Assigning a name to an output file with a string variable - python-3.x

Question

I'm writing a program to concatenate 3 csv files into a single new output csv file. As part of this I have to ask the user for the name they want to use for the output filename. My problem is that my output filename is declared in the arguments of the function and therefore, I get an error on the first line of this function because myFile is not declared (see 2 lines later for the declaration).
def concatenate(indir="C:\\Conc", outfile="C:\\Conc\\%s.csv" %myFile):
os.chdir(indir)
myFile = input("Please type the name of the file: ")
fileList=glob.glob("*.csv")
dfList=[]
colnames=["Symbol", "Name", "LastSale", "MarketCap", "ADR TSO", "IPOYear", "Sector", "Industry", "Summary Quote", " "]
for filename in fileList:
print("merging " + filename + "...")
df=pandas.read_csv(filename, skiprows=1,header=None)
dfList.append(df)
concatDf=pandas.concat(dfList, axis=0)
concatDf.columns=colnames
concatDf = concatDf.sort_values(by=["Symbol"], axis=0, ascending=True)
concatDf.to_csv(outfile, index=None)
print("Completed. Your merged file can be found at: " + outfile + "\n")
This function is called from a menu function (as below) so I was wondering if it's possible to pass it from the menu??
if choice == "1":
myFile = input("Please type the name of the file: ")
concatenate(myFile)
menu()
But neither options seem to work because I always get an error saying that myFile is not declared. I know this is basic stuff but I'm scratching my head trying to figure it out.

Answer 1

Your reputation suggests you're new to stack overflow and your name suggests you're new to python. So welcome to both! Here's a quite verbose answer to hopefully make things clear.
There are a few issues here:
concatenate() takes two arguments, indir and outdir, but you're only passing one argument when you're calling it: concatenate(myFile). Your arguments are what is called keyword arguments, since you've given them names (i.e. indir and outdir). When you're only passing one argument (myFile), without using the keyword, the myFile variable is passed as the first argument, which, in this case, is the indir (this doesn't happen yet in your code, as the error you're getting precedes it and stops it from being executed). However, you seem to want the myFile variable assigned to your outfile argument. You can achieve this by being explicit in your call to concatenate() like so: concatenate(outfile=myFile). Now, your indir argument remains the default you've set (i.e. "C:\\Conc"), but your outfile argument has the value of myFile. This would, however, not fix your problem (detailed in point 2). I suggest you change the outfile argument to represent an output directory (see full example below).
This is where you get an error. Your concatenate() function has no knowledge of the myFile variable you've declared in your if-statement, as the function and your if-statement are different scopes. That's why function arguments exist, to pass data from one scope (your if-statement) to another scope (your function). So you get an error in your function saying that myFile is not declared because, to the function's eyes, it doesn't exist. It can only see indir and outfile, not myFile. Also, don't do string manipulation in arguments. I'm not sure how it works, but it surely doesn't work how you expect in this case.
You're asking the user for myFile twice. Once in the if-statement and once in your concatenate() function. You only need to ask for it once, I suggest to keep it in the function only.
Small correction: You should combine directory paths and filenames in a platform-independent manner. The os module has a function for that (os.path.join(), see example below). Paths in Windows and paths in Linux look different, the os module handles those differences for us:)
Here's a final suggestion, with comments, addressing all points:
# Import your modules
import pandas
import os # <-- We're gonna need this
# Your code
# ...
# The concatenate function, with arguments having a default value, allowing to optionally specify the input directory and the output *directory* (not output file)
def concatenate(indir="C:\\Conc", outdir="C:\\Conc"):
os.chdir(indir)
myFile = input("Please type the name of the file: ") # <-- We ask for the output filename here
# Your code
fileList=glob.glob("*.csv")
dfList=[]
colnames=["Symbol", "Name", "LastSale", "MarketCap", "ADR TSO", "IPOYear", "Sector", "Industry", "Summary Quote", " "]
for filename in fileList:
print("merging " + filename + "...")
df=pandas.read_csv(filename, skiprows=1,header=None)
dfList.append(df)
concatDf=pandas.concat(dfList, axis=0)
concatDf.columns=colnames
concatDf = concatDf.sort_values(by=["Symbol"], axis=0, ascending=True)
# Saving the data
# First, we need to create the full output path, i.e. the output directory + output filename. We use os.join.path() for that
output_path = os.path.join(outdir, myFile)
# The rest of your code
concatDf.to_csv(output_path, index=None) # <-- Note output_path, not outfile
print("Completed. Your merged file can be found at: " + output_path + "\n")
# The if-statement calling the ´concatenate()´ function
if choice == "1":
# We're calling concatenate() with no arguments, since we're asking for the filename within the function.
# You could, however, ask the user for the input directory or output directory and pass that along, like this:
# input_directory_from_user = input("Please type the path to the input directory: ")
# output_directory_from_user = input("Please type the path to the output directory: ")
# concatenate(indir=input_directory_from_user, outdir=output_directory_from_user)
concatenate()
menu()