Question

I’m using VBA to save an excel worksheet as a CSV file. The macro saves it as a CSv and then opens it in excel. I have a python code that reads the file, by the user selecting it (using tkinter) to open.

Is there a way to modify my python code to detect an open csv file, so that I can skip the user select stage? I want it to return the filepath or read open csv file. This is the current version of my code, which works if the csv file is open in excel:

import numpy as np
from Tkinter import Tk
from tkFileDialog import askopenfilename
tk=Tk()
tk.withdraw()
filename = askopenfilename()
data = np.genfromtxt(filename, dtype=[('x',float),('y',float)],comments='"', delimiter=',',skip_header=1,missing_values=True)
tk.destroy()
x=data['x']
x = x[np.logical_not(np.isnan(x))]
y=data['y']
y = y[np.logical_not(np.isnan(y))]
print x
print y

Answer 1

Which OS? If it is Windows, take a look at Get name of active Excel workbook from Python. The code in the question itself may do what you’re looking for:

import win32com.client

xl = win32com.client.Dispatch("Excel.Application")
print(xl.ActiveWorkbook.FullName)

As mentioned in that question’s answer, this only prints the path to the most recently opened file, but in your case it sounds like the macro is opening the file, so perhaps it would work? Otherwise, take a look at the actual answer – it seems very complete.