files = os.listdir(file_dir)
file_renamer(list_of_files, new_file_name_list)
Pre-assumptions:
newnames.txt - comma
0000.1.0000.1234567,1234567
0000.1.0000.1234568,1234568
0000.1.0000.1234569,1234569
0000.1.0000.1234570,1234570
0000.1.0000.1234571,1234571
Files
1111.1.0000.1234567
1111.1.0000.1234568
1111.1.0000.1234569
were renamed to
0000.1.0000.1234567
0000.1.0000.1234568
0000.1.0000.1234569
The terminal output:
>python file_renamer.py
enter file with new names: newnames.txt
The list of files - ['.git', '.idea', '1111.1.0000.1234567', '1111.1.0000.1234568', '1111.1.0000.1234569', 'file_renamer.py', 'newnames.txt.txt']
Successfully renamed 1111.1.0000.1234567 to 0000.1.0000.1234567
Successfully renamed 1111.1.0000.1234568 to 0000.1.0000.1234568
Successfully renamed 1111.1.0000.1234569 to 0000.1.0000.1234569

Why is my write function not creating a file?

According to all the sources I've read, the open method creates a file or overwrites one with an existing name. However I am trying to use it and i get an error:
File not found - newlist.txt (Access is denied)
I/O operation failed.
I tried to read a file, and couldn't. Are you sure that file exists? If it does exist, did you specify the correct directory/folder?
def getIngredients(path, basename):
ingredient = []
filename = path + '\\' + basename
file = open(filename, "r")
for item in file:
if item.find("name") > -1:
startindex = item.find("name") + 5
endindex = item.find("<//name>") - 7
ingredients = item[startindex:endindex]
ingredient.append(ingredients)
del ingredient[0]
del ingredient[4]
for item in ingredient:
printNow(item)
file2 = open('newlist.txt', 'w+')
for item in ingredient:
file2.write("%s \n" % item)
As you can see i'm trying to write the list i've made into a file, but its not creating it like it should. I've tried all the different modes for the open function and they all give me the same error.

It looks like you do not have write access to the current working directory. You can get the Python working directory with import os; print os.getcwd().
You should then check whether you have write access in this directory. This can be done in Python with
import os
cwd = os.getcwd()
print "Write access granted to current directory", cwd, '>', os.access(cwd, os.W_OK)
If you get False (no write access), then you must put your newfile.txt file somewhere else (maybe at path + '/newfile.txt'?).

Are you certain the directory that you're trying to create the folder in exists?
If it does NOT... Then the OS won't be able to create the file.

This looks like a permissions problem.
either the directory does not exist or your user doesn't have the permissions to write into this directory .

I guess the possible problems may be:
1) You are passing the path and basename as parameters. If you are passing the parameters as strings, then you may get this problem:
For example:
def getIngredients(path, basename):
ingredient = []
filename = path + '\\' + basename
getIngredients("D","newlist.txt")
If you passing the parameters the above way, this means you are doing this
filename = "D" + "\\" + "newlist.txt"
2) You did not include a colon(:) after the path + in the filename.
3) Maybe, the file does not exist.

creating a output file when the output file exists in the path

How do I code in python the option that when the output file exists in the path, the output file will automatically be "originalname"+"_1" / "originalname"+"_2" and so on ?

Something like
import os.path
def getnewfilename(filename):
testfile = filename
i = 0
while os.path.exists(testfile):
i += 1
testfile = "%s_%s" % (testfile, i)
return testfile
This should generate
filename
filename_1
filename_2
if you use %s_%3i" you should get
filename
filename_001
filename_002
filename_003
which will then list alphabetically (but have problems when i>=1000)

You can use os.path.exists to check if a file already exists. The rest is a simple loop that tries new filenames.

isfile checks for the existence of a file and goes down simlinks as well; you can use the full filepath.
if os.path.isfile(filename):
do_something()